text:
What is the penalty for failing to follow suit?
Assume the mistake is unintentional, what course of action should be taken when this occurs and is discovered?
If the person discovers themselves that they earlier made the mistake, is the penalty different than if an opponent discovers it?

In Duplicate Bridge the penalty for a revoke (failing to follow suit) is very clear.
The official rules handle revokes in Laws 61 to 64.  Wikipedia's summary of 61-64 is

A revoke may be corrected (correct card substituted) without trick penalty before any player of the offending side plays to the next trick; otherwise, it becomes established. If a revoke is corrected, the exposed card becomes a penalty card and the opponents may change their played cards as they see fit (however, the revoking side may not take advantage of those seen cards—see Alcatraz coup). When a revoke is established, in general, one of subsequent tricks won by the offending side is transferred to the opponents. If the revoke card has won the trick, that trick is transferred too. (If the offending side did not win any subsequent tricks, no penalty is assessed). Additional tricks can be transferred if the revoke has caused more damage to the opponents than was redressed by those penalties.

If the revoke is not immediately corrected, there is no difference in penalties based on the person who establishes the revoke.  It makes no difference whether the mistake was intentional or unintentional.

It is important that this rule is the penalty to be applied given an assumption of innocence. If evidence exists that the revoke is not an innocent mistake, then much more severe penalties, for unethical behaviour, come into play. The Rules always assume innocent error unless explicitly specified otherwise.

The Wikipedia summary is also in error in one detail - it matters not whether the card played in revoke has won the trick, only whether the revoking side has won the trick for the additional trick penalty to apply. Update That is a new change in the 2017 Rules - I stand corrected here, and the Wikipedia summary is correct.

Failing to follow suit is called a "revoke" or "renege" when that person subsequently plays a card from that suit (assuming that the mistake was not discovered and corrected in time). The governing law is the 2017 amendment (to the earlier 2007 version) of Law 64 of the ACBL.
The penalty is normally one trick. But there are two major exceptions.
a) There could be a second trick per Law 64:

1, "When a revoke is established, and the trick on which the revoke occurred was won by the offending player*, at the end of the play the trick on which the revoke occurred is transferred to the non-offending sidetogether with one of any subsequent tricks won by the offending side.
  2. and the trick on which the revoke occurred was not won by the offending player* then, if
  the offending side won that or any subsequent trick, after play ends one trick is transferred tothe non-offending side.

Note: 2), above, is effected by placing the reneged card face up on the table.
b) If "there is no rectification" (e.g. because the offending side loses the reneged trick, and all tricks subsequent to the discovery of the renege), the penalty is zero. (The offense didn't cause any damage.)

According to what Rule and circumstance would "The "reneged" card is placed face up on the table" ever happen?

@ForgetIwaseverhere: I replaced a stale link, and cited the 2017 version of Law 64 to clarify these issues.

The statement "Note: 2), above, is effected by placing the reneged card face up on the table.' is a grossly incorrect reading of the Law. The phrase "failing to play any card faced on the table" refers to penalty cards , both major and minor, previously required to be faced on the table at the time of the revoke. Also, in the absence of specification otherwise in the OP, I believe the Laws of Rubber Bridge should be referenced rather than the Duplicate Bridge Laws.

The revoke (renege) laws are complicated!
If a player discovers that they have revoked before the side has led to the next trick OR if the revoke occurred at trick 12 OR both sides have revoked at the same trick and only one has led to the next, then the revoke is NOT established. It must be corrected and, if a defender, the revoke card becomes a Major Penalty card. The player after the revoker can change their card without penalty (information about the card withdrawn is Authorised for the non offending side but unauthorised for the offending side): if they do then the partner of the revoker can change their card - but the original one remains a major penalty card.
If the revoke IS established then there may be an automatic trick adjustment - this does not happen if the card that was used to revoke was already visible i.e in dummy or a major penalty card, at trick 12 (see above), if it is a second revoke in the same suit, or if both sides have revoked on the same trick, if the round has ended or the non-offending side has called on the next hand. Law 64 B
The automatic trick adjustment is:
If the revoke card won the trick - this almost always means that it trumped in the trick (although it could be that a player was required to lead a suit because of a previous infraction and failed to do so when they could) - then THAT trick is transferred over to the other side.  ONE more trick is transferred over but ONLY if the offending side won the trick AFTER the revoke.
Finally there are some situations where the non-offending side are not adequately compensated for the effects of the revoke - in which case the Tournament Director restores Equity.
In summary - when attention is drawn to a revoke- call the director! Do not accept or agree to any statement made by the opponents as to the correct rectification.

Unless specified in the question, assume Rubber Bridge rather than Duplicate, and reference the appropriate rules.


text:
In Axis and Allies, what US strategy has the highest win rate? Why?
Everyone knows how the United States REALLY won World War II. It concentrated first in North Africa, then the Mediterranean (Italy), and closed with the cross-channel invasion at Normandy that ended in Berlin. That is a viable strategy for the U.S. player in Axis and Allies.
A number of U.S. players build a factory in Sinkiang to stop the Japanese occupation of Asia. The logical follow-up to that is to build (or re-build) a Pacific Navy as a hammer to the anvil of Chinese troops. This plan ends with America's "spoils of war" being Japan, China, and the "ASEAN" countries, while the UK and Russia take down Germany.
About the one thing the US (and UK) DIDN'T try to do in World War II was to try to link up with Russia via Norway. But it is the basis of the simple "shuck-shuck" strategy devised by Don Rae.
This appears to be the preferred option of many AA aficionados: Dump American infantry into Norway in "assembly line" fashion after Russia (or the UK) has captured it, send them through Karelia and Eastern Europe to Berlin if necessary (otherwise transport them across the Baltic for an amphibious assault on the enemy heartland with air support).
Is the real life "cross channel" (and cross-Med) strategy the best one in AA? Or do our strategists know something that America's real life military didn't "know" in favoring one of the others?

Downvote for "Everyone knows how the United States REALLY won World War II".  The Americans entered the war late after significant work had been done by other allied nations.

@sixtyfootersdude: I was talking about METHODOLOGY, not contribution. What you said is true, that other nations did most of the work. But in this context, the discussion was about the historical sequence of events, in contrast with proposed game strategies.

Fair enough.  Down-vote removed.

This may do better on the History StackExchange phrased as "Why didn't the allies uses this strategy?"

What Is the Preferred Strategy For the US in Axis and Allies?

You answered your own question - the simple "shuck-shuck" strategy devised by Don Rae. The reason it works is because infantry are cheap and defend at 2, but only attack at 1. Therefore you can build an impenetrable infantry force that can withstand a large German offensive push and keep Germany hemmed into mainland Europe, freeing Russia and the UK to focus on pushing back Japan and tipping (or keeping) the IPC balance in favor of the Allies.

Do our strategists know something that America's "real life" military didn't know?

There were many reasons why "shuck-shuck" was not implemented in real life. Among them include:

Geopolitical reasons - remember, the Allies only grudgingly allied with Russia. They were viewed as the lesser of two evils and a necessary ally to help end the war. Also, in real life Finland was attacked by Russia in 1939 in the Winter War. Then, from 1941-1944, Finland allied with Germany and declared war on Russia and aided Germany in attacking Leningrad. Given that they were at war with Russia, I doubt Finland would have permitted the US and UK to march troops through their country to assist in the defense of Russia.
Logistical reasons - moving troops from Norway to Karelia takes "one turn" in the game, but would be more labor and time intensive to accomplish in reality given the topology and climate of that region. Also, I presume the infrastructure to move millions of troops, equipment, and foodstuff was not in place and would need to be built.
Sociological/psychological reasons - how positively would the American public respond to a war strategy that boiled down to, "We send millions of our own boys into the arctic and down into Russia to have a staring contest with the Germans. We'll liberate France when we get around to it. Also don't forget that this was not long after World War I, which was notorious for brutal trench warfare that resulted in pointless, high casualty offensive actions and long standing stalemates. It would be hard to sell the public on a repeat of a "dig in and wait"-type strategy.
Economic reasons - the US has the economic means and raw materials to produce modern weaponry at a scale that couldn't be matched by the Axis powers. The US had not been engaged in warfare for years, suffered bombing runs on their mainland factories, and so forth. It wouldn't make much sense to have the US focus on building a large defensive troop position when they could focus on building battleships, bombers, planes, tanks, and so on. Also, in real life the value of a human being is worth more than 3 IPC. :-)

This goes to what I call "weather" rules. Weather (and similar situations) caused real life campaigns to be easier or harder than they look on the map. So North Africa/Med/Normandy is preferable in real life, while Norway is preferable on the game map. (Unless you alter the map to penalize movement at higher latitudes, etc.)

