text:
When the top card of my library is permanently revealed, when is more than one card revealed?
I have an effect like Courser of Kruphix that is revealing the top card of my library. I'm trying to avoid revealing extra hidden information about what cards I'm playing in my deck to my opponent while resolving effects that manipulate cards in my library. The relevant Oracle rulings on it state:

If the top card of your library changes while you’re casting a spell, playing a land, or activating an ability, you don’t reveal the new top card until you finish doing so. This means that if you play a land from the top of your library, you can’t look at the next one until you’ve handled any replacement effects that modify that land entering the battlefield, such as that of Vesuva.

While playing with the top card of your library revealed, if you draw multiple cards, reveal each one before you draw it. If you put multiple cards on top of your library at once, reveal only the one new top card.

It's clear drawing multiple cards will reveal more hidden information to my opponent than just the top card of my library before and after the spell's resolution.
Are there any other deck manipulating actions that I can take that would reveal more cards than just the top card of my library before and after the spell's resolution? Some effects I'm thinking of are:

Cards that shuffle multiple times during their resolution like Green Sun's Zenith. Is the top card of my library after the first shuffle but before the second shuffle revealed?
Cards that search a library for a card like Polluted Delta's activated ability. Assuming that the card I search for and put on the battlefield is the top card of my library, is the new top card of my library revealed before I shuffle?
Cards that can be played while searching your library, which I think is exclusively Panglacial Wurm. If Panglacial Wurm is the top card of my library and I cast it while searching my library, is the next top card of my library revealed after moving Panglacial Wurm to the stack? Is the next top card of my library revealed before or after I complete searching?

Alternatively, are there deck manipulating actions that I can take that change the top card of my library don't end up revealing the intermediate cards?

The first Oracle ruling points out some instances where manipulating the library won't have intermediate cards revealed. Are these the only effects of that type?

I asked about the interaction with Green Sun's Zenith in the Magic Judge Chat:

I control a Courser of Kruphix which is revealing the top card of my library. I cast Green Sun's Zenith to find a creature, and during resolution, the card's text says that I shuffle twice. After the first shuffle but before the second, do I reveal the top card of my library? Or do I wait until after the second shuffle?

I was told:

Yes, you reveal the top card of your library between the two shuffles.

The important distinction here is that if multiple cards become the top card of the deck as the result of separate actions, you need to reveal all of them, but if multiple cards on top of the library get manipulated simultaneously as the result of a single action, you only need to reveal the one that ends up on top.
Drawing cards makes you reveal each card because drawing multiple cards actually results in drawing each card as a separate action, because of rule 121.2:

Cards may only be drawn one at a time. If a player is instructed to draw multiple cards, that player performs that many individual card draws.

The first ruling applies to playing lands because that is a single self-contained action, and to casting spells and activating abilities because of rule 401.5:

Some effects tell a player to play with the top card of their library revealed, or say that a player may look at the top card of their library. If the top card of the player’s library changes while a spell is being cast, the new top card won’t be revealed and can’t be looked at until the spell becomes cast (see rule 601.2i). The same is true with relation to an ability being activated.

None of the other subrules in the "Library" section specifies other times that cards would not be revealed, so that ruling can't be broadened from there.
It seems clear from this that you reveal two cards while resolving Green Sun's Zenith. The two shuffles are separate actions, written in separate sentences. As an interesting note, Green Sun's Zenith is the only card ever printed that instructs you to shuffle twice in a single ability.
The questions about Polluted Delta and Panglacial Wurm both mostly boil down to the question of whether searching and shuffling constitute one single action or two separate actions. I would lean towards arguing that they are separate, because they are separate keyword actions in the rules (701.19 and 701.20, respectively). That would mean that if the top card of the library changes while searching, you should reveal the new top card after you have finished searching, before you shuffle. This is not explicitly stated in the rules though, and judges may rule the other way.
Panglacial Wurm in particular is a bit of a rules nightmare that doesn't always have the clearest rules interactions. It is the only effect ever printed that allows you to stop in the middle of an action and go do something else, and then come back to that action when you're done.

When the top card of my library is permanently revealed, when is more than one card revealed?

Never: There is only a single card revealed by this effect. You're overthinking it. As the various oracle rulings try to clarify, you finish all deck manipulation (such as searching, revealing, drawing, shuffling) as part of spell resolution. Then after doing all of that, you reveal the top card of your deck (typically by physically flipping the card over). Here's one such ruling (emphasis mine):

If the top card of your library changes while you’re casting a spell, playing a land, or activating an ability, you don’t reveal the new top card until you finish doing so. This means that if you play a land from the top of your library, you can’t look at the next one until you’ve handled any replacement effects that modify that land entering the battlefield, such as that of Vesuva.

There is only one "exception" which is really just a clarification:

While playing with the top card of your library revealed, if you draw multiple cards, reveal each one before you draw it.

The reason for the clarification is that the rules actually specify that drawing multiple cards is actually drawing cards one at a time as many times as necessary.

121.2. Cards may only be drawn one at a time. If a player is instructed to draw multiple cards, that player performs that many individual card draws.

What makes an effect like "Draw 2" which is equivalent to "Draw a card, draw a card." different to Green Sun's Zenith's effect of "Shuffle, shuffle."? Here, the effects aren't batched together. I asked about specifically the Green Sun's Zenith interaction in the Judge Chat and updated my post - are you able to help clarify? What makes "Draw a card, draw a card" special?

I think you have misunderstood the first ruling you quoted. It applies specifically to the process of casting a spell or activating an ability, because it is essentially rewording rule 401.5. Neither that rule nor that ruling applies to resolving a spell or ability, which is the focus of this question.


text:
Ticket to Ride Nordic Countries: endless destination cards harvesting loophole
For the moment we have had fun playing Ticket to Ride Nordic Countries, however, the fun has been spoiled by now due to the following incentive of which I don't know whether it is allowed.
Consider the following situation:

Two players build a network across the map based on the initial destination cards they got from the start.
Then each player extends their network up to the point that they have almost run out of trains, but keeps enough trains apart to not let the game finish automatically.
Now the players can harvest destination tickets with 3 at a time, where at least 1 destination card should be kept.
With only a handful of trains left in your stock and a reasonably well-connected network, chances are pretty high that at least one of the three cards you draw will be already covered by your network (at least this is my experience). So one does not have to be afraid of negative scoring for unfinished destination tickets.

As far as I understand the rules, unlimited harvesting of destination tickets is allowed. The only punishment that exists in the game is that any destination ticket you fail to complete will be counted negatively in your final scoring.
If this is true, TTC NC changes from a strategy game into a lottery.
So my question is:
Is unlimited destination ticket harvesting allowed in the way I described before?
I am aware of a similar question asked here, but in contrast to that question, my question is specifically about the balance of luck against strategy and planning in this game, not about the influence of the luck itself.
I could think of house rules where the number of destination tickets you are allowed to draw is depending on your available stock of trains: the fewer trains you have left, the fewer cards you may draw. But I'd rather stick with official rules.

At least in other variants of TTR, the game ends as soon as one player has almost depleted their stock of trains, so if my opponent is doing this I'll just finish the game and win while their score goes very negative. Is that different in TTR Nordic? (I don't see anything in the BGG summary which says this is different)

@PhilipKendall, I added an extra bullet point to my list. The thing is that the destination tickets in TTR Nordic Countries are very similar to each other and destinations are easily added to your network. So you can make a long stretch like Oslo-Kuopio, and draw a card for Oslo-Stockholm or Helsinki-Kuopio later on while these two trajectories were already included in your network when you were working on your earlier destination ticket. In that way, there is hardly a chance that you will get negative scorings from your destination tickets

From the rules here with emphasis mine

A player can use his turn to draw more Destination Ticket cards. To do
so, he draws 3 new cards from the top of the Destination Ticket deck.
He must keep at least one of the cards, but may also keep two or all
three if he chooses. If there are less than 3 Destination Tickets left
in the deck, the player only draws what cards are available. Any
returned cards are removed from the game and put back in the box

Most Ticket to Ride game tickets untaken are returned to the bottom on the deck. In some variants such as Swiss and Nordic untaken tickets are removed from the game.  So there is not 'unlimited ticket harvesting'. the deck will run out if lots of players are just taking tickets.
As the comment says if your opponent is taking lots of extra tickets then just make the game end quicker by placing your remaining trains quicker.
Players are also taking the risk that they draw tickets where none are completed or that they won't be able to complete them.   Obviously you don't find this our until the end of the game as completed tickets are not revealed until the game ends.

Right, indeed the deck shrinks but given the sheer size of the deck and the maximum of 3 players for this game, the deck won't likely run out, making it supposedly endless. But you are right in the sense that the opponent should probably just end this behavior by 'killing' the game via finishing his/her trains.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Are knights a worthy investment in Catan's Card Game?
I'm still a proud owner of the 1996 Catan Card Game and have two expansions as well. While I like the artwork and the horizontal layout of the game, the gameplay doesn't flow well oftentimes.
One aspect of the game is the knights. Knights are added to your dominion in the same way buildings are added. Every knight has two values one for strength points and one for tournament points These points serve to:

Strength points: Compete for the wooden knight figure (worth 1 victory point)
Tournament points: Compete for the knight-event on the events-die (equals the 'good year' event but exclusively to the player owning the most tournament points).

From my experience, it may come in handy to invest in an occasional knight just to gain an edge over your opponent, especially early on in the game. But as the game progresses, the rewards of the knights seem rarely to outweigh the costs as knights are not only expensive, they also occupy space needed for buildings. This is even more apparent when playing with the Wizards and Dragons expansion, where the dragons are even more expensive than the knights. So all in all, the competition for knights doesn't seem to span the whole game through.
So my question is: are investments in knights worth it in Catan's Card Game?