Asking "preferred strategy" questions opens the door to individual opinion.  I will try to avoid this by employing simple mathematics.
"Germany First" is the most logical path to choose, but not by ignoring the Pacific theatre.
The U.S. can typically build with impunity during the entire game, outside of the reach of German or Japanese strength.  Once a large enough force is gathered on the eastern coast, (turn 3), they can head on over to Africa, and from there they can threaten either southern or western Europe. Meanwhile, each turn the U.S. can build a couple of transports and units to load, thereby forming a continuous loop of troops to Europe. And, in a short time, the U.S. can stop building transports and focus on the ground forces (and some air), thereby making the situation even more acute for Germany. Add to this the U.K. building its own fleet up, and Germany is soon a non-participant in the Atlantic. If Germany tries to keep pace with the U.K. and U.S. in the Atlantic, Russia will never be threatened.  Germany simply doesn't have the economy to seriously engage all three allies at once.
The German player cannot ignore this viable threat, and must draw land forces from his eastern attack on Russia to protect his flanks.  This, in turn, will allow Russia a measure of calm before the scrap at the gates of Moscow.
If the U.S. forsakes the Atlantic, the U.K. doesn't have the forces to threaten Germany with any great strength, and Germany can pour its malice directly upon the Soviet Union.  And, although the U.S. has a strong economy, it cannot bring many ground troops to bear against Japanese holdings for a few turns, and when it does, the Japanese are in a far better position to fight back.
Mathematically, about three turns in, the Germans will have a revenue that is close to the combined Soviet Union/U.K. income.  The problem for the Germans is that they only get one turn out of five to employ this apparent bonus.  The Allies can bring three turns out of five against Germany.  The Soviets and British have similar advantages of being able to trade space for time, although the push on Moscow is a real and imminent threat.
One 'loop-hole', if you will, in the original A&A is that Japan is out of position for most of the war.  Unless the U.S. forsakes the Pacific entirely, and Britain doesn't fight for India, Japan will poke around Asia for a while, and maybe take some islands here and there, but for the most part, there are no large gains to be made as the Japanese.  True, they have an impressive naval and air presence, but there are simply no big ticket targets for them.  So, they build and build and build, and slowly spread around their area of the board, but unless the S.U. player is completely careless, Japan can never truly threaten Moscow, and, by the time they do get tanks to Moscow, Berlin is either a smoking ruin, or the Allies were inept and Germany and Japan are putting the pincer movement on Moscow.  And the idea of Western U.S. falling is ludicrous, unless, of course, the U.S. player unaware of the impending doom of 10 Japanese transports slowly making their way across the Pacific Ocean.
So, to answer the question in one sentence: send as many troops as possible eastward and keep constant pressure on Germany.  If the Axis player still wants to play after Berlin falls, well, enjoy nuking Tokyo.

The game map does NOT emulate correctly the historical distances!
Therefore it is far more feasible in the game to bring allied troops/equipment through Norway to Russia than it would in reality.
E.g. unlike in the boardgame historically allied operations in Norway had no land-based fighter support coverage.
The germans capitalized on it ensuring their Norway based fighter coverage would provide the required coverage for its surface fleet... among them the Tirpitz... which was operational past D-Day landings.
Notice the Germans didn't try to bring their surface fleet to the english channel to challenge D-Day operations. They knew far too well Allies possessed total air superiority, if not supremacy, there...
Historically the Allies still tried(as "correct" strategy, like in the boardgame, dictates) to link up with Russia through artic convoys though...at times Germany successfully disrupted them too...see convoy PQ17. And there was a flow of supplies arriving to Russia trough Persia too.
Board game distances versus real distances...

Welcome to the site. An upvote for a good answer. See my comment under Scott Mitchell's answer about "weather rules."

Agree with Kirt that U.S. must send most/all its resources into the Atlantic, simple math says Russia/U.K./U.S. dollar spend focused solely on Germany ensures Germany's upfront IPC advantage is quickly trumped and overtaken before Japan can mount enough pressure on Russia to relive pressure.  Problem is I think Kirt has too often played weak Japanese players - a well played Japan has troops marching into Eastern territories of Russia, and all U.S. forces/land taken by the end of their 2nd round, leaving the only the last battle for India (which depends on if UK built an IC or not).

Here's how I think the game creators saw it: Russia is just cannon fodder. The Japs can easily attack from the East, and Germany has a good 2-3 turns before it has to worry about anything besides Russia and strategic bombing raids/rocket attacks. England/U.K. starts off with a probable SBR from Germany, and might as well return the favor. They have also probably lost Anglo-Egypt to the Algerians and/or Lybians, thus shutting down the Suez Canal. They have a lot to defend, but are without the means to do so. Japan, again, will most likely try to attack Russia in the Soviet Far East. China is also a probable hotbedthat they may want to take (because, hey, 4 I.P.C.s per turn is nothing to sneeze at. Also, since AA guns hit 1/6 of the time, may as well take India off of Britain's hands with the generous force that starts off in French Indochina.

You say a lot about the strategies of "other" countries, but nothing about the U.S.

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text:
Chances to win battles in Risk
I want to estimate the probability of winning Risk battles.
It doesn't need to be a perfect calculation, I just want to get a general idea of whether to attack or not. It needs to be easy to memorize.
I'm sure there is a trade off between accuracy and easy of use, so I'm hoping there is more than one method proposed. I would be using this to estimate after each die roll so I can stop attacking when my winning probability drops below some threshold.
Bonus: also estimating how many armies are left.

All of the detailed probability calculations and Markov analysis posted by Eric P. and ire_and_curses can be distilled into two simple heuristics:

Large battles favor the attacker but only very slightly.
For small battles, attack if you have more armies, stop if you don't.

The rationale for these guidelines is outlined below.
Definition: a large battle is a battle of 3 or more attackers vs 2 or more defenders.
For large battles the army count has nothing to do with the outcome of any single battle, which means you only need to remember a single number: the net attrition rate, defined as the expected value of the difference between defender and attacker losses. This rate can be calculated from data in Table 2 of the Jason Osborne paper in Eric P.'s answer:

The expected value of the defenders lost is 1.08. (This is the number of defenders lost by events π_32x weighted by their probability.)
The expected value of the attackers lost is 0.922.
This means the net attrition rate is 0.158 (i.e. over 10 rounds of combat, the defender will lose 1.58 more armies--on average). Over 20 rounds 3.16.

Leading us to our first heuristic:

Large battles favor the attacker but only very slightly. As the attacker, it takes on average 20 rounds to make up a three army deficit. Whether you should attack in any given situation is a strategic not a tactical decision, but tactically the attacking advantage is on average very slight. Remember as well that past events have nothing to do with future rolls.

For small battles, the full probability matrix is supplied above, but our second rule distills this knowledge:

For small battles, attack if you have more armies, stop if you don't. When outnumbered your probability of winning is no more than 0.417 and probably less. Otherwise your probability of winning is at least 0.656 but as high as 0.916.

I upvoted several answers, but I think this one actually does the best at giving a usable, memorable heuristic.

NO, I this heuristic neither brief, nor memorable, nor correct. In 1v1 games (at least) the concept is to always attack if you have 3 attackers (4 or more troops on the attacking territory). There is no intrinsic difference between one large and many small battles.

@TheChymera -- I don't think what you said is inconsistent with this answer: Adam defined a "large battle" as 3 or more attackers and 2 or more defenders.

Perhaps it's worth noting that "attackers" means the number of armies actually attacking; this would not include the 1+ armies that must remain on the attacker's original territory.  If you have 3 armies and your opponent has 2, you have only 1 or 2 attackers, not 3.

A good paper by Jason Osborne can be found here. (It's a correction to an earlier paper by Tan.) He uses Markov chain calculations to get the exact probabilities. You'll especially want to look at Table 3 on page 6, which has these probabilities rounded to three decimals for up to 10 armies per side. I've reproduced it below:

For quicker visual scanning, here's a colorized version using conditional formatting in Excel and percentages:

As to ease of use: just print it out and look up your odds after every roll!

I've often found players don't like to use such aids during play (thinking it cheating), this is why I prefer mental calculations.

Looks like you get a reasonably good approximation by taking 0.42 + 0.11 * A - 0.09 * D. So, start with 0.42, add 0.11 for every attacker, subtract 0.09 for every defender. (I tried to get a good fit for the results between 0.1 and 0.9; if you get outside of that range it's less important anyway.)

@barrycarter - interesting work. Essentially there is a third parameter at play here: the minimal probability of winning at which you're willing to continue. You set that at 0.5, I think (from a quick glance over your code). I think in different situations, you'd want to use different probabilities here: sometimes it's worth to risk a few armies for a long shot that would help you a lot. So we would need different tables for different values of this parameters... interesting, interesting.

You have a typo at 4 vs 4 - it should read more like 0.488 than 0.477.

If anyone is interested, there was a related question on CodeReview, and I wrote a Python answer, which, among other results, can output the above matrix: https://codereview.stackexchange.com/a/254915/139491 & https://codereview.stackexchange.com/a/255211/139491
Your answer helped me a lot for the theory.

@ForgetIwaseverhere: 7 years later : no, 0.477 seems correct AFAICT.