It looks like your second bullet point got cut off

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why did a triggered ability trigger after the creature holding it died?
In Arena, I had Fynn, the Fangbearer and another creature with deathtouch in play. My opponent had one creature with at least 3 power. I attacked with both creatures; my opponent blocked Fynn with their creature, killing them both, while my other creature damaged the opponent. Immediately afterward Fynn's ability triggered and my opponent received two poison counters.
Why did that happen? The closest I can find is rule 603.10 relating to looking back in time, but a creature dealing combat damage to a player is not listed in the exceptions.

The triggered ability actually triggered before Flynn died, not after.
This happened because Fynn doesn't die immediately; it only dies when state-based actions are checked. Simultaneously, Fynn was dealt damage and your opponent was dealt damage by a creature with deathtouch. This event caused Fynn to trigger; as Fynn was still on the battlefield and was an "object that exists immediately after an event" to quote rule 603.10. In other words, Fynn triggered because Fynn was still on the battlefield at the moment the damage was dealt and immediately afterwards.
Before the next player gets priority, state-based actions are checked, and Fynn will die because it has at least as much damage as toughness. But it's ability has already triggered by that point, it just hasn't gone on the stack or done anything yet.

603.2. Whenever a game event or game state matches a triggered ability’s trigger event, that ability automatically triggers. The ability doesn’t do anything at this point.

704.3. Whenever a player would get priority (see rule 117, “Timing and Priority”), the game checks for any of the listed conditions for state-based actions, then performs all applicable state-based actions simultaneously as a single event. If any state-based actions are performed as a result of a check, the check is repeated; otherwise all triggered abilities that are waiting to be put on the stack are put on the stack, then the check is repeated. Once no more state-based actions have been performed as the result of a check and no triggered abilities are waiting to be put on the stack, the appropriate player gets priority. This process also occurs during the cleanup step (see rule 514), except that if no state-based actions are performed as the result of the step’s first check and no triggered abilities are waiting to be put on the stack, then no player gets priority and the step ends.

This makes sense, thanks! Arena could do better at showing what happens, because visually the creatures dissolve first and then the trigger appears with its origin as Fynn in the graveyard.

@TheWreckersCompanion Yeah I'm guessing Arena is only showing triggers as existing in any form once they are going on the stack.

@TheWreckersCompanion The digital MTG games (DotP, MD, MTGA) have all been pretty bad at showing things the way they really work. They make it look like damage reduces toughness, they 'destroy' objects immediately not as a SBA when they should be, etc. It feels more natural but leads to questions where what actually happens and what they make it look like happens don't match.


text:
When restarting the game, do players shuffle all their cards into their libraries?
Karn Liberated has an ultimate ability that says

Restart the game, leaving in exile all non-Aura permanent cards exiled with Karn Liberated. Then put those cards onto the battlefield under your control.

At first I interpreted this effect as saying that players shuffle all cards they own into their libraries, then carry out the start of game procedure, thereby making a fully and completely reset game and board state (except for the exiled cards). But reading the comprehensive rules only mystified me, because there's nothing that explicitly says that, with some rules seeming to require it and others requiring the opposite.
Rule 721.1 states

...All players in that game when it ended then start a new game following the procedures set forth in rule 103, “Starting the Game”...

And there is nothing in rule 103, nor rule 721, that states players shuffle all their cards into their libraries, only "shuffle their deck" in rule 103.2. So now I interpret the effect as all permanents remain on the battlefield and players' existing libraries become their decks in the restarted game. This is supported by rule 721.3:

Because each player draws seven cards when the new game begins, any player with fewer than seven cards in their library will lose the game when state-based actions are checked during the upkeep step of the first turn, regardless of any mulligans that player takes.

If players shuffled all their cards into their libraries before starting the game, they couldn't lose due to mill in the opening hand. I was comfortable with this state of affairs until I saw the example rule 721.2, which boggled my mind.

Example: A player casts Living Wish, bringing a creature card into the game from outside the game. Then that game is restarted. The creature card will be part of that player’s library when the new game begins.

Huh?? So suddenly a creature card does go back into the library? Wouldn't that apply to other cards as well? This example doesn't say what zone the card was in, seeming to make it unconditional.
The other argument I can think of is: if cards aren't shuffled back into libraries, why would Karn's effect need to specify that cards are left in exile?
I'm so confused.

Murgatroid99's answer certainly is the proper answer, but I just wanted to point out something in your question.

Because each player draws seven cards when the new game begins, any player with fewer than seven cards in their library will lose the game when state-based actions are checked during the upkeep step of the first turn, regardless of any mulligans that player takes.

If players shuffled all their cards into their libraries before starting the game, they couldn't lose due to mill in the opening hand. I was comfortable with this state of affairs until I saw the example rule 721.2, which boggled my mind.

There are a number of situations where this is possible.  The simplest example involes Karn, Liberated himself.
Imagine that you have slowly and methodically exiled each card from an oppenent's deck to the point where 54 cards are currently exiled by Karn.  Those cards, per the abilities on Karn, would not get shuffled back into your opponent's deck when the game is restarted.  Assuming your opponent only has 60 cards in their deck, they would only shuffle 6 cards back into their library, then attempt to draw 7 at the beginning of the game.
This would trigger the 719.3 rule, although technically this ruleisn't necessary because 719.3 is just a reminder to apply 704.5b even when drawing your opening hand.

Other scenarios where a player would restart the game with less cards in their library than when they started playing include ante cards (think token copies of Tempest Efreet).  Just because Ante cards are illegal in all formats doesn't mean that the comprehensive rules don't need to cover them.

The comprehensive rules very much do not cover silver-bordered cards. Most silver bordered card do not actually work according to a strict interpretation of the rules.

The CR does NOT cover silver bordered cards. There's no 702 entry for Last Strike, Undeathtouch, or Super Haste for example. Rulings (I think you meant rules tough) in general are just ways of stating the general rules in the CR for a specific case - none of them are "necessary" in that sense, they just simplify and specify what's already correct. As for rules, the CR says the same thing multiple times in multiple places - wherever it applies worded to focus on that part of the rule.

I've removed the bit about silver bordered cards, and you were right, I meant to say "rule" instead of "ruling" when I was mentioning 719.3

While murgatroid's answer is indeed the direct answer, yours is the one I prefer. The only obstacle to shuffling everything for me was rule 721.3 and you explained why that happens.

Yes, players shuffle all of their cards into the library when starting a new game.
The full text of rule 103.2 says

After the starting player has been determined, each player shuffles their deck so that the cards are in a random order. Each player may then shuffle or cut their opponents’ decks. The players’ decks become their libraries.

Your deck is not the same thing as your library. Your deck is the cards that you start the game with. When you restart the game, you pick up every card in every zone that you own to reconstruct your deck, then you shuffle it, then it becomes your library.
This is reflected in Karn Liberated's second ruling:

The player who controlled the ability that restarted the game is the starting player in the new game. The new game starts like a game normally does: Each player shuffles their deck (except the cards left in exile by Karn’s ability). Each player’s life total becomes 20 (or the starting life total for whatever format you’re playing). Players draw a hand of seven cards. Players may take mulligans. Players may take actions based on cards in their opening hands, such as those of Leylines.

"Deck" must refer to more cards than just the library for the exception to make any sense.

How do companions figure into this?  Or is that a separate question?

Similar to the card you find with wish cards, companions also start the game in the sideboard and later enter the game, so I would assume that the conclusion is the same as with the Living Wish example in the question and you shuffle them into your deck. The rest of the companion procedure happens "Before the game begins", so I assume that you don't get to repeat it when restarting the game, the same way that you can't re-sideboard when restarting the game.


text:
ID 1980s kids' astronomy book featuring a board game about asteroid mining
I am searching for a book that contained a playable board game within its pages.
About the book
In the UK in the late 1980s, many schools participated in the Puffin Book Club, where pupils could buy at a discounted rate from a monthly selection of books by Puffin (Penguin's imprint for kids).  I'm not sure, but I think that other, non-Puffin books were included in the selection as well.
I remember getting a large-format paperback book about astronomy and space science (quite similar in style to the Osborne books), possibly focusing on our solar system.
I can't find a back catalogue of Puffin books online, I'm not even sure it was a Puffin book, and the title could have been almost anything containing "space", "astro" or "solar", so finding it is a pretty tough task.  I'd imagine it's likely it may have been republished internationally as well, so I'm hoping it may seem familiar to someone somewhere.
About the board game
The  book included four (I think) board games, each of which took up a double-page spread in the book, where the board and rules were printed across the two pages.  I'm not sure if you just drew the pieces on paper yourself, or if they were supplied in pull-out form.
The only game I remember was themed around mining in the asteroid belt.  There was a large circular track of locations occupying most of the double-page spread, representing the asteroid belt, and there were pieces for the asteroids Ceres, Pallas, Juno and Vesta.  Each turn, you would advance each of the asteroid counters one space around in its orbit.  Each player also had a spaceship counter, which you would move slowly between the asteroids as you collected minerals by mining.
SE suggests "Solarquest" as a related topic, and the theme and ideas seem similar, and I suspect this game was influenced by Solarquest, but it's not the same thing.

I remember a similar themed book with games. But it was in Dutch.

I'm pretty sure this is the game "Space Pirates" from Battlegame Book 3: Galactic War. It's an Usborne book from the 1980s. The book does indeed contain four games.

There is a detailed review here.

This is a game of mineral prospecting in the asteroid belt. In other
words its a 'fetch and carry' type of game in which you pick up
natural resources from the major asteroids and deliver them back to
earth. The game is set in 2076. This probably seemed a safe bet in
1975, although the current NASA finances leave the prospect a little
more shakey. The game was not designed by series author (and Kingmaker
designer) Andrew McNeil, but by Dave Rotor. He had (I learn from his
BGG page) previously put out an SF tabletop game called 'Galactic
Warfare'.
Background
The game requires players to collect one resource token from each of
four major asteroids- Silicon from Ceres, Zirconium from Juno,
Titanium from Pallas and Magnesium from Vesta.