As stated in the Risk FAQ the expected losses per attack for standard Risk rules is about 6 to 7.  This means the attacker is expected to lose 6 armies for every 7 defender armies destroyed.   Since we're talking about expected values, this represents the mean (average), which is most akin to a 50th percentile or 50% chance that that is what will happen.  (this is not techically correct in terms of statistics, but I'm trying to explain it in more layman's terms).
As for 'chances to win a battle' this is very difficult to produce a rule of thumb for, since the numbers vary wildly depending on # of armies in play.  For example, an 'even match' of 100 to 100 is won with over 85% chance, but 10 to 10 is only about 50%-50%.
It would be easier if you specified a particular odds at which you want to attack or not, then a simple matrix of attacker/defender army counts is possible to create (perhaps with an easy formula), but when you start with N attackers and M defenders the best 'answer', statistically, is a probablilty distribution function which isn't easy to calculate.
Assuming you want to attack at, at least 50% expected value (as above), use the 6 to 7 rule.  This will also tell you how many you are expected to have left, so, using this rule, if you have 20 armies (to attack with, so 21 in the country), and he has 21, the 6 to 7 rule would say that you're expected to lose 18, killing his 21 (6*3, 7*3), thus leaving 2 left over.

Although the full calculation to discover whether you will win a sequence of battles is difficult to make, it is easy to calculate the chances of winning any particular combination of attacker and defender dice. I reproduce here the table of expected losses described in this paper.
                       Defender Dice
                       1           2
Attacker Dice
1                  0.58/0.42   0.75/0.25
2                  0.42/0.58   1.22/0.78
3                  0.34/0.66   0.92/1.08

The first number in each cell is attacker losses, the second defender losses. So, if you attack with 1 die against a 2 dice defence, you have a 3/4 chance of losing the battle, which in this case will cost you 1 army. Remember also that if each of you have 2 or more dice, then exactly 2 armies will be lost in the battle. This is why in 2 dice vs 2, the attacker will lose on average more than one army.
So, you could easily memorise this table, and use it to decide at each step of a battle whether to proceed or not. Or you could just remember the broad rule of thumb described in the paper:

When both attacker and defender have a large number of armies, the attacker
  will, on average, lose armies at a 15%
  slower rate than the defender.
Towards the end, when either the attacker or the defender must shake
  fewer dice, the advantage swings more
  strongly toward the player with the
  most armies.

The details of the full calculation are explained in the paper. The author also provides an online Javascript implementation of the calculations that you might find interesting to play around with.

If you are online, the easiest way to determine whether you have a good chance of winning a battle is to use this calculator:
http://armsrace.co/probabilities
It emphasizes a non-trivial conclusion: if you have the choice, always attack the big guys first in your sequence!
For instance, if you have 6 on a territory, and want to attack a 2 and a 1 (and you have the choice to start with either one), you have:

73,60% chances of winning if you attack 2 then 1
69,56% if you attack 1 then 2

That's a fairly large difference!
The reason why we're seeing this is basically because the more armies there are in front of you, the more armies you need yourself. If you start by attacking the 1 with 5 armies, you'll be at your best with 4 armies left to attack the remaining 2. And there is a chance that you have in fact only 3 or 2. Meaning that if you lose only one army (in attacking the 1 or the 2) you'll be in the disadvantageous situation where you have an equal number of dice against the 2.
On the other hand, if you start by attacking the 2, you have more cases where you get 3 dice against 2.
This is a very litteral answer, but I think it helps getting the idea why you should attack the most powerful territories first.

The link is now broken.

Here is an incredibly simple way to figure this out without graphs or calculators.
Simply count one of your soldiers defeated for every other soldier you intend to defeat & one of your soldiers left behind for every territory your conquer.
Example: If you wish to eliminate another player that has 12 soldiers spread over 5 connected territories & that player is right beside the territory that you will place your soldiers on, you will need 12+5=17 soldiers in addition to the one that must stay behind from the original attacking territory.  If there are other soldiers or territories in the way of your intended goal then, after figuring out the least expensive route, simply count one of your soldiers defeated for every other soldier you intend to defeat & one of your soldiers left behind for every territory your conquer.  This method slightly favours the attacker in big numbers & slightly favours the defenders in small numbers which brings me to my next point.  From time to time you will have to split your forces in different directions in the path you have chosen to pursue.  This event that I call a fork will force some of your soldiers to stop being able to attack usefully.  For every dead end fork add 2 soldiers as they will be required to ensure you can always roll 3 dice right to the end of each fork.

Analyzing Dice Rolls in the Game of Risk
The Goal of this research is to answer this question: Who has the probable advantage in the game of Risk? Does the Attacker or the Defender have this advantage?
Initial Thinking:
With two dice rolled, what is the probability that at least one six will appear?