Awesome, I remember this from a long time ago. The cover was different but the game is exact the same.

The cover isn't familiar at all, but the Space Pirates game is exactly the one I was thinking of!  I wonder if it got republished elsewhere.  In any case, that's an excellent pointer, many thanks!


text:
What is this tic-tac-toe like game with green and yellow balls?
I found recently this board game in my parents' home (Spain); it seems like it might possibly be a version of tic-tac-toe.
Does anybody know the name of the game or the rules?

The 24 spaces (counting the holes) and the line connections make me thnik that it is of the morris family. But I've never seen a morris variation that used two boards.

The board looks pretty home made, but how many of the green/yellow counters are present?
It's possible the holes are ways to remove already placed pieces

I'm thinking could there be more boards?  Maybe it can be arranged different way.  there are thin lines coming off some circles to edge of blocks.

Looks like a variation of Chinese Checkers to me.

It looks almost like a two-player version of pachisi: https://en.m.wikipedia.org/wiki/Pachisi - the lines on the board could be the path around the board; in that case green has two home already. What argues against that are the lines disappearing off the sides of the board. (Two on one side, one on the other). Are both boards identical? Are there any markings on the other side of the boards?


text:
What is the minimal number of permanents which, in a finite number of turns, can create a copy of each of themselves?
Consider the following (unlikely) scenario in a game of Magic the Gathering:

You have X permanents on the battlefield, none of which have any counters on them (initially).
There are no static effects in play which interfere with the abilities of your permanents (aside from any caused by your permanents themselves, if relevant).
Neither you nor your opponents are casting any spells or playing any abilities from cards in hand.
Your libraries are any finite size...you and your opponents can take any finite number of turns without the concern of milling yourselves out.

Given this scenario, what is the minimal number of permanents, X, which are collectively capable of creating a copy of each of themselves within a finite number of turns? (X must be greater than zero)
As an example, and an upper bound, considering the following set of permanents, giving X=10 :
Helm of the Host
Karn, Silver Golem
Mirage Mirror
Mirror Gallery
4x Blinkmoth Nexus
2x Inkmoth Nexus

With 6 mana at our disposal, any of the lands can animate themselves and have the Helm equipped. So in 6 turns we've copied all our lands.
Using Karn's animate ability, Mirage Mirror and Mirror Gallery can be copied the same way.
Karn can be copied without needing to animate.
Finally, to copy the Helm itself, we

Have Mirage Mirror become a copy of the Helm (which does not die because the Mirror Gallery is suppressing the legend rule),
Animate the original Helm with Karn, and
Equip the Mirage-Mirror-Helm onto the original Helm.

I'm certain we can do better by being more creative with the mana base. But how low can we bring it? And perhaps there's an entirely different mechanic that would meet the conditions?

Is it acceptable if, during one repetition, some permanents are copied more than once? Must each permanent be copied, or is it acceptable as long as you have twice as many as when you started?

Clarification: I meant "twice as many of each", not just twice as many.

What would your rules be regarding pre-existing copy effects? For example, say I want to have a Phyrexian Metamorph that is a copy of a particular creature, but the original creature is no longer on the battlefield. Could I could that as just one permanent?

@murgatroid99 I would allow that, since I have no rule that precludes it. Although I will be obliged to accept mreux's answer with X=1, I am still very interested in your setup!

@Arthur: Yes. As long as, in some finite number of turns, each of the original permanents has been copied at least once, you're free to operate as you wish in the meantime. I am interested in seeing what you come up with! :)

Does Splinter Twin + Deceiver Exarch count, since you can make infinite Deceiver Exarch tokens?

@nick012000 I don't think it would for this question. Firstly, that would give you copies of Deceiver Exarch; how would you get a copy of Splinter Twin? Secondly, copies generated by Splinter Twin don't last past end of turn, right? I suppose I didn't explicitly rule out the case where all the copies are done within one turn, but the idea was that they should be able to self-perpetuate like cells...

X=1
Chronozoa completes the first iteration in 0 turn thanks to the 1st rule, then each other iteration in 3 turns.

I am disappointed that the answer is this simple.

Kind of same, but not surprised!

I don’t think this works according to the technical rules in the setup. According to the rules, your Chronozoa doesn’t have any time counters on it... Vanishing only triggers when the last time counter is removed, which isn’t happening here. Either it was already removed before the starting point of the puzzle (and the triggered ability countered), or something else prevented the time counters from ever being placed on it. Either way, you can’t just assume that the last counter removed trigger is already on the stack during the puzzle setup.

Despite @GendoIkari's accurate point about the technicality, this answer was the quickest and gave the lowest possible number, so I'm accepting it. (It's still in the spirit of the question, too--the "no counters, initially" rule was intended to maintain the self-sufficiency of the permanents in their action, and Vanishing creatures supply their own counters...)

A solution with X=3:

Mycosynth Lattice
2 x Mechanized Production

One Mechanized Production is attached to Mycosynth Lattice, and the other is attached to the first Mechanized Production, which is legal because it is an artifact thanks to Mycosynth Lattice. At the beginning of the next upkeep, the two triggered abilities create a new copy of each of those permanents.

But the second mechanized production never gets a copy.

You can get two copies of Mechanized Production, by waiting two turns. In game, there's no tracking of which specific object something is a copy of, so I don't think it makes sense to make that distinction here either.

Alternatively, the first copy of Mechanized Production can be attached to the second Mechanized Production, and then on the following turn, each original Mechanized Production is copied.

What I love about this one is that it would explode really fast!
Although, I suppose the game would end really fast, too.

I'm pretty sure it just ends up doubling each turn, except a little less because one Mechanized Production is stuck making Mycosynth Lattices. Other solutions have the same or greater growth rate.

X=1, but with two cards:
Mirror-sigil sergeant with Sea-Dasher Octopus mutated onto it to make it blue.
Works with any blue mutate creature.

X=1 solutions can be accomplished in a couple of ways:
Giant Adephage, Mist-Syndicate Naga, and Spawnwrithe can each simply attack and deal damage to the opponent, and they will create another copy of themselves.
Progenitor Mimic that is a copy of any vanilla creature (and many other creatures) will create a copy of itself each upkeep.

From the puzzle setup, there are no vanilla creatures on the battlefield for Progenitor Mimic to copy. You control X permanents, and after a number of turns, you need to have a copy of all of them.

In the comments on the question, it was clarified that pre-existing copy effects are allowed. The setup I'm describing isn't "a Progenitor Mimic entering the battlefield as a copy of a vanilla creature", but rather "a Progenitor Mimic already on the battlefield, having copied a vanilla creature, which is no longer present."

A solution with X=4:

Mirrorworks
Sol Ring
Mirage Mirror
Phyrexian Metamorph (artifact) that is a copy of a Spark Double (non-legendary) that was a copy of a Brago, King Eternal

Each turn, Brago attacks and exiles one of the permanents, then when it returns, pay for the Mirrorworks trigger using the mana from Sol Ring. Specifically, the turns go as follows:

Copy Sol Ring.

Copy Mirage Mirror.

Turn a Mirage Mirror into a copy of Mirrorworks, then flicker and copy the original Mirrorworks. That all costs 4 mana, paid for with the 2 Sol Rings.

Turn a Mirage Mirror into a copy of the Brago, then flicker the Phyrexian Metamorph Brago, have it enter as a copy of the Mirage Mirror copy of Brago, and then copy it with Mirrorworks.
Alternatively, flicker the Mirage Mirror, have it copy Brago in response to the Mirrorworks trigger, then copy it with Mirrorworks.

This solution relies on rule 706.9b:

Some copy effects modify a characteristic as part of the copying process. The final set of values for that characteristic becomes part of the copiable values of the copy.

This means that we can chain Phyrexian Metamorph and Spark Double to get an artifact non-legendary copy of Brago, and any further copy of that permanent will have those same characteristics.


text:
Can you tap three colored-mana-producing lands to produce one mana of any color?
I don't recall where I first heard of this being a thing, whether it just being from a friend mentioning it or reading it off of some random website. Regardless, my friend group and I have been playing with this rule for a good while now, though none of us can remember where we first heard of it, and we can't find any information about speculation of this online anywhere.
I know technically tapping a land produces one mana of the color on it (assuming it's your normal land that only produces one mana upon being tapped, such as a basic land), so technically it would be more accurate to ask if three colored mana can be swapped out with one mana of any color. If this is a thing, do the three mana need to all be one color, or can they be of any color, or even colorless? If this isn't, then I can only wonder as to where I first heard of this rule (though I guess nothing is stopping the rule being homebrewed in during casual play).

It's definitely an obscure house rule, nothing official like that ever existed. It does sound a lot like Settler of Catan's 3:1 trading port.

No, this is not a rule. I don't think there is a succinct way to provide a quote to prove it, other than "Read the entire rule set and see that it's not there".
But as you say, there is nothing stopping you from using that rule in casual play.

There are, however, cards such as Prismite that give you the ability to turn any 2 mana to 1 mana of any color.


text:
Is the 1" rule completely or only partially ignored on a successful charge in Necromunda N18?
When my fighter does a Charge (Double) action, is he allowed to move into the 1" space of any fighters he passes or is he only allowed to move into the 1" space of those fighters he intends to end up being in base-to-base contact with?
I am talking about the latest edition of Necromunda, started through Necromunda Rulebook in 2018, sometimes referred to as N18.
Page 58 of the Necromunda Rulebook in section "Charge (Double)" states the following:

A charging fighter can move to within 1" of one or more enemy fighters
… but if they do move to within 1" they must have sufficient movement
to get into base to base contact with at least one enemy fighter …

And page 61 in section "The 1" rule" states also:

The only exception to this rule is when … makes a Charge (Double)
action, in which case they may move within 1" of one or more enemy
fighters, provided that they end their movement in base to base
contact with one or more enemy fighters

For comparison, in the predecessor edition N17's rulebook, Necromunda: Underhive page 46 the following is written:

… they can move within 1" of a standing or Pinned enemy (or more than
one if they wish) but if they do, they must move into base contact,
becoming Engaged.