There are 36 combinations of dice numbers possible when you roll two dice in one instance.  There are 11 instances where 6 appears once or more in this set of 36 possibilities.  The answer is: 11/36=0.305.
What follows is a Permutation Set covering the combinations of dice rolls that are possible when 3 attack dice are used against 2 defense dice:
The Number Headings are used to number a logical structuring of the permutation set.  The permutation set of numbers contains 6 defence instances (where the defence is assumed to have rolled at least 1 instance of a 1) times 6 respective defence instances (where the defence rolls a number from to 1 to 6) times 1 instance (where the Attacker only rolls ones) + (6-1) (this accounts for whats left (1 to 5) after the attacker rolls only ones) times 3 (number of Attacker dice used).  Each of the number headings represent a number block from 1 to 16 based on the attackers total set of possible rolls.
Total Possible Rolls when the attacker uses 3 dice and the defence must use 2 dice.
S=6 x 6 x (1 + 15) = 6 x 6 x 16 = 576
If you study this rather large set of permutations, you should be able to understand the logical structure in this enumerating approach to answering who has the advantage in Risk, the attacker or the defender.
A means Attacker
Dr means Draw
D means Defence
Enumerating the Possible Dice Combinations in Risk
The defender has 2 dice and the attack 3 dice.
[D1, D2, A1, A2, A3]
*Sorry if the tables are hard to read, the original formatting could not
be preserved for this post.  I can reformat the tables using a color coding
system that will make them easier to read.  Let me know if you are interested
and I will provide them as quickly as possible.
1
D wins 36 times resulting in A losing 2 units in each instance
[1,1,1,1,1]
[1,2,1,1,1]
[1,3,1,1,1]
[1,4,1,1,1]
[1,5,1,1,1]
[1,6,1,1,1]
[2,1,1,1,1]
[2,2,1,1,1]
[2,3,1,1,1]
[2,4,1,1,1]
[2,5,1,1,1]
[2,6,1,1,1]
[3,1,1,1,1]
[3,2,1,1,1]
[3,3,1,1,1]
[3,4,1,1,1]
[3,5,1,1,1]
[3,6,1,1,1]
[4,1,1,1,1]
[4,2,1,1,1]
[4,3,1,1,1]
[4,4,1,1,1]
[4,5,1,1,1]
[4,6,1,1,1]
[5,1,1,1,1]
[5,2,1,1,1]
[5,3,1,1,1]
[5,4,1,1,1]
[5,5,1,1,1]
[5,6,1,1,1]
[6,1,1,1,1]
[6,2,1,1,1]
[6,3,1,1,1]
[6,4,1,1,1]
[6,5,1,1,1]
[6,6,1,1,1]
2
D wins 35 times resulting in A losing 2 units in each instance.  In 1 instance, D and A both lose 1 unit.
[1,1,2,1,1] Draw
[1,2,2,1,1]
[1,3,2,1,1]
[1,4,2,1,1]
[1,5,2,1,1]
[1,6,2,1,1]
[2,1,2,1,1]
[2,2,2,1,1]
[2,3,2,1,1]
[2,4,2,1,1]
[2,5,2,1,1]
[2,6,2,1,1]
[3,1,2,1,1]
[3,2,2,1,1]
[3,3,2,1,1]
[3,4,2,1,1]
[3,5,2,1,1]
[3,6,2,1,1]
[4,1,2,1,1]
[4,2,2,1,1]
[4,3,2,1,1]
[4,4,2,1,1]
[4,5,2,1,1]
[4,6,2,1,1]
[5,1,2,1,1]
[5,2,2,1,1]
[5,3,2,1,1]
[5,4,2,1,1]
[5,5,2,1,1]
[5,6,2,1,1]
[6,1,2,1,1]
[6,2,2,1,1]
[6,3,2,1,1]
[6,4,2,1,1]
[6,5,2,1,1]
[6,6,2,1,1]
3
A wins two units in 1 instance,
5 draws + 5 draws = 10 draws,
D wins 25 times
[1,1,2,2,1] A wins twice
[1,2,2,2,1] Draw x5
[1,3,2,2,1]
[1,4,2,2,1]
[1,5,2,2,1]
[1,6,2,2,1]
[2,1,2,2,1] Draw
[2,2,2,2,1] D wins x 5
[2,3,2,2,1]
[2,4,2,2,1]
[2,5,2,2,1]
[2,6,2,2,1]
[3,1,2,2,1] Draw
[3,2,2,2,1]
[3,3,2,2,1]
[3,4,2,2,1]
[3,5,2,2,1]
[3,6,2,2,1]
[4,1,2,2,1] Draw
[4,2,2,2,1]
[4,3,2,2,1]
[4,4,2,2,1]
[4,5,2,2,1]
[4,6,2,2,1]
[5,1,2,2,1] Draw
[5,2,2,2,1]
[5,3,2,2,1]
[5,4,2,2,1]
[5,5,2,2,1]
[5,6,2,2,1]
[6,1,2,2,1] Draw
[6,2,2,2,1]
[6,3,2,2,1]
[6,4,2,2,1]
[6,5,2,2,1]
[6,6,2,2,1]
4
A wins two units in 1 instance,
5 draws + 5 draws = 10 draws,
D wins 25 times
[1,1,2,2,2] A wins twice
[1,2,2,2,2] Draw x 5
[1,3,2,2,2]
[1,4,2,2,2]
[1,5,2,2,2]
[1,6,2,2,2]
[2,1,2,2,2] Draw
[2,2,2,2,2]
[2,3,2,2,2]
[2,4,2,2,2]
[2,5,2,2,2]
[2,6,2,2,2]
[3,1,2,2,2] Draw
[3,2,2,2,2]
[3,3,2,2,2]
[3,4,2,2,2]
[3,5,2,2,2]
[3,6,2,2,2]
[4,1,2,2,2] Draw
[4,2,2,2,2]
[4,3,2,2,2]
[4,4,2,2,2]
[4,5,2,2,2]
[4,6,2,2,2]
[5,1,2,2,2] Draw
[5,2,2,2,2]
[5,3,2,2,2]
[5,4,2,2,2]
[5,5,2,2,2]
[5,6,2,2,2]
[6,1,2,2,2] Draw
[6,2,2,2,2]
[6,3,2,2,2]
[6,4,2,2,2]
[6,5,2,2,2]
[6,6,2,2,2]
5
A wins two units in 3 instances,
4 + 5 x 1 draws = 9 draws,
D wins 24 times,
[1,1,3,2,2]  A wins 2 units
[1,2,3,2,2]  A wins 2 units
[1,3,3,2,2]  Draw
[1,4,3,2,2]  Draw
[1,5,3,2,2]  Draw
[1,6,3,2,2]  Draw
[2,1,3,2,2] A wins 2 units
[2,2,3,2,2] Draw
[2,3,3,2,2] D wins 2 units
[2,4,3,2,2] D wins 2 units
[2,5,3,2,2] D wins 2 units
[2,6,3,2,2] D wins 2 units
[3,1,3,2,2] Draw
[3,2,3,2,2] D wins 2 units
[3,3,3,2,2] D wins 2 units
[3,4,3,2,2] D wins 2 units
[3,5,3,2,2] D wins 2 units
[3,6,3,2,2] D wins 2 units
[4,1,3,2,2] Draw
[4,2,3,2,2] D wins 2 units
[4,3,3,2,2] D wins 2 units
[4,4,3,2,2] D wins 2 units
[4,5,3,2,2] D wins 2 units
[4,6,3,2,2] D wins 2 units
[5,1,3,2,2] Draw
[5,2,3,2,2] D wins 2 units
[5,3,3,2,2] D wins 2 units
[5,4,3,2,2] D wins 2 units
[5,5,3,2,2] D wins 2 units
[5,6,3,2,2] D wins 2 units
[6,1,3,2,2] Draw
[6,2,3,2,2] D wins 2 units
[6,3,3,2,2] D wins 2 units
[6,4,3,2,2] D wins 2 units
[6,5,3,2,2] D wins 2 units
[6,6,3,2,2] D wins 2 units
6
A wins two units in 4 instances,
4 x 2 + 4 x 2 draws = 16 draws,
D wins 16 times,
[1,1,3,3,2] A wins 2 units
[1,2,3,3,2] A wins 2 units
[1,3,3,3,2] Draw
[1,4,3,3,2] Draw
[1,5,3,3,2] Draw
[1,6,3,3,2] Draw
[2,1,3,3,2] A wins 2 units
[2,2,3,3,2] A wins 2 units
[2,3,3,3,2] Draw
[2,4,3,3,2] Draw
[2,5,3,3,2] Draw
[2,6,3,3,2] Draw
[3,1,3,3,2] Draw
[3,2,3,3,2] Draw
[3,3,3,3,2] D wins 2 units
[3,4,3,3,2] D wins 2 units
[3,5,3,3,2] D wins 2 units
[3,6,3,3,2] D wins 2 units
[4,1,3,3,2] Draw
[4,2,3,3,2] Draw
[4,3,3,3,2] D wins 2 units
[4,4,3,3,2] D wins 2 units
[4,5,3,3,2] D wins 2 units
[4,6,3,3,2] D wins 2 units
[5,1,3,3,2] Draw
[5,2,3,3,2] Draw
[5,3,3,3,2] D wins 2 units
[5,4,3,3,2] D wins 2 units
[5,5,3,3,2] D wins 2 units
[5,6,3,3,2] D wins 2 units
[6,1,3,3,2] Draw
[6,2,3,3,2] Draw
[6,3,3,3,2] D wins 2 units
[6,4,3,3,2] D wins 2 units
[6,5,3,3,2] D wins 2 units
[6,6,3,3,2] D wins 2 units
7
A wins two units in 4 instances,
4 x 2 + 4 x 2 draws = 16 draws,
D wins 16 times,
[1,1,3,3,3]  A wins 2 units
[1,2,3,3,3]  A wins 2 units
[1,3,3,3,3] Draw
[1,4,3,3,3] Draw
[1,5,3,3,3] Draw
[1,6,3,3,3] Draw
[2,1,3,3,3] A wins 2 units
[2,2,3,3,3] A wins 2 units
[2,3,3,3,3] Draw
[2,4,3,3,3] Draw
[2,5,3,3,3] Draw
[2,6,3,3,3] Draw
[3,1,3,3,3] Draw
[3,2,3,3,3] Draw
[3,3,3,3,3] D wins 2 units
[3,4,3,3,3] D wins 2 units
[3,5,3,3,3] D wins 2 units
[3,6,3,3,3] D wins 2 units
[4,1,3,3,3] Draw
[4,2,3,3,3] Draw
[4,3,3,3,3] D wins 2 units
[4,4,3,3,3] D wins 2 units
[4,5,3,3,3] D wins 2 units
[4,6,3,3,3] D wins 2 units
[5,1,3,3,3] Draw
[5,2,3,3,3] Draw
[5,3,3,3,3] D wins 2 units
[5,4,3,3,3] D wins 2 units
[5,5,3,3,3] D wins 2 units
[5,6,3,3,3] D wins 2 units
[6,1,3,3,3] Draw
[6,2,3,3,3] Draw
[6,3,3,3,3] D wins 2 units
[6,4,3,3,3] D wins 2 units
[6,5,3,3,3] D wins 2 units
[6,6,3,3,3] D wins 2 units
8
A wins two units in 8 instances,
3 x 2 + 1 + 3 x 2 draws = 13 draws,
D wins 15 times,
[1,1,4,3,3] A wins 2 units
[1,2,4,3,3] A wins 2 units
[1,3,4,3,3] A wins 2 units
[1,4,4,3,3] Draw
[1,5,4,3,3] Draw
[1,6,4,3,3] Draw
[2,1,4,3,3] A wins 2 units
[2,2,4,3,3] A wins 2 units
[2,3,4,3,3] A wins 2 units
[2,4,4,3,3] Draw
[2,5,4,3,3] Draw
[2,6,4,3,3] Draw
[3,1,4,3,3] A wins 2 units
[3,2,4,3,3] A wins 2 units
[3,3,4,3,3] Draw
[3,4,4,3,3] D wins 2 units
[3,5,4,3,3] D wins 2 units
[3,6,4,3,3] D wins 2 units
[4,1,4,3,3] Draw
[4,2,4,3,3] Draw
[4,3,4,3,3]  D wins 2 units
[4,4,4,3,3]  D wins 2 units
[4,5,4,3,3]  D wins 2 units
[4,6,4,3,3]  D wins 2 units
[5,1,4,3,3] Draw
[5,2,4,3,3] Draw
[5,3,4,3,3]  D wins 2 units
[5,4,4,3,3]  D wins 2 units
[5,5,4,3,3]  D wins 2 units
[5,6,4,3,3]  D wins 2 units
[6,1,4,3,3] Draw
[6,2,4,3,3] Draw
[6,3,4,3,3] D wins 2 units
[6,4,4,3,3] D wins 2 units
[6,5,4,3,3] D wins 2 units
[6,6,4,3,3] D wins 2 units
9
A wins two units in 9 instances,
3 x 3 + 3 x 3 draws = 18 draws,
D wins 9 times,
[1,1,4,4,3] A wins 2 units
[1,2,4,4,3] A wins 2 units
[1,3,4,4,3] A wins 2 units
[1,4,4,4,3] Draw
[1,5,4,4,3] Draw
[1,6,4,4,3] Draw
[2,1,4,4,3] A wins 2 units
[2,2,4,4,3] A wins 2 units
[2,3,4,4,3] A wins 2 units
[2,4,4,4,3] Draw
[2,5,4,4,3] Draw
[2,6,4,4,3] Draw
[3,1,4,4,3] A wins 2 units
[3,2,4,4,3] A wins 2 units
[3,3,4,4,3] A wins 2 units
[3,4,4,4,3] Draw
[3,5,4,4,3] Draw
[3,6,4,4,3] Draw
[4,1,4,4,3] Draw
[4,2,4,4,3] Draw
[4,3,4,4,3] Draw
[4,4,4,4,3] D wins 2 units
[4,5,4,4,3] D wins 2 units
[4,6,4,4,3] D wins 2 units
[5,1,4,4,3] Draw
[5,2,4,4,3] Draw
[5,3,4,4,3] Draw
[5,4,4,4,3] D wins 2 units
[5,5,4,4,3] D wins 2 units
[5,6,4,4,3] D wins 2 units
[6,1,4,4,3] Draw
[6,2,4,4,3] Draw
[6,3,4,4,3] Draw
[6,4,4,4,3] D wins 2 units
[6,5,4,4,3] D wins 2 units
[6,6,4,4,3] D wins 2 units
10
A wins two units in 9 instances,
3 x 3 + 3 x 3 draws = 18 draws,
D wins 9 times,
[1,1,4,4,4] A wins 2 units
[1,2,4,4,4] A wins 2 units
[1,3,4,4,4] A wins 2 units
[1,4,4,4,4] Draw
[1,5,4,4,4] Draw
[1,6,4,4,4] Draw
[2,1,4,4,4] A wins 2 units
[2,2,4,4,4] A wins 2 units
[2,3,4,4,4] A wins 2 units
[2,4,4,4,4] Draw
[2,5,4,4,4] Draw
[2,6,4,4,4] Draw
[3,1,4,4,4] A wins 2 units
[3,2,4,4,4] A wins 2 units
[3,3,4,4,4] A wins 2 units
[3,4,4,4,4] Draw
[3,5,4,4,4] Draw
[3,6,4,4,4] Draw
[4,1,4,4,4] Draw
[4,2,4,4,4] Draw
[4,3,4,4,4] Draw
[4,4,4,4,4] D wins 2 units
[4,5,4,4,4] D wins 2 units
[4,6,4,4,4] D wins 2 units
[5,1,4,4,4] Draw
[5,2,4,4,4] Draw
[5,3,4,4,4] Draw
[5,4,4,4,4] D wins 2 units
[5,5,4,4,4] D wins 2 units
[5,6,4,4,4] D wins 2 units
[6,1,4,4,4] Draw
[6,2,4,4,4] Draw
[6,3,4,4,4] Draw
[6,4,4,4,4] D wins 2 units
[6,5,4,4,4] D wins 2 units
[6,6,4,4,4] D wins 2 units
11
A wins two units in 15 instances,
3 x 2 + 1 +  2 x 3 draws = 13 draws,
D wins 8 times,
[1,1,5,4,4] A wins 2 units
[1,2,5,4,4] A wins 2 units
[1,3,5,4,4] A wins 2 units
[1,4,5,4,4] A wins 2 units
[1,5,5,4,4] Draw
[1,6,5,4,4] Draw
[2,1,5,4,4] A wins 2 units
[2,2,5,4,4] A wins 2 units
[2,3,5,4,4] A wins 2 units
[2,4,5,4,4] A wins 2 units
[2,5,5,4,4] Draw
[2,6,5,4,4] Draw
[3,1,5,4,4] A wins 2 units
[3,2,5,4,4] A wins 2 units
[3,3,5,4,4] A wins 2 units
[3,4,5,4,4] A wins 2 units
[3,5,5,4,4] Draw
[3,6,5,4,4] Draw
[4,1,5,4,4] A wins 2 units
[4,2,5,4,4] A wins 2 units
[4,3,5,4,4] A wins 2 units
[4,4,5,4,4] Draw
[4,5,5,4,4] D wins 2 units
[4,6,5,4,4] D wins 2 units
[5,1,5,4,4] Draw
[5,2,5,4,4] Draw
[5,3,5,4,4] Draw
[5,4,5,4,4] D wins 2 units
[5,5,5,4,4] D wins 2 units
[5,6,5,4,4] D wins 2 units
[6,1,5,4,4] Draw
[6,2,5,4,4] Draw
[6,3,5,4,4] Draw
[6,4,5,4,4] D wins 2 units
[6,5,5,4,4] D wins 2 units
[6,6,5,4,4] D wins 2 units
12
A wins two units in 16 instances,
4 x 2 + 2 x 4  draws = 16 draws,
D wins 4 times,
[1,1,5,5,4] A wins 2 units
[1,2,5,5,4] A wins 2 units
[1,3,5,5,4] A wins 2 units
[1,4,5,5,4] A wins 2 units
[1,5,5,5,4] Draw
[1,6,5,5,4] Draw
[2,1,5,5,4] A wins 2 units
[2,2,5,5,4] A wins 2 units
[2,3,5,5,4] A wins 2 units
[2,4,5,5,4] A wins 2 units
[2,5,5,5,4] Draw
[2,6,5,5,4] Draw
[3,1,5,5,4] A wins 2 units
[3,2,5,5,4] A wins 2 units
[3,3,5,5,4] A wins 2 units
[3,4,5,5,4] A wins 2 units
[3,5,5,5,4] Draw
[3,6,5,5,4] Draw
[4,1,5,5,4] A wins 2 units
[4,2,5,5,4] A wins 2 units
[4,3,5,5,4] A wins 2 units
[4,4,5,5,4] A wins 2 units
[4,5,5,5,4] Draw
[4,6,5,5,4] Draw
[5,1,5,5,4] Draw
[5,2,5,5,4] Draw
[5,3,5,5,4] Draw
[5,4,5,5,4] Draw
[5,5,5,5,4] D wins 2 units
[5,6,5,5,4] D wins 2 units
[6,1,5,5,4] Draw
[6,2,5,5,4] Draw
[6,3,5,5,4] Draw
[6,4,5,5,4] Draw
[6,5,5,5,4] D wins 2 units
[6,6,5,5,4] D wins 2 units
13
A wins two units in 16 instances,
4 x 2 + 2 x 4  draws = 16 draws,
D wins 4 times,
[1,1,5,5,5] A wins 2 units
[1,2,5,5,5] A wins 2 units
[1,3,5,5,5] A wins 2 units
[1,4,5,5,5] A wins 2 units
[1,5,5,5,5] Draw
[1,6,5,5,5] Draw
[2,1,5,5,5] A wins 2 units
[2,2,5,5,5] A wins 2 units
[2,3,5,5,5] A wins 2 units
[2,4,5,5,5] A wins 2 units
[2,5,5,5,5] Draw
[2,6,5,5,5] Draw
[3,1,5,5,5] A wins 2 units
[3,2,5,5,5] A wins 2 units
[3,3,5,5,5] A wins 2 units
[3,4,5,5,5] A wins 2 units
[3,5,5,5,5] Draw
[3,6,5,5,5] Draw
[4,1,5,5,5] A wins 2 units
[4,2,5,5,5] A wins 2 units
[4,3,5,5,5] A wins 2 units
[4,4,5,5,5] A wins 2 units
[4,5,5,5,5] Draw
[4,6,5,5,5] Draw
[5,1,5,5,5] Draw
[5,2,5,5,5] Draw
[5,3,5,5,5] Draw
[5,4,5,5,5] Draw
[5,5,5,5,5] D wins 2 units
[5,6,5,5,5] D wins 2 units
[6,1,5,5,5] Draw
[6,2,5,5,5] Draw
[6,3,5,5,5] Draw
[6,4,5,5,5] Draw
[6,5,5,5,5] D wins 2 units
[6,6,5,5,5] D wins 2 units
14
A wins two units in 24 instances,
4 draws + 2 draws + 4 draws = 10 draws,
D wins 2 times,
[1,1,6,5,5] A wins 2 units
[1,2,6,5,5] A wins 2 units
[1,3,6,5,5] A wins 2 units
[1,4,6,5,5] A wins 2 units
[1,5,6,5,5] A wins 2 units
[1,6,6,5,5] Draw
[2,1,6,5,5] A wins 2 units
[2,2,6,5,5] A wins 2 units
[2,3,6,5,5] A wins 2 units
[2,4,6,5,5] A wins 2 units
[2,5,6,5,5] A wins 2 units
[2,6,6,5,5] Draw
[3,1,6,5,5] A wins 2 units
[3,2,6,5,5] A wins 2 units
[3,3,6,5,5] A wins 2 units
[3,4,6,5,5] A wins 2 units
[3,5,6,5,5] A wins 2 units
[3,6,6,5,5] Draw
[4,1,6,5,5] A wins 2 units
[4,2,6,5,5] A wins 2 units
[4,3,6,5,5] A wins 2 units
[4,4,6,5,5] A wins 2 units
[4,5,6,5,5] A wins 2 units
[4,6,6,5,5] Draw
[5,1,6,5,5]  A wins 2 units
[5,2,6,5,5] A wins 2 units
[5,3,6,5,5] A wins 2 units
[5,4,6,5,5] A wins 2 units
[5,5,6,5,5] Draw
[5,6,6,5,5] Draw
[6,1,6,5,5] Draw
[6,2,6,5,5] Draw
[6,3,6,5,5] Draw
[6,4,6,5,5] Draw
[6,5,6,5,5] D wins 2 units
[6,6,6,5,5] D wins 2 units
15
A wins two units in 25 instances,
5 x 1 + 5  draws = 10 draws,
D wins 1 times,
[1,1,6,6,5] A wins 2 units
[1,2,6,6,5] A wins 2 units
[1,3,6,6,5] A wins 2 units
[1,4,6,6,5] A wins 2 units
[1,5,6,6,5] A wins 2 units
[1,6,6,6,5] Draw
[2,1,6,6,5] A wins 2 units
[2,2,6,6,5] A wins 2 units
[2,3,6,6,5] A wins 2 units
[2,4,6,6,5] A wins 2 units
[2,5,6,6,5] A wins 2 units
[2,6,6,6,5] Draw
[3,1,6,6,5] A wins 2 units
[3,2,6,6,5] A wins 2 units
[3,3,6,6,5] A wins 2 units
[3,4,6,6,5] A wins 2 units
[3,5,6,6,5] A wins 2 units
[3,6,6,6,5] Draw
[4,1,6,6,5] A wins 2 units
[4,2,6,6,5] A wins 2 units
[4,3,6,6,5] A wins 2 units
[4,4,6,6,5] A wins 2 units
[4,5,6,6,5] A wins 2 units
[4,6,6,6,5] Draw
[5,1,6,6,5]  A wins 2 units
[5,2,6,6,5] A wins 2 units
[5,3,6,6,5] A wins 2 units
[5,4,6,6,5] A wins 2 units
[5,5,6,6,5] A wins 2 units
[5,6,6,6,5] Draw
[6,1,6,6,5] Draw
[6,2,6,6,5] Draw
[6,3,6,6,5] Draw
[6,4,6,6,5] Draw
[6,5,6,6,5] Draw
[6,6,6,6,5] D wins 2 units
16
A wins two units in 25 instances,
5 x 1 + 5  draws = 10 draws,
D wins 1 times,
[1,1,6,6,6] A wins 2 units
[1,2,6,6,6] A wins 2 units
[1,3,6,6,6] A wins 2 units
[1,4,6,6,6] A wins 2 units
[1,5,6,6,6] A wins 2 units
[1,6,6,6,6] Draw
[2,1,6,6,6] A wins 2 units
[2,2,6,6,6] A wins 2 units
[2,3,6,6,6] A wins 2 units
[2,4,6,6,6] A wins 2 units
[2,5,6,6,6] A wins 2 units
[2,6,6,6,6] Draw
[3,1,6,6,6] A wins 2 units
[3,2,6,6,6] A wins 2 units
[3,3,6,6,6] A wins 2 units
[3,4,6,6,6] A wins 2 units
[3,5,6,6,6] A wins 2 units
[3,6,6,6,6] Draw
[4,1,6,6,6] A wins 2 units
[4,2,6,6,6] A wins 2 units
[4,3,6,6,6] A wins 2 units
[4,4,6,6,6] A wins 2 units
[4,5,6,6,6] A wins 2 units
[4,6,6,6,6] Draw
[5,1,6,6,6] A wins 2 units
[5,2,6,6,6] A wins 2 units
[5,3,6,6,6] A wins 2 units
[5,4,6,6,6] A wins 2 units
[5,5,6,6,6] A wins 2 units
[5,6,6,6,6] Draw
[6,1,6,6,6] Draw
[6,2,6,6,6] Draw
[6,3,6,6,6] Draw
[6,4,6,6,6] Draw
[6,5,6,6,6] Draw
[6,6,6,6,6] D wins 2 units
Summary of the Permutation Set
1
D wins 36 times
2
A wins two units in 1 instance,
1 draw,
D wins 35 times,
3
A wins two units in 1 instance,
5 draws + 5 draws = 10 draws,
D wins 25 times,
4
A wins two units in 1 instance,
5 draws + 5 draws = 10 draws,
D wins 25 times,
5
A wins two units in 3 instances,
4 + 5 x 1 draws = 9 draws,
D wins 24 times,
6
A wins two units in 4 instances,
4 x 2 + 4 x 2 draws = 16 draws,
D wins 16 times,
7
A wins two units in 4 instances,
4 x 2 + 4 x 2 draws = 16 draws,
D wins 16 times,
8
A wins two units in 8 instances,
3 x 2 + 1 + 3 x 2 draws = 13 draw,
D wins 15 times,
9
A wins two units in 9 instances,
3 x 3 + 3 x 3 draws = 18 draws,
D wins 9 times,
10
A wins two units in 9 instances,
3 x 3 + 3 x 3 draws = 18 draws,
D wins 9 times,
11
A wins two units in 15 instances,
3 x 2 + 1 draw+ 3 x 2 draws = 13 draws,
D wins 8 times,
12
A wins two units in 16 instances,
4 x 2 draws + 2 x 4  draws = 16 draws,
D wins 4 times,
13
A wins two units in 16 instances,
4 x 2 draws + 2 x 4 draws = 16 draws,
D wins 4 times,
14
A wins two units in 24 instances,
4 draws + 2 draws + 4 draws = 10 draws,
D wins 2 times,
15
A wins two units in 25 instances,
5 x 1 draws + 5 draws = 10 draws,
D wins 1 times,
16
A wins two units in 25 instances,
5 x 1 draws + 5  draws = 10 draws,
D wins 1 times,
Conclusion
576 total number of combinations as stated earlier
Let's now total the results we found in the permutation set
A wins = 1+1+1+3+4+4+8+9+9+15+16+16+24+25+25 = 161
Draw = 1+10+10+9+16+16+13+18+18+13+16+16+10+10+10 = 222
D Wins= 36+35+25+25+24+16+16+15+9+9+8+4+4+2+1= 193
Number of Sucessful Attacks/ Total Combinations = 161/576 =0.279 = 28% of the time Attacker will win 2 units in a 3 dice on 2 scenario.
Draws/Total Combinations = 222/576 = 0.385 = 39% of the time both sides will lose 1 unit in a 3 dice on 2 scenario.
Number of Successful Defenses/Total Combinations = 193/576 = 0.335 = 34% of the time the defense takes 2 units in a 3 dice on 2 scenario.
So to answer the primary question, the Defender is 6% more likely to win 2 units rather than that for the attacker.
Here are some more analyses, they should be fairly self-explanatory:
A+Dr/T=(Attacks + Draws) / Total Combinations = (161 + 222) / 576 = 0.665
D+Dr/T=(Defences + Draw) / Total Combinations = (193 + 222) / 576 = 0.72
Advantage in Favor of the Defense = (D+Dr/T)/(A+Dr/T) = 0.72/0.665 = 1.083
D/A = 193/161= 1.198
A/Dr = 161/222 = 0.725
D/Dr = 193/222 =  0.869
A Rule for Calculating the Number of Attacking Units Required
The following explanation covers the nature of the formula I provide as a rule to use when attacking with large forces in Risk.  So, 5/4 relates to D/A = 1.19 (approx. = 6/5).  I chose 5/4 because it is slightly larger than 6/5.  This was chosen, because it should counteract any bad luck the user of this method may experience in most cases.  The exponential function provides another additional buffer to ensure a win on the attacker's side.  This buffer minimizes to e when DF is equal to infinity.  This component of the formula accounts for the advantage that the attacker gains when both him and the defender wage battle with forces that are very large. The 1 added to the end of the equation accounts for a single unit being left behind on the territory from which the attacker originated.
AF means number of Attacker's Forces
DF means number of Defender's Forces
AF=(5/4)(DF-2)+e^(1+(1/DF))+1       [The number of Attacking units required]
Table showing the results for the above formula:

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text:
What progress has been made in computerized bridge play?
Computer programs have been produced for games such as Chess, Go, Poker, StarCraft 2, Dota. The best ones, Deep Blue  and AlphaGo , AlphaZero, Pluribus,...  are now considered better than the best human players. More to the point, the computers' game results have been influencing human play.
Apparently, computers are not yet better than human players in Bridge. There can be computer simulations of various hands and hypothetical opposing hands. But what progress have computers made in playing human players in tournaments? Have any new theories of bidding or play evolved as a result of computer-human interaction in Bridge?

I understand that to get the results we got for chess/go a lot of commitment/funding is required. Moreover, the prior result from chess/go are not easily translatable to something else, such as bridge. As a result, you might not see the same level of advancement in a bridge computer player until such time some one with the necessary prerequisites takes interest in it. This is not to say this is not possible with the current level of technology.

see https://en.wikipedia.org/wiki/Computer_bridge

There's a fundamental problem that the rules of bridge are not well-defined for computers. Specifically, the rules require that partnerships communicate their agreements - including implicit agreements arising from experience playing with their partner - to their opponents.  No one has really figured out what it means for a computer to communicate its partnership agreements to other humans or to a different computer program.
(Note that it does happen in human bridge where one partnership explains their agreements to their opponents, there is misunderstanding, and a director or an appeals committee has to sort everything out, so these rules are not perfectly well-defined for humans either.)
This is less of a problem if the computer is playing a human designed bidding (and carding) system, but a machine-learning designed bidding system could quite possibly (in my opinion would very likely) be incomprehensible to humans.