Which seems less complex and easier to grasp at first. Especially because the "base-to-base with at least one" condition is missing here.
I tend to believe that the "one or more" fighters that one can come closer than 1" to should be the same as those the fighter intends to come to base-to-base contact with. Together with the rule that a charge should always use the shortest path possible, this would allow fighters to protect fighters from being charged by blocking tight corridors or at least make it a lot harder to charge someone by increasing the distance to be moved around them.
I believe this mostly because in other Games Workshop games like Kill Team (second edition) which shares a lot of similarities to N18 and was released in the same year, it is even explicitly stated (Kill Team Core Manual page 24):

It cannot move within 1" of an enemy model that was not a target of
its charge.

I think this is supposed to simulate that if a fighter moves along an enemy fighter too close they might interfere with their charge.
This also appears in much simpler games of Games Workshop like Warhammer Quest: Blackstone Fortress, where a characters movement ends as soon as it comes into adjacency to a hostile model.
Is there any official source that make interpretation more clear? What would be your interpretation and why?

Not an answer, but I read the newer rule as less ambiguously allowing you to be within 1" of other models and not in base contact, so long as you reach the charge target


text:
Magic: The Gathering - Difference between card prints of the same set and same collector number, but different letter suffix
I have noticed that there are cards in Magic: The Gathering that belong to the same set, but appear under slightly different set number.
For example, in scryfall, for those 2 cards:

https://scryfall.com/card/sta/73/ja/%E5%89%A3%E3%82%92%E9%8D%AC%E3%81%AB
https://scryfall.com/card/sta/73e/ja/%E5%89%A3%E3%82%92%E9%8D%AC%E3%81%AB

They seem identical, the only difference I can see is that one of the images is a bit blur. But those 2 cards have different (?) set numbers: 73 and 73e (this is the 'blur' one).
From the same set, another one is:

https://scryfall.com/card/sta/62/lightning-helix
https://scryfall.com/card/sta/62e/lightning-helix

with set numbers 62 vs 62e. Here, it's slightly more obvious that there is some difference in those cards, if you pay attention to the grey-coloured part on the card 62e, but there is no other difference.
I have also noticed the same thing in other sets as well, there are cards with same collector number (in the number part) but different suffix, ie. 70 vs 70e, 80 vs 80p etc, but they seem identical cards.

Are those different card prints (from collector perspective)?
How can they be distinguished?
Are all of those card print versions included in booster packs?
Is there any difference in the collector value of the card, for such cards?

Ignore the difference in the art quality on Scryfall - that's just a matter of the quality of picture they happened to use for each printing.

Scryfall (and other similar sites) are using the "e" suffix to denote the "foil-etched" cards first introduced in Commander Legends (see the section "Foil-etched legendary creatures", although this treatment was obviously extended to non-creatures in the Strixhaven Mystical Archive) - these are to some extent "partially foiled" cards where the frame or other areas are highlighted, but not the full art.
Are those different card prints (from collector perspective)?
Yes - but then collectors are remarkably precise in their categorisations, so this shouldn't be a surprise.
How can they be distinguished?
Quoting from How to Tell the Difference Between Mystical Archive Cards on TCG Player:

[Etched foils] look very similar to the non foils, BUT when you tilt them, there will be a shiny texture in the framing of the card. The English version has the etching in the gold parts of the frame. The Japanese version's texture is around the textbox (in the colors that correspond to the card's colors!).

The video in the Commander Legends article linked above illustrates this nicely.
Are all of those card print versions included in booster packs?
This time quoting from Collecting Strixhaven: School of Mages:

The only place to get foil-etched Mystical Archive cards is in the Collector Booster.

i.e. you will not get any foil-etched cards if you buy set or draft boosters.
Is there any difference in the collector value of the card, for such cards?
There is always a difference in value between different treatments. As of July 2021, it seems that for the regular art of Revitalize, the foil-etched treatment is more valuable than the non-foil, but for the Japanese alt-art, the foil-etched treatment is less valuable than the non-foil (but they're all under $1, so you're not going to get rich on any of them). Make of that what you will.

Edited the description to define "different cards" bit better. Thanks for the detailed explanation Kendall! One question that comes up from this, what is 'foil etched'? Is it different than regular foil that we are used to find in boosters? Or is it the same thing? If it's the same thing, then the 'e' suffix would just mean that the card is a foil version of the normal card then?

All this is pretty much covered by the Commander Legends and TCG Player articles I've linked, but I've updated the answer a bit to include some information.

I'm both a many years Magic player and, as luck would have it, an antique dealer - I have to say that collectors (of all things) find the most inane details to draw distinctions. Finding which of those distinctions (especially when no one has noticed such a distinction before) is critical to identifying value!


text:
Why does the effect of Enchantress not stack in Dominion?
In Dominion, if my opponent plays the Enchantress, this prevents my first action card from executing its instructions and gives me +1 Card and +1 Action instead.
My question is, what happens if my opponent plays two Enchantresses?
Will I get +2 Cards and +2 Actions on my first action card? I guess No, but I need an official source or convincing reasoning on this.
Initially, it might sound like the Enchantress is "replacing the instructions of the action card with +1 Card, +1 Action", and replacing it twice still leaves a single +1 Card, +1 Action.
But this is not the phrasing on the card:

Until your next turn, the first time each other player plays an Action card on their turn, they get +1 Card and +1 Action instead of following its instructions.

The crucial word here is "instead". I do not know how to interpret this exactly
One interpretation is as follows: the second Enchantress encounters the first action card, prevents its instructions from happening (even though they would not have happened anyway because of the first Enchantress) and gives me +1 Card and +1 Action, stacking with the first Enchantress,
The Wiki is not helpful on this, not addressing this point at all. Is there any official source telling me that I will get only +1 Card, +1 Action? Has anyone encountered this situation in the online game?

Enchantress cannot stack; the first action played by an opponent will just give +1 card +1 action.
A quote from the creator/designer of Dominion confirms this:

Throne / Enchantress does not give other players +2 Cards +2 Actions for their first Action played.

Source
The reason it works this way is that Enchantress only does something when you would be otherwise following a card’s instructions; that comes from “instead of following it’s instructions”. So after the first enchantress has done it’s thing (whichever Enchantress you choose to follow first when playing a card); the second Enchantress no longer has anything to do, because there is no time at which you are about to follow the instructions that can be replaced.
The same idea applies to Trader (for original wording Trader; it has been updated to no longer use “instead”). When you would gain a card; you can reveal Trader to gain a Silver instead. Once you do that; you no longer “would gain” the card you would have gained; so you can’t reveal Trader again to gain yet another Silver. Only one Silver per card you would have gained.

One interpretation is as follows: the second Enchantress encounters the first action card, prevents its instructions from happening (even though they would not have happened anyway because of the first Enchantress) and gives me +1 Card and +1 Action, stacking with the first Enchantress,

This interpretation disconnects the preventing of following instructions from the new thing you get instead. It would be correct if Enchantress said “the first time each other player plays an Action card on their turn, they do not follow its instructions. They get +1 Card and +1 Action.” With this; second Enchantress wouldn’t care if the instructions were going to be followed anyway. But as worded; Enchantress doesn’t fo anything if you were not going to follow the card’s instructions; because it needs the thing to replace.

Because it is a replacement effect. The first one you play replaced it with +1 card and +1 action. The second one you play would replace that with +1 card and +1 action but that is the same so there really isn't a change.
Each additional one you play replaces the effect of the last one meaning it will always be +1 card and +1 action. This is important in case there is another card that also has a replacement like effect.
If they wanted them to stack they would have added text that indicates that and not just replace what happens.

This answer probably matches designer intent, but note that card text doesn't effect that. The replacement effect says that the player gets +1 card and +1 action instead of following the instructions on the card rather than the card's instructions being replaced with the new instruction  "+1 card and +1 action".

At first I thought this was the correct interpretation and had almost included it in my answer. But reading through various posts on the Dominion Strategy Forums; Enchantress is always referred to as making it so that instead of following instructions you do something different; which is subtly different than replacing the card’s instructions with new instructions. The card still has its original instructions; you just don’t follow them. Same conclusion; different reasoning.


text:
What is this game involving white disks, fish, and wavy pieces?
I'm wondering if someone might be able to help me out with an odd request. Living in Ireland, someone approached me to identify a game. They found games pieces in a very old wooden box (100 years or so old). There's three types of pieces, each made of a white stone (possibly marble):

Very thin circular white disks, maybe an inch and a half in diameter. Maybe 20 or 30 of these.
White fish pieces, maybe 5 to 10 of these.
What I think are river pieces (wavy lines), 5 to 10 of these too

There were no instructions for play or title or anything. I'm not sure if it was a published game or just a cobbled together family game that was ever in any way well known.
So the pieces are made of something that almost looks like marble,could be some form of white stone. I dont have a photo but will try to get one. The pieces dont look that old, but i think the box kept them in good condition
Does anyone have any idea as to what this could be?

What were the pieces made of? Wood?

Are you able to post pictures? Those are a lot of help in identifying games.

Is there a board or just pieces?

I remember my Grandparents having a game like this when I was young.   The pieces were cardboard, not stone.  It was a sort of fishing game of chance.  You "fished" for a piece and got a fish, nothing or another turn.  I don't remember any name for it.

The material, age, lack of instructions, and the fact that some of them are shaped like fish, suggest that these are 19th-century gambling chips.
I am guessing based on the fact that some of them are in the form of fish, which was historically a common design for a gaming chip.
For example, see here for a reproduction of such chips.
Perhaps you could see if there is anything similar to your pieces on antiquegamblingchips.com?

Brandubh could be a good possibility.  It's a variant of Tafl games.
https://en.m.wikipedia.org/wiki/Tafl_games


text:
How can we make Just One more difficult/competitive?
Our friend group has been playing Just One a lot but it gets a bit boring quickly as 95% of the time the person can guess their word, even if we follow all the rules strictly (only dictionary words, no punctuation, no capitalization, singular form only). Is there a way to make the game more competitive and challenging?

How many people are you typically playing with?

@PatLudwig 5 to 8 people

@Allure um, we’re playing it as a group of friends, not in an online tournament. No one is going to do prior agreements.

@Allure It does feel that is players are going to prior agreements about types of clues to give then are kind of missing the fun of the game of trying to not to the same any anyone else.  Is having that kind of system in place not a bit joyless to play?

One simple idea has been posted here. In short:

Each turn, select two words to be guessed. If using the provided cards, this can be done by selecting two numbers, as described in the linked post. However, if you have played a lot, you might start to remember some combinations of words found on the same card; in that case, it may be better to take two cards and select one word from each.
Each player other than the guesser still writes a single clue.

To further adjust the difficulty, you can choose whether or not to allow players to indicate which word or words their clue is intended for.

Here is one idea I had.

Each turn, two players will be guessers (instead of one).
As usual, the other players each write a clue.
After removing duplicates, divide the remaining clues between the two guessers. (This decision could be made by the other players together, or if it proves impractical to do that without leaking information, a single player could be designated to do the division.)
Each guesser writes down a guess, seeing only the clues assigned to them. (If playing in person, have the guessers positioned opposite each other, to make it easier to show clues to only one of them.)
Score a point only if both guesses are correct. (This avoids having it be optimal in some situations to assign all the clues to one guesser, unless that is unavoidable because less than two clues remain.)

Add a traitor?
Find a deck of cards, take (n-1) hearts and 1 spade, and everybody draws a card.  The one who draws the card is the traitor, and his job is to sabotage your attempts without being identified.

Too easy to identify the traitor as they would have dissenting clues.


text:
What's name of of this Jenga game?
I once played a game in which players need to make a chain of plastic rectangular blocks by hooking them together. The plastic blocks are coloured and semitransparent. First block hooks to the top of a tower and player continues attaching blocks.
A player would lose the game if the tower topples due to heavy weight of chain. Player can start as many chains as they like from the tower by adding a new piece at top of tower as long as there's space available.
Note: The name of the game includes Jenga word in it but I'm unable to recall the entire name. I have drawn a sketch of how the game looks like.

This appears to be Jenga Max. Here's the description from the overview on Board Game Geek:

Jenga Max is a dexterity game similar to but different from Jenga. Players take turns attaching plastic pieces to the top of a tower. There are different ways of attaching pieces together, and if an added piece disturbs the balance of the tower, it will fall down and the player loses. The winner is the last play who successfully placed a playing piece.


text:
King of Tokyo: Poison Quills
The Poison Quills card says
When you roll 2 2 2 or more, add 2 damage to your Roll.

Does that mean 2 2 2 2 or also 3 3 3?

usually more implies more dice of the same type... no over the number...

Can you give the complete text of the card?  That doesn't seem like the full text of the card.

Added the exact text in question to make it more clear.

Wow what version is this from? Not only is the first part wording different, the effect itself is actually completely different!

Now I'm wondering if poison quills can be interpreted in some printings to be able to contribute to activating berserk mode (requires rolling 4 claws) from the Cybertooth expansion...  If quills adds two claws, I'd think it does.  If quills just adds two damage, I'd think it doesn't.

You need to roll at least three 2s; it doesn't matter if you roll more 2s.
There are a couple different wordings used depending on your version/printing; other versions do not say "or more". One version I found just states "when you score 2 2 2".
So the text applies when you are scoring three 2s.

I have the Spanish version and it says 2 2 2 o superior which makes it sound more like 3 3 3 than 2 2 2 2 (otherwise they would have translated it to o más.
Also if it was 2 2 2 2+ then I don't think they would need to mention the or more. Just 2 2 2 would be enough.
Same for other cards that say 1 1 1 o superior so 2 2 2 or 3 3 3.


text:
Should investigators take Set Aside Cultists when playing with the Epidemic prelude?
When playing with Epidemic prelude

After resolving setup, if Abhoth is the Ancient One, each Investigator spawns 1 Cultist Monster on a Wilderness space that does not contain a Cultist Monster.

and an Ancient One have Set Aside Cultist Monsters should these be drawn from

the Set Aside Cultists?
the monster cup?
first the monster cup if able, then the Set Aside cultists?
first the Set Aside cultists if able, then the monster cup?

Draw from the set aside Cultists only.
Rule Reference, Page 10

Components that are set aside cannot be used except by effects that specifically name them. For example, a Monster token that has been set aside cannot be spawned except by an effect that specifically names that token.

By that - if you have set aside Cultists, you can not randomly or purposely draw a Cultist from the monster pile even if you have more cultists in the available pool. If you draw a cultist randomly from the pool, it should be discarded and another monster draw in it's place (if you haven't removed all Cultists from the pool.
Component Limits - the Core and Under the Pyramids sets have a limit of 8 cultists (5 core, 3 UtP). If you play with only the UtP expansion you would be removing all of the cultists. It can be inferred from this that the intention is that all cultists are removed from the general pool and only drawn from the set aside. In general when you set aside a monster the game specifies the maximum amount available but Cultists are prevalent in other expansions and this is not clearly dealt with in the rules.
Thematically when you set aside a monster it has a specific role, method or story around how they're coming into the world so it makes sense you would not encounter them randomly or in other ways.

Where did you find "if you have set aside Cultists, you can not randomly or purposely draw a Cultist from the monster pile even if you have more cultists in the available pool."?
I was only able to find "Monsters cannot be spawned from the Monster cup if the Monster cup is empty. Set aside Monsters cannot be spawned if all set aside Monsters of the named type are on the game board.".
The "Set Aside" section in the reference guide doesn't say that when setting a Monster aside then other Monsters of the named type should be returned to the game box.

The quote is directly from the rules reference. The rest is my inference based on that rule.
In general when you set aside a monster you are removing all of those monsters from the pool. Expansions have added additional tokens of the same types which has clouded the rule a bit but look at it this way.
The core game comes with 5 cultists.
The UtP expansion adds 3 cultists.
If you play the core game with only the UtP expansion there are ONLY 8 cultists available, therefore you would ONLY be able to draw from the Set Aside cultists.


text:
Clarification on "shuffled back into deck" for double-sided cards
The reference guide states for Double-Sided Cards that

When a double-sided card is discarded, it is immediately shuffled back into its respective deck.

The phrase "shuffled back into deck" is never elaborated on and I couldn't find any discussion on this related to Eldritch Horror.
The mechanic "shuffled back into deck" can mean two things

insert the card at a random location in the deck
add the card to the deck then shuffle the deck. This would also trigger the rule "another player cuts the deck"

The only explanation of this mechanic I could find is for HearthStone.

Shuffle into deck effects place cards into a player's deck, with its placement randomly determined. While not stated in card text, the randomness of the card's placement is achieved through subsequently shuffling the entire deck.
https://hearthstone.fandom.com/wiki/Shuffle_into_deck

This would suggest that the deck should be shuffled when a player discards a double-sided card

The Eldritch Horror page on the Fantasy Flight website has all the rules and reference guide PDFs for free. If we look at page 5 of the reference guide on Discarding it says:

Double-sided cards, such as Spells or Conditions, are immediately shuffled back into their respective decks when discarded.

This means that these Spells and Conditions are never permanently discarded from the game. Cutting the deck is not necessarily integral to the gameplay. The important thing is that whenever a card is re-added, all cards in the specific deck (including the re-added card) are randomly shuffled together so that the re-added card ends up in an unknown spot in the same deck.

so you also agree that the "shuffled back into deck" mechanic in Eldritch Horror means that the deck should be shuffled

Absolutely. It says to shuffle the deck; the entire deck gets shuffled.

The important part is that next time you gain this condition you may gain the same exact back side effect, but don't know about it. You can achieve this by whichever method is more convenient, as long as it provides enough randomization


text:
2 players for 2 nobles
If player 1 got enough resources for noble A and B, because the game only lets players obtain 1 noble per turn, player 1 got noble A.
Then player 2 got enough resources for noble B.
Who will get noble B, is it player 1 because it is a "noble in waiting" or player 2 because of turn player priority.

Player 2 gets noble B if they have enough for that noble on their turn it doesn't matter if Player 1 had enough for it as they chose Noble A on their turn.
It doesn't matter if you have enough to get that noble if you are unable to take it on your turn due to take a different noble.

From the rules:

At the end of their turn, each player checks the noble tiles in
order to determine if they’re receiving a visit from one of them.
A player can be visited if they have (at least) the quantity and
type of bonuses indicated on the noble tile.
It is impossible to refuse the visit from a noble, which is not
considered to be an action.
If a player has enough bonuses to be visited by more than one
noble at the end of their turn, that player chooses the noble to
be received.

So player 2 will receive the noble in this case. It does not matter whether player 1 also has the requirement; you only receive a noble when it is your turn.
Note that if it were the other way, and noble B also were "in waiting" for player 1, then there wouldn't be much point to limiting it to 1 noble per turn. It would only make a difference in the rare case that the second noble would cause the game to end. The point of limiting the nobles to one per turn is to give the other player(s) the chance to get the other noble for themselves.


text:
Smallest Scrabble board with all letters in play
Was just doing a mental exercise and thought I'd ask people with actual knowledge and expertise in the game's theory:

What is the "smallest" legal Scrabble board on which all letters have been played at least once?

"Smallest" is defined here as "smallest occupied tiles-squared area".
The area is absolute: no credit for any empty spaces within the occupied area.
A blank is not required to satisfy the solution.
While not restricted from use, blanks do not count as "playing" their respective letter either.