Not just when there's partnerships, but even when it comes to multiple separate players, good AIs are harder to design - that's part of why all the examples given are 2 player games, there's less activity between the AIs turns to calculate based off of (and thus a lot less possible scenarios to evaluate)

The first year Al Levy ran the computer bridge championship, Ginsberg entered GIB playing a wild transfer opening, relay strong club variant (which works better for computers than for humans as computers don't forget).  It was very good straight up (for the time), but since the opponents couldn't "understand" it, it got lots of advantage from "incomprehensible bidding".  After that (and even now), the allowed system is very restricted.


text:
Where to go to self publish a game?
I am interested in a commercial printing company to print my game.  I need a US-company where I can print the entirety of my game and have it assembled already when shipped.
So far I found:

The GameCrafter seems like a good solution but its costly.
Business card printing places or Playing card printing places: neither of these will work since they only allow 1 design per order and I don't want to make 30 different orders since my game won't have the same amount of every card in them.
Other option is Self producing cards of my game but I don't want to go this way, I prefer to pay someone to do that for me.

Does anyone know about a US-based commercial printing company that can print  my game?

We're in touch with Cartemundi at the moment, their core business is card games but they do board games too. For us they'll be doing cards, rules, box, sourcing and packing counters and completing the boxes (wrapping and all) ready to ship.
They do a minimum order size of 500 and they're prices get substantially better the more you print. John is extremely helpful, and has helped ensure that our game is as cheap as possible to print, while retaining maximum quality.

In the US there's the company that print Fluxx (for the life of me I can't find their name)
Or these chaps: http://360manufacturingservices.com/

I did some digging and Fluxx is printed by DeLano.

Yup, was going to pop back on and add that (I checked the box on my Cthulhu Fluxx when I got home) DeLano seem ok, though close study of the quality of the box makes them seem a little low quatlity

I used cartamundi when I did this and I'd endorse the recommendation. However, my comment is to note that for them, and probably any other card printer, if you use a cheaper/recycled card make sure your card design has a lot of colour area on it.  I'd relied on the plan white card substrate too much, and the first batch was a bit spotty.

I really like The Game Crafter.  They have a lot of parts in stock, they'll sell your game in their store if you'd like or you can just use them for printing/publishing.  And their prices are very reasonable compared to the other sites I've looked in to.

http://www.drivethrucards.com/ sells card games as print-on-demand physical card decks (in addition to print-at-home PDFs). No up front publishing investment, nothing gets printed until your players place an order.
(I used to work for this company)

I published 5 games, three of them made in professional quantities (3,000, 2,000 and 3,000). For the first one I used Carta Mundi in the UK, but I was pretty unhappy with the customer service and the finished product quality. The other two I used LudoFact in Germany (http://www.ludofact.de/). They were excellent, both in customer service, delivery schedule and finished product quality. They are however, expensive.
A cheaper option would be to go Chinese, Panda Manufacturing (http://pandagm.com/) are one example of that. I have no experience with Chinese manufacturers however, so you would have to  approach them and get your own feel for them as a partner.

FWIW, Panda is arguably the most expensive popular printer in China. They do good work, but have higher MOQ's and rates. Whatz Games is a rising star.

As mentioned, I have no experience with Chinese manufacturers so are unable to comment on their costs or quality.

One tip is not just to look for specialist 'game' printers but to look for fancy box manufacturers.  You may get a better range of prices/options depending on the nature of your game.
This doesn't fulfil your requirement to have the game made in one place, but, having worn the self-publisher T-shirt myself, I'd say sourcing the different parts from different places can give you an edge on costs - and flexibility when one sub-component doesn't come out as expected.

Printer's Studio will do entirely custom-printed decks. I haven't used them personally, but have heard good things from others.


text:
How does Rook compare in difficulty to other trick-taking games?
How does Rook compare in difficulty to other trick-taking games?
It is commonly played by older people in my area, and so my siblings and I learned it as small children.
What games would be good to pick up, having that background?

You might also find the question "Getting started with trick-taking games (whist, euchre, bridge, pinochle, etc.)" helpful.

For an added challenge, try the 5-player CMU/MIT version: http://www.georgejas.com/rook/

@NealTibrewala That sounds like some weird bastardization of normal Rook and "Zhao Pengyou"

Thanks for that link!  I hadn't heard of that game before, but I'll try it.

Of all the other card games I know, Rook is most like partnership Whist, with only a few exceptions:

Tournament Rook uses the rook-bird card, which serves as the highest trump card (not a wild card: the rook-bird play still must follow rules of following suit)
Bidding in Tournament Rook is done by point counting, not by trick counting (as with Whist): fives are 5, tens and fourteens are 10, and the rook-bird is 20. (As with Whist, trump is not named until after the bidding has passed out.)
Tournament Rook sets aside a "nest" of five cards on each deal-out that is not used by the players to play the hand, nor are the cards visible. This gives each hand 10 cards. The winning partnership counts the points in their won tricks, plus any tricks in this "nest".

The two aspects of the game I like most (and which make it more interesting than Whist) are the "nest" and the separation of bidding and measure of "successful play" from the number of tricks captured.
There are plenty of other card games that assign point values to cards and both bid and measure success based on those cards (Mü, Skat, Tichu). However, I can't off the top of my head think of other games that also incorporate a kitty of cards taken out of play that get assigned to the winning partnership at the end of play.

This explains the game, but doesn't really address difficultly.  I like the answer, perhaps enhance it with saying if the bullets make it easier or harder?

Because there's more disconnection between play action and value result (that is, the tricks themselves aren't directly valuable, it's what they may contain that's valuable), I think Rook provides more depth during play. Some folks might see this as "harder", but I don't really see either game as significantly harder to learn.

There was an answer deleted that had a good link with a complete answer,  I'll repeat it here.
Is the Rook Game Similar to the Bridge Card Game?

Try mini-bridge:
An advantage of mini-bridge over rook/spades is that it's deeper, but equally easy to learn the rules.  It also gives you a lot of upside in that you could move onto playing Bridge, which is the classic among trick taking games: more complex, more fun, and more rewarding.
Mini-bridge is being taught as an alternative to Chess in many schools among younger children, with the idea that they'll move on to play the real thing.

+1: Mini-Bridge is a good recommendation. btw, the number of bridge questions on this site has increased suddenly! You might want to chime in :-)

I'd say that it's among easier trick-taking games, but not too simple. Games of similar difficulty would be Spades, Euchre, and Pinochle. Hearts also works, but you'd need to remember it's about avoiding taking points. None of the games have anything similar to the rook card. Spades, Euchre, and Pinochle each have team play with the option to "go it alone". Spades has a set trump suit. Euchre and Pinochle use modified decks, and both change card ranks depending on the trump. Pinochle is the hardest of the three, with a melding phase.

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text:
Resources for Nine Men's Morris
I just started playing Nine Men's Morris. I have tried playing with both the computer and on online multiplayer settings and have been doing abysmally. I tried finding some good resource on the net that could give me some in depth study of useful strategies for this board game but I found very basic guides that were basically just a rehash of the rules. Could somebody help me in my dilemma, I want to start winning at least a few of the games.

There's some information available in German. The game's called "Mühle" (= mill) in German,
and there exists at least one book on this subject.
The book can be found online here: http://www.hubbie.de/
Click on "Mühle", the book can be viewed in html or as an applet.
It's a beginner's book, but contains even some advanced strategies.
It should be possible to understand a lot of the strategies even without knowledge
of German, but with "Google translate", or dict.leo.org you should be able to make
sense of most of the text.
There's even an organization which organizes (European) championchips, see
http://www.muehlespiel.eu/
It might be possible to get information from there, too.

I realize you are searching for a human strategy, but it might be useful to know that it was solved for a computer play. Nine Men’s Morris was solved by Ralph Gasser in 1993. The game is a Draw.

With respect to basic strategy, I'm assuming that you've found this, which seems to have scraped the text from this page (or vice versa).
In the early '90s, at least one person believed that the strategy you are seeking was not possible to describe.

What is an optimal opening strategy (for humans)?

I must admit that I have not examined the opening moves for any easily describable patterns. However, the examination of mid- and endgame databases has repeatedly shown optimal play to be beyond human comprehension. Therefore, I very much doubt that any such
simple-to-describe strategy exists.

I've not found anything to contradict him, so unless someone else has more recent experience to share, you may be out of luck.

I recommend to start with

The book Nine Men's Morris Strategy: A guide for beginners to intermediate players, it goes into a bit more depth regarding plays.

A perfect play repository by jochen-hoenicke

Once you have some basic plays, follow them through on as many different pathways as you can. Really good players have great memory, and know how to respond to many plays.

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.


text:
Are there any English servers where I can play Mahjong online?
I recently stumbled upon the Asian game Mahjong and came to like it.  I already play Go and Chess online, but had difficulties finding a good, English-language Mahjong server.
Are there any English servers, and if so what are they?

You're more likely to find a good answer for this in Gaming's chat.

Voted to close as off-topic, see http://meta.boardgames.stackexchange.com/questions/34/should-questions-purely-about-online-versions-of-a-game-be-allowed/59#59

@C. Ross: This is a question that may be interesting to many people, so it should be asked on the site and be visible. Imo, Chat is definitely the wrong place for this kind of question.

As per the question on meta, I believe this question should NOT be closed.

Nominated to reopen, as per what seems to be the consensus on meta. If you play a board game online, it's still the same game; thus it's still on topic for this SE. If the online game is substantially different, then it would probably be off topic. In this case, the mahjong solitaire that is generally played only on a computer would be off-topic, but mahjong the multi-player gambling game would count as a board or card game and asking about playing it online should too.

Also voted to reopen, in agreement with @Brian.

We've learnt a lot in the past 14 years, and this question is a great example of why "lists of things" questions don't work in the Stack Exchange model - about half the links no longer work, and the question is not just attracing spam answers.

I've played Mahjong online through Yahoo. But, there usually aren't that many players on their server. Yahoo also have solitaire Mahjong games, if that's what you're referring to.
You could also try MahjongExpress. I've never played on there, but it looks like it requires you to install software, kind of like the online poker games.

I tried Mahjong Express, it's a nice program with a limited but reasonable set of features. Few players (only around 5-10 players online usually).

A friend notified me of a site he occasionally plays on: http://tenhou.net/
According to him, it's a nice site. Problem is that it's entirely in Japanese. He's lucky enough to understand Japanese but most people don't so I added this answer just for reference.
Update: I found a nice site where you can play (only) against the computer. English language, no setup, completely free and easy to use - a great introduction into Mahjong.
Update 2: This page lists several clients. I did not try them yet.

DragonKong Mahjong
Janryumon (雀龍門)
Japanese Mahjong by Game Design
Kinma Mahjong Online Battle (近代麻雀オンラインバトル)
Kong 88
MahjongTime
Maru-Jan
Pairon (ぱいろん)
Ron2 (ロン2)
Tenhou (天鳳)
Tetsuya@Tonpuusou (哲也＠東風荘)
Tonpuusou (東風荘)
Toupaiou (闘牌王)
Tsumo (ツモ)
Yahoo! Games
Yahoo!ゲーム - 無料ゲーム - 麻雀

I've also fell in love with this game recently. To play against players I am trying out MahjongTime, they give you a certain number of chips to play a day and you can play off of that. (four times a day if you don't win any games) and play for 0 mp
They have different styles of play, Hong kong, Taiwanese, American... etc.
And different rules. Not sure I would trust them for cash games but it is a good relief for quick games.
I don't find it very social though.
http://www.mahjongtime.com/

Could you add a link?

mafutrct has a link above but I will edit my post in a few seconds

Just had a look, nice site though few players

The National Mah Jongg League has an "official" version of American Mah Jong online with a PC based client. It cost an additional $40/year to your membership/card for the league. I have never tried it but it does allow for live online play with the current year's card.
http://www.nationalmahjonggleague.org

There's a new site that supports Chinese (MCR rules): https://playmahjong.io/
It's free to play and you can play bot and multiplayer 4-player games. The in-game UI is quite nice and it's being actively developed.

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text:
Term for a "connected" set of tiles in grid based games
This is a terminology/nomenclature question in the context of grid based games, specifically the game of Pentomino. I'm not a native English speaker, so I feel like I would easily miss something.
In Pentomino, each play piece covers exactly five cells of a grid. These cells/tiles then belong to the same play piece; they are "connected".
I am looking for a name for a set of tiles that is known to be connected without knowing which piece they belong to. For instance, four connected tiles in a row may belong to an L-, Y- or I-piece. So these tiles form an incomplete piece. How should I call this set of tiles?
As an example from a different game: In the game of Go, stones placed on the board grid by the same player that are (transitively) adjacent are "connected". They are often called a "group" or (more science-y) a "string".
So far, my best guess is to call a connected set of tiles in Pentomino a "string". Would this be understandable? Are there better options?

In David Parlett's The Oxford History of Board Games, pages 165-166, when discussing variants on Pentominoes, the author states

The other [variation], which has just occurred to me, is a linear
version of Pentominoes played with Cuisenaire rods, which are used for
teaching numbers and lengths to young children.The linear equivalent
of Pentominoes would be a set of rods each consisting of a number of
squares joined in a straight line, there being one of five squares,
two of four, three of three, four of two, and five of one.

I therefore suggest that you adopt the term rod to identify any straight sequence of squares within a pentomino.
Reference :

If the tiles are known to all be part of the same piece, without forming the entirety of the piece, then that would be a subpiece.

It's hard to prove a negative, but I don't think there's an unambiguous term for this, especially one that is readily understood in multiple different boardgames.
If you call it a "string" after explaining that concept to your audience, it'll work fine. Others may use two words, e.g. "connected group" and that would work too. In the case of Pentomino, you might call it an "incomplete piece"; that obviously doesn't apply to go.


text:
Know any games designed for old game systems like Orion or Sly
A Game System is a set of components and/or rules that are intended to be used to create games.  That is, one set of components that allows to play several diffrent games.
Recent game systems like piecepack, Decktet, or Stonehenge have tons of games to try, but older game systems like Sly and Orion only seem to have the official rules (PDF) that came with them.
Somehow Icehouse managed to survive from the 1980's until the web could attract a community for it, does anybody know of any additional rule sets written for other game systems published before 1995?

Since nobody else seems to have made any new games for Sly, I did it myself. Well, I guess some credit should go to Eric Solomon.

Seven years after I asked, Chris Goodwin designed Tingle to be playable on Sly, piecepack, and Looney Pyramids.


text:
Can every game of Klondike-Solitaire be solved?
I play a lot of Solitaire games on my Android phone and love to keep an eye out for the statistics.
Given that the Solitaire version lets you restart the game endlessly, I usually play until I solve it. But I never managed to solve more than 80% of the games played (1000+).
So now I wonder, is every Solitaire game solvable?

I presume you mean Klondike solitaire?

What a solitaire addict! Forever alone :)

No.
Example: If all of your cards face up on the board are red, and the cards that come up every third card are also red, and none of them are aces. You lose. Do not pass go, do not collect $200.

As a fact, I have come up with almost this exact setup on the computer version of Solitaire (but one card was black, just completely impossible to place anywhere).

Yeah, there's other variations on this. But I've played several games, where no cards ever got moved.

Another example that just happened to me:
All cards shown are even.

http://en.wikipedia.org/wiki/Klondike_%28solitaire%29#Odds_of_winning

Even simplier: all aces are on the same column and 2 is above them.

@Oltarus Aces in same column and 2 above them is still winnable. Its annoying and probably a loss but doable.

Oops, my bad. You are right.

A column with  will be unwinnable, since there's no way for the 2 to move.

Literally just played a game in which one of the stacks (the one containing 4 cards) was lead by the 9 of diamonds, and the cards inside of it were the King of Spades, the 5 of diamonds, the 10 of spades, and the 10 of clubs (I know this because I had the entire field solved except for this stack and used process of elimination). As far as I can see this makes the game impossible. I have a 9 of diamonds in which can never be moved, as the two 10s that it's eligible to rest upon are trapped underneath it in the stack face down. Attempting to get rid of the 9 by moving it to the diamond stack would also be fruitless, as the 5 of diamonds is stuck underneath it too. Unless someone can tell me some way that this could be solved, I'm pretty darned certain that if a card that is leading a stack is covering a stack that contains the two cards it is capable of resting on, and a lower number of it's own suit, then the game is made impossible right from the get-go.

Solitaire is a game that precedes its computer version, and that means that all the cards are truly shuffled, without the computer peeking in to verify the game is solvable.
And like McKay mentioned, with a random shuffle you can definitely end up with an unsolvable game.
I'm sure it is possible to design a Solitaire variant in which each game is solvable, though.

@Arda, there are some conditions that could be easily tested - for example, a card other than a King can be played on only three other cards in the deck (the next-lowest card in its suit, or the foundation for an Ace, and the next-higher cards of the opposite color). If all three of those cards are face down below that card on a pile, the game isn't winnable. Unfortunately I think that's a small percentage, and testing for other conditions might require a ton of recursion.

@Arda - that's true, that's what I was thinking with respect to recursion. Without some way to demonstrate that any given position is unwinnable, you'd essentially have to play a certain series of cards until you were blocked, back up to the last decision point, and repeat ... it's an intriguing concept, but I've never seen a solitaire program do that.

That shouldn't be too difficult.

@Arda Could simply work backwards from the solution, randomly moving cards into the deck and onto the board from the four suit piles, always using the reverse of a legal play. Probably won't have the same probability distribution as shuffling and checking for winnability, but I doubt that matters to most players.

I know I'm late to this, but a "reverse" play should have an identical winnability distribution. There are a finite number of winnable layouts, and the direction of search doesn't change that. Instead of looking for a single goal state, from a random start you're searching for a more complex goal state from the same start. The intuition behind why this is equivalent is reaching a state where every legal move returns you to a state that has already been visited.

No. Eric Sink decided that he would start a micro-ISV to create a version of solitaire that is always winnable. This was mostly just an experiment to see what it would be like running a software company with one person, but he eventually sold the product which is still available for purchase.
There have been some estimates about the number of Klondike Solitaire games that are unplayable (no moves possible, about 1 in 400), and several guesses about how many games are unwinnable, although this percentage varies wildly from 30%-10%.
The difficulty of this problem stems from the sheer number of initial deals 54! that would need to be evaluated to determine which were winnable and which were not.

would the number of initial deals be 52! ? (unless you expect the jokers to be dealt, too)

Fortunately, one does not need to use the brute force method (look at all possible deals) to compute the odds of winning (since that computation would take longer than the age of the universe - 8x10 to the 68th power decks).  An analysis