(Other variations would be "fewest tiles in play", or even just "lowest total board score".)

As always, I did run searches first—here and elsewhere—before posting this. But if I still missed an answer to this same question, please direct me there with my apologies. I'm also only an amateur Scrabble player, so if there are any "official" terms that I should have used in this post, please let me know.

Does it have to be a square; or are rectangles allowed? If a square, then 36 would be a simple minimum baseline; and I wouldn’t be at all surprised if that were the actual answer.

Rectangles are fine. I'd also like to see solution boards if possible…

6x6=36, 7x4=28, 9x3=27, 10x3=30. The question comes down to which have legal solutions. Specify dictionary (may be relevant). My gut feeling is 6x6, because that allows the most extra vowels, but it may not have a viable solution.

I suppose a script could just brute force the problem, but I thought maybe a solution (or solutions) were already known. As for Dictionary, I'm a complete amateur for that: whatever is generally used for US tournament play.

If you don't find a definitive answer, the folks at Puzzling would probably find this an interesting challenge

When you say "legal Scrabble board": are you just asking that the words in the final configuration are legal? Or do you require that it's possible to reach that final position via a sequence of legal Scrabble moves?

Sorry to be tardy with the response, but I was implying a configuration reachable via normal/legal gameplay.


text:
Can you identify this John Conway board game?
A colleague of mine received the item pictured below as a gift from Diana Conway, the widow of the mathematician John H. Conway. It was found among Conway's many recreational possessions, but was not accompanied by any identifying information.  Presumably it is some kind of game, though possibly it is some kind of puzzle. In case it is not clear from the photo, there is a 7x7 wooden board accompanied by some number of (but far fewer than 49) tiles, each of which fits snugly into one of the squares of the board. The tiles are dark on one side and light on the other, which is reminiscent of the game Reversi/Othello. The game was found in a plastic bag, so it's possible that some pieces were lost.
Someone suggested to me that this might be a version of
Pathagon, but in Pathagon, there is no reason to have tiles that are dark on one side and light on the other side.  Tiles that fit snugly in the squares are also a bit strange if the tiles are meant to be removed or flipped easily when surrounded by other tiles.
On my own website you can view a
short video
that demonstrates that the tiles can be slid along the board (as in the game
Wąż), although not especially smoothly.

Most likely it’s Infection (aka Ataxx), which Conway was at least aware of, having referenced it in Winning Ways for Your Mathematical Plays, Vol 1

See http://www.pressibus.org/ataxx/gen/gborigines.html

Thanks for this suggestion. Do I understand correctly that there should be 49 tiles, or close to that? This isn't impossible but it would mean that a lot of pieces were lost.

I've never played, but the production board there indicates yes. Maybe Conway didn't complete his set build, or maybe he just used it for testing variants, or (insert other wild speculation here) ...?

I hate to point out the obvious but, if there are indeed missing pieces, couldn't this just be a small board on which to manually conduct a Game of Life simulation?

This does seem like a possibility that should be considered.  Perhaps a one off gift from a fan that he kept?

@PatLudwig While it could be a possibility it seems there should be more to this answer to explain why it could be to conduct said simulation.

@JoeW - I would agree that this isn't a perfect answer, but as it stands, I feel this answer has merit and should be considered rather than summarily deleted.

I would have added it as just a comment, but for some reason as a new contributor I can only comment on my own posts.

nb: the game of life wouldn't have flippable pieces. Flippable pieces mean each cell has three possible states: empty, dark-side-up, and light-side up. In Conway's Life, cells have only two states. "dead" cells would simply be empty.

As suggested, if there were indeed missing pieces to cover the whole board, that's binary conditions on each cell. Was just a thought given the previous owner.

@L.ScottJohnson - actually two colors was the original way to play! 1. Start with dark counters. 2. put a second dark counter on all counters that will die.  3. Put a white counter on all birth cells. 4. cleanup - eliminate squares with doubled counters, replace white with black.  From Martin Gardner's column in Scientific American, October 1970.

@PatLudwig OK. But still not calling for flippable pieces.

@L.ScottJohnson - If you don't feel there is a similarity between two different colored pieces and pieces that have two differently colored sides AND that "this answer is not useful" then vote your conscience and move along.

@PatLudwig Non-sequitur, but I don't see the option to vote not to delete, so I'm not able to vote my conscience (except by omission).


text:
Pulling double after Weak 2 and competition?
This hand is discussed in the April 2021 ACBL Bulletin (p. 8) as a tempo issue, but I am interested in the underlying bidding issues
Bidding

West
South
East
North

2S
3H

3S
Pass
Pass
4C

Dbl
Pass
4S
Pass

Pass
Pass

East hand

S
H
D
C

KTxxxx
xx
Txxx
Q

West hand

S
H
D
C

Jxx
AQx
AQJx
xxx

Is West's double a reasonable action?
Is East correct to pull the double?

Both sides vulnerable, duplicate match points, assume Am Std or 2/1 bidding.

@Mycroft  I object to the recent edit, which has the IMO quite undesirable feature of not putting the dealer in nth first column, and also if I am not mistaken, changes the form in which the hand was presented in the Bridge Bulletin, but I have not yet double checked that source, so i have not, yet, rolled the edit back.

It's your question, revert if you wish. I was surprised that you thought the ACBL Bulletin would do that though, as I remembered it being the way of RIchard's style guide (one of the first online, and almost universally followed these days),  So I checked - and in the April 2021 Bulletin, bidding diagrams are universally "west first". The item in question is a Letter to the Editor, which has text bidding, not a diagram, so it's not an example.  My original phrasing of this comment was inappropriate, and I truly apologize.

I would never pull a double on that auction, regardless of my hand. "The book of rebids by preempter is 1000 pages long, and they're all blank," as the saying goes. Responder is the captain on this auction, and the double is not cooperative. If I'm not willing to sit this double, I shouldn't have opened.
West's double is a bit aggressive, but not terribly so at matchpoints. The AQx in hearts is a great holding, and the three clubs are at least trumps not held by the opponents and a good opening lead. The diamonds are also likely worth a couple of tricks. You're counting on a favorable layout or perhaps a trick from partner, not an unreasonable hope at these colors. The diamond king in dummy would be very unwelcome.

I agree with ruds - sort of.  I play with my partners that double by preempter is "I want to sacrifice"; explicitly so that preempter's partner can say "I don't".  Preempter doesn't get to pull (playing any standard system).  If their contract makes, it's partner's misguess.  So that's question 2 solved.
Having said that, the answer to "is the double reasonable", like every answer to questions about preempts, depends critically on what a weak 2 looks like vulnerable in first seat for this pair, not on whether anything is "right".
With one regular partner, this is not a good opener, but it's not a minimum either.  In that pair, we don't double many games; and we miss a fair number of our own.  But they're figuring out the hand at the 3 level a lot more often than the field.  I wouldn't dream of doubling this auction; partner has a zero-entry hand and KD or stiff heart in dummy means they make it.
With my other regular partner, this is not only too weak to open, it's not even close.  Our partnership philosophy is that we can defend better than the field, so don't lose the hand in the bidding.  We bid (and double) very constructively after our weak 2s.  But we pass several hands that others are preempting with, and the opponents get to better contracts, too.  Here, I expect I'd double, and I know my partner would.  And it's basically going to need a spade void, or Axx opposite stiff, and both kings in dummy to make.
Obviously, what you can expect from partner determines if this is a reasonable double.  I expect to get both my aces and one of my queens (likely the HQ); and 200 is the magic matchpoint score.  On the other hand, +100 beats -100 just as well as +200 does, and is 3S making?  I think it depends on the red kings (as well as how good opener's suit is).  If 3S makes, then 4C goes down; if 3S goes down, 4C likely makes.
Note: that they're willing to play 4Cx is scary.  It sure looks like hearts are a source of tricks and a pitch for diamonds; and you need partner to lead diamonds even if the king is right.  Will partner ever get in?
Another interesting question for the double - are you willing to double 4H if LHO goes back there?  On this hand, yes - you're happier doubling 4H than 4C; at least you're not getting your heart Q ruffed out.  But with a different hand - "Don't double the only contract you can set."
But preempters don't pull partner's penalty double.  They have no idea what partner has.  Unless, of course, partner makes it clear their double isn't a lock, say by tanking before doubling.  Then preempter looks at their hand, sees no help, and maybe pulls it...triggering a TD call and an adjusted score if pull was right.

It seems to me that the double by W is reasonable. West would anticipate at least three tricks in the red suits, and since west is over the probable strength in the north, the red finesses have an above-average chance of working,so there might easily be 4 or even 5 red-suit defensive tricks.. One spade trick is plausible, so the double seems a fair choice.
As for East, well first of all I would not open this with 2 spades, particularly not vulnerable. If their (EW) agreement is to open a suit as poor as that, then the double doesn't look as good.  There is surely less defensive value here than partner might have expected, so pulling the double might be the winning move, but for all  East can tell West has this set cold. I would go with "trust your partner" here.

With 9 Spades assumed for West and East. the chance of Opponents getting a discard on a third round of Spades is small. Despite the very weak hand, West has the defense promised by his opening preempt and should be pleased by East's double - they stand to make more defending than Declaring. His preempt has worked.

@forget, I don't quite follow your point about  A spade discard. NS have a 9-card club fir including the AKJT, an 8-card Heart fit including the KJ9, and the AQ9 of spades. I would expect them to have 4 or 5 Club tricks, 1D trick  3 H tricks, and 2 spade tricks. We don't know the distribution, but the original letter said that both 4Cand 4s were cold .as the cards lay. With the sort of spade holding I would expect from my partner as W, (2 of top 3 or 3 of 5) NS would get at most 1 S trick, if any.

If EW have 2 Spade losers and a Club loser then the red suits are solid with both finesses onside. Even with 2 Spade tricks and 6 Club tricks how is 4 Clubs cold with no red suit tricks available? 2) I never said I was happy with East's Double - I said i was happy to sit for it as West. 3) "Original letters" are rarely correct, and only serendipitously so, in their statements of what can and cannot be made. if they could assess playing prospects well they'd be writing answers instead of questions.


text:
Duplicate bridge rules for Revoke, and any penalty that should be imposed
In a recent game we found 3 players had  4 cards left and one player had 3 left. This was noticed on the 9th trick and no one knows when this happened.
What, if any, is the penalty?

Did the missing card turn up in one of the already-played tricks, on the floor, or not at all? This will affect the answer.

The Short Answer: The Director finds which card is missing and restores the player's hand. This card is considered to have belonged continuously to the owner's hand, and as a result may have caused an accidental revoke, which is penalized as if it was any normal revoke.
The long answer:
The first step of a Director in this circumstance is to review each of the tricks, starting with the most recent, until a irregularity is detected. Most likely, the player with 3 cards left will eventually find a trick where they played 2 cards, at which point law 67: Defective Trick handles the rest:

(a) When the offender has played more than
  one card to the defective trick, the Director
  inspects the played cards and requires the
  offender to restore to his hand all extra
  cards, leaving among the played cards the
  one faced in playing to the defective trick
  (if the Director is unable to determine which
  card was faced, the offender leaves the
  highest ranking of the cards that he could
  legally have played to the trick). Ownership
  of the defective trick does not change.
(b) A restored card is deemed to have
  belonged continuously to the offender’s
  hand, and a failure to have played it to an
  earlier trick may constitute a revoke.

The Director should avoid, when possible, exposing
  a defender’s played cards, but if an extra card to be
  restored to a defender’s hand has been exposed, it
  becomes a penalty card (see Law 50).

Should the rewind result in what appears to be normal play, then the card must be either on the ground/otherwise hidden, or it doesn't exist. Law 14 gives the solution for this problem:

if the card is found elsewhere, it is restored
  to the deficient hand. Rectification and/or
  penalties may apply (see 4 following).
if the card cannot be found, the deal
  is reconstructed using another pack.
  Rectification and/or penalties may apply
  (see 4 following).
a card restored to a hand under the provisions
  of Section B of this Law is deemed to have
  belonged continuously to the deficient hand.
  It may become a penalty card, and failure to
  have played it may constitute a revoke.

The same concept is used if a card is missing during the auction/ just prior to the play: The hand is re-constructed by finding the card or using a new deck, and the hand continues as if the offender had the card all along. However, in any case, the fact that there was a missing card is unauthorized information should it happen after the first call, and Director should make sure that partner doesn't make any inferences based on the fact that some/all of the offender's calls or plays were made with 12 cards.
There's also the highly unlikely but amusing possibility that the problem isn't that one player is missing a card, but three players have an extra card. This results in a similar process, but a much harsher one, as a player who played no card to a trick would "complete" it by playing a card (following suit if possible,) but counting that play as an automatic revoke regardless. If it turns out they had an extra card before the play of the cards, then it's entirely up to a Director judgment call as to whether the situation can be rectified (usually no) or if adjusted scores are required.

Then of course, there is the ubiquitous auction "Pass" "Pass" "Pass" "Director!  I played this hand last week in 1NT making 2. LHO's hand is ... and RHO's hand is ...."

@ForgetIwaseverhere Bridge player memories are very good :) (also: the reason I am not a good one)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Can one reasonably make inferences about partner's first bid based on the second bid?
With no one vulnerable at matchpoints, my (occasional) partner dealt and bid one diamond with the following:
(s) 752 (h) QT7 (d) AQJ52 (c) A8.
I bid one heart. She re-bid two diamonds, and I reluctantly bid three clubs, a suit in which I was as long as hearts. She rebid three diamonds, and I passed, even though my distribution was 3-5-void-5 (with seven high card points).
Partner said, the she didn't want to raise my (presumed) four card heart suit with only three, and she never dreamed that I had zero diamonds. I said that I had bid two suits, and therefore was more likely to be at least 5-4 hearts-clubs than 4-4, wanted her to choose a round suit and implied shortage in her suit, since I didn't raise two diamonds. Was this (verbal) reply reasonable given the previous bidding?
And was I right to pass based on my mistaken premise that partner had seven diamonds for her bid, on the grounds that 7-0 in diamonds would be better than a (presumed) 5-2 fIt (which was actually 5-3) in hearts?

You really have to teach your partner to ALWAYS rebid 1 NT with a balanced or semi-balanced minimum after opening 1 of a minor. The auction 1D 1H; 1NT 2C; 2H Pass; will work in any system at all that plays 15-17 NT, regardless of what 2C means. The moment the minimum misfit is revealed - the object is to settle quickly in a playable denomination that won't get doubled and as a chance of making.

Egads.
Your partner should have bid 1N over your 1H bid.  The 2D bid promises a 6th diamond (at least not in competition over a 1H response).
You should have passed 2D - your 3C bid promises more strength, at least close to game forcing (some partnerships will play it as game forcing).
Your partner should bid 4H over your 3C bid.

Tom doesn't tell us his hand. Why do we believe that the hand was weak enough to pass 1NT? It could well be invitational. Frankly, based on past questions it may well have merited a game force.

He says it was 3-5-0-5 with 7 hcp - I wouldn't consider T98 AKT98 - T9876 invitational

Ah, I missed the 7 hcp part

My 3 clubs was a "rescue" bid. I put partner on 13 high card points and six diamonds for her two diamond rebid, so we would have 20 between us. My three club bid meant "1) I can't stand diamonds even if you have six so I have 0 or 1; 2) I am 5-5 in the "round" suits so pick the one in which you have three cards. With 20 hcps and a 5-3 fit, we stand to go down one, versus down two in two diamonds with a 6-0 fit (law of total tricks). If we're lucky, our spade shortage might allow us to make three clubs or hearts."

In standard american, your 3C bid is forcing for one round and invitational to game.  How else would you bid Kx AKxxx xx Axxx?

@TomAu Rescue bids are very rare in well-designed bidding systems because constructive bidding is more important -- when you have a misfit, you should stop bidding, rather than raising the level. In this particular instance, partner may have 7 or even 8 diamonds, and 0 clubs. If you feel you must rescue, you can rebid your hearts, which shows more hearts than you have but at least doesn't promise values.

@AlexanderWoo nails it, but I'll show the inferences.
Re/Bid NT

Almost all systems have a "NT ladder"; a large number of sequences of bids that show a balanced opening hand of a limited (2 or 3 HCP) strength.  It is just easier to bid if you use them if possible.This example: opener can treat her hand as balanced (even with a strong 5-card minor) and show "12-14 balanced, 4 or 5 diamonds, not 4 hearts"(1); or she can treat her hand as a one-suited diamond hand and show "12-16/bad 17, 6 diamonds, reason to avoid 1NT (probably shortness, likely in hearts)". Which one will it be easier to evaluate your hand (whatever it is) against?
This advantage is so strong that players over time are lumping more and more hands into "balanced".  5hearts-3-3-2?  Sure.  6-diamonds-3-2-2?  Absolutely.  2=2=2=bad 7?  Some will. 2=4=2=5 16 count?  Many will open 1NT because 1♣-1♠ leaves them without a comfortable call.  Open suit? Two?  Sure, partner has a stopper, or they don't and it might not get found, or partner will put us somewhere where it doesn't matter.
Who wants to play in a minor anyway?  Especially if game is available, it's likely in NT (2).  So use the NT ladder to show both parts of one's hand rather than focussing on one.
Frankly, I wrote something in the last century that said "your rebids show 6-card suits".  Yes, there are exceptions, and there are times where every other call is a bigger lie, but it's a good thing to assume in general.  What about this hand wants one to promise 6 diamonds when they can show 12-14 balanced instead?

I play a weak (12-14) 1NT in my most regular partnership.  Playing that, I wouldn't dream of making any call but 1NT with that hand.  ♦AQJ52?  So?  The "NT ladder" is that strong.
If she rebids 1NT as she should, you can place the contract.  If 2♣ is natural for you, great, try it (partner will correct to hearts).  It isn't for me, so I'd bid 2♥ and hope to survive.  Magic this time.
Rescue
Don't (at least undoubled). Only rarely is it possible, and it erodes partnership confidence even when it works.  If you pull partner's NF call (as opposed to showing extras), it's not because you're afraid of leaving her there, it's because your pull is better, whether it's "my suit is longer than yours (even with bad support)", or "if we're going to play a misfit at the 2 level, let's at least score +110 instead of +90 if it makes".
Especially avoid running when it raises a level.  As A.Woo points out, 3♣ should be forward-going, not "2♦ is bad, a random suit at the 3 level will be better".
The time to rescue?  1♦-1♥; 2♦-p-(X); p-(p).  Now, if you play it, redouble for rescue.  But know that at least once, I've seen this auction get "rescued" into 3♦ doubled, which was the best result!
2♦ probably doesn't play well; but the opponents might come in.  They might double; they might bid spades and partner might compete in hearts.  Still better than the table result.
3♦
This is an even more awful call than 2♦.  You have not raised partner's "6-card" suit; how many diamonds can you have?
Also, the way to "not have to rebid 5-card suits" is for partner to show secondary support "the second time around".  After 3♣, opener has what could possibly be the best possible heart holding: QJx.  Surely you didn't decide this is the right time to show 3=4=1=5 or 2=4=1=6?  Show partner your hearts, on the way.  I'm not as strong as A. Woo - I think a poor 2♦ hand should show it, so I'd bid 3♥.  He might be right - showing clubs might upgrade this hand to a "good" 12-16; but even 3♥ is better than rebidding those diamonds again!
Even 4♥, assuming a not-awful break, will score better than 3♦, especially if they don't find a trump lead.
Summary
So, lots of ways to get to a better contract than you did, from both sides.
Why does partner do this?  Because their partners will never raise their diamonds, even after 1♦-1♥; 2♣ because "but partner, you could have 3" (it's even worse when they open 1♣ instead!) So they have to "show they have a real suit" when they have 5 - especially this strong a 5.  Because their usual partners, too, will not show delayed support "in case your suit is longer".
Why do you do this?  Because you can't trust your partners to have a suit playable (even if uncomfortably) opposite a void when they rebid it, either.
Trust your partner; show trust in your partner; bid such that partner can trust you, and expect them to show it.  Take your bad results when the RNGod gives them to you - even if, this time, there was a better place to land.  Take your bad results when, this time, it turns out you couldn't trust partner (yet.  That's where we can have a useful post-mortem). Your results will improve, and so will their confidence (and their bidding, at least in time!)
It's not easy; and some never learn.  And it will absolutely hurt while you're both unlearning the habits.  And it will absolutely hurt after you've unlearned them, and you play again with a partner who you can't trust.  But it is best for your game.
(1) this is close to a 2♥ bid for some players - double-honour third, a ruffing value, and an empty suit?  I'd expect many to bid 2♥ with (s) 72 (h) QT7 (d) AQJ52 (c) A85.
(2) Even more so at matchpoints.  Ever heard "at matchpoints, 5 of a minor is a slam try?"  I describe the choice as "If you don't know, bid 3NT.  If 5♣ is the better option, you'll know."  Remember that 3NT= is two tricks less and the same score as 5♣; and 3NT+1 is two tricks less and beats 5♣+1.

Note: Even if you did decide that this was the time to show 3=4=1=5 or 2=4-1=6, you must have good values and so your next call over partner's 3H showing secondary support will be 3S (I don't have a spade stopper) or 3NT (I do have a spade stopper), and you will find 3NT or 5C as appropriate (or even a Moysian 4H).

I would certainly expect partner to have at least 6 diamonds, at most 3 spades, and likely not 3 hearts. Certainly they should have a minimum opener. Example hands would be Kxx xx AQJxxx Qx, or KQx x KQJxxx xxx, or x Axx KQJxxx xxx. With partner's actual hand, their first rebid should be 1NT.
With a balanced hand with opening values, you should endeavor to show your balanced shape and point range with your first two bids, unless it is necessary to show a fit instead. With 12-14 hcp, your plan is to rebid 1NT at round 2.
You don't tell us what your hand is, but I can infer it has invitational values since you rebid 3C (a forcing bid) but passed on the third round (so you don't have a game force). Probably after partner's 1NT rebid, you should bid 2D (new minor forcing, a common conventional agreement). Partner should bid 3H (showing 3-card support and a maximum), so you should end up in 4H.

You ask:

Can one reasonably make inferences about partner's first bid based on the second bid?

and then produce a fairly specific example of one particular sequence. (Ruds and Alex have dealt with that sequence in detail).
The answer to the question as asked is of course, emphatically, "yes".
The classic example is when partner opens and subsequently shows a second suit. You can immediately deduce that the first suit is as long as the second and in almost all cases (drat those pesky 4-4-4-1 hands), that the first suit is at least five cards in length. Playing Acol, where an opening of one of a major only promises four cards, that is a valuable inference.
Another, more contrived example would be partner opening a weak 1NT (12-14) and, when asked for aces, showing 3. You can conclude that opener has exactly one suit containing no card higher than the queen.
Bidding tells a story. Each new page can reveal information about a character introduced in the first chapter.


text:
How is the "base" score calculated, for the online game?
In the online Ticket to Ride game, after the last cards are played, a "base" score is shown for each player.

How is this score calculated?

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

Just the same as in the tabletop version, these are the points gained for claiming routes - i.e. 1/2/4/7/10/15 points for a route of length 1/2/3/4/5/6 (in the original game, slight variation in different versions). It's where your scoring marker would be in the tabletop game.

I've never played the tabletop version.
In the online game, immediately after the last card is played, each team is shown as having "base" points. This is before the animated pictures of the trains move forward to show how much each team scored for each route.
Do you mean that these "base" points are calculated just from the lengths of the routes in the tickets that one took, whether or not one was successful in completing the routes?

No, they are from the routes you claim; assuming you're playing the Days of Wonder implementation (the official one), you can see the points adding up as you play through the game. Your tickets (other than the number you have taken) are hidden information until game end.


text:
Must the infinite loop rule be triggered on a Grindstone and 4 Progenitus
Suppose that my deck contains a Grindstone and 4 copies of Progenitus, and that every other card in my deck shares a color.
I use Grindstone's ability, targeting myself.
Because every card in my deck shares a color, Grindstone's ability will always repeat. Whenever it mills Progenitus, I shuffle my deck. Eventually (after much milling and much shuffling), my deck is reduced to 4 cards. I know that my deck contains only Progenitus, and therefore that all future iterations of Grindstone will have exactly the same result, but my opponent only knows that I have at least 2 copies of Progenitus - the other two cards in my deck could be anything (but are highly likely to be Progenitus).
Hypothetically from my opponent's perspective, I could conceivably mill a card other than Progenitus, which (ignoring the fact that no such cards have ever been printed) could conceivably have a replacement effect which would interrupt the loop. Announcing that this is an infinite loop would require announcing privileged information about my deck.
How do the rules handle this? Am I required to announce that the loop is infinite, even though it contains hidden information, or would the rules allow me to keep milling and shuffling until one of us resigns?

Intentional Draws have always been a part of mtg.

The only way every other card in your deck shares a color is if someone has a Painter's Servant in play, otherwise your lands have no color and will end the loop.

@Andrew - Not entirely true; there are cards which are dual-faced with spells on the front and lands on the back. If every land in your deck was such a card, all of the same color, and every other card in that deck was also that color, then that deck would consist of all cards of the same color.

@TimC Fair, Oops all spells is the exception (though I've never seen an oops all spells that would ever want prog, still is possible)

@Andrew It would never want Progenitus. Spell-lands are all monocolor, so it would need to generate 8 colored mana from treasures and other mana-fixing to be able to cast Progenitus. This deck has no gameplan except to draw the game.

This scenario is not covered by the rules.
Loops of mandatory actions are already an edge case in the rules. This scenario, in which determining whether the loop of mandatory actions is endless depends on hidden information, is an edge case of an edge case. Remember that Magic: the Gathering is Turing Complete, which means that even without invoking hidden information, it is in principle not always possible to determine whether a sequence of mandatory actions will end.

Practically speaking, in a tournament, both players can agree that a game is a draw, even if the game rules do not cause a draw (MTR 2.5). In this scenario, the player performing the combo knows that the game can make no progress, and their opponent can infer that this is the case, so both would reasonably decide to end the game in a draw.
If either player proposes a draw and the other refuses, a judge could look at the library and confirm for both players that the loop is in fact infinite and the game is a draw. In addition, if the player performing the combo refuses a draw, they could get a Slow Play penalty because they know that the game cannot make progress. I can't see a reason that neither player would want a draw, but in that case the game would go to time and end in a draw anyway.

If there are two gennies in your deck it is a mandatory loop. One on the other hand and the grindstone ability resolves with your opponent having one gennie in there library.

I do remember in 2005 there was an infinite life gain deck in standard where the standard strategy was if you cannot stop the combo play for a draw. The lifegain deck wanted to win so did not want an ID and you could not win so a draw was better than a loss.

The moment a single grind stone reveals two progenitus, there's no need to look at hidden information to determine the status of the loop as something that cannot be ended, that there are two cards that will continue the loop indefinitely is public information at that point, though proposing a draw as the player who knows they are still in his deck and having a judge confirm is an easy way to shortcut the long process to get to that point.

Part of the hypothetical in the question was that some other card may be able to break the loop. A card like that does not currently exist, but this answer accounts for it.


text:
Probability of drawing duplicates of a card in magic
Was trying to figure this out, and realized I don't know the exact math:
In the card game magic you have a deck of 60 cards. You can have 4 copies of any individual card, and you are dealt a starting hand of 7. Each turn, one additional card is drawn.
What are the odds of drawing more than one of a card (2, 3, or 4 of the same card) on turns 1, 2, 3 etc...
If anyone knows how to break this down, I'd love to see the math behind it.
Cheers

@bautista That's about having at least 1, not at least 2, and only in the opening hand.

@Arthur the top answer also links to the hypergeometric distribution tool, which can calculate any starting hand. If you want the cards after n turns, you just make your starting hand 7+n

This is not math.stackexchange, so I don't have access to any formula formatting. I'll try to make it readable regardless. The basic mathematical construct used here are the so-called binomial coefficients:

(a choose b) means a!/(b!*(a-b)!), it says how many different possible hands of b cards there are when you have a unique cards to choose from. I will pretend that you can see the difference between the duplicates, say they are reprints from different sets or something. It's just easier that way.

Say you have seen k cards of your deck (the seven you start with, plus any you have drawn). And you have one particular card you're looking for duplicates of (if you're asking about having any duplicate at all, then this becomes messy). I am going to pretend that you just keep drawing cards, never playing any and never discarding, so that all the cards you've seen are in your hand. It's easier to phrase this way.
The probability of having drawn none of your cards is (56 choose k)/(60 choose k). The numerator is the number of hands of size k that do not contain your card, and the denominator is the total number of hands with k cards. The probability of having drawn exactly one is 4*(56 choose k-1)/(60 choose k). The probability of not having duplicates is the sum of these two. The probability of having duplicates is 1 minus that:

Probability of duplicates: 1-(56 choose k)/(60 choose k) - 4*(56 choose k-1)/(60 choose k)

Here is a table of the first few values.

would it be true to say you are equally likely to draw any non-basicland card in an edh deck?

@NeilMeyer If your shuffling is good, yeah. And the same can be said about the basic lands, if you can tell them apart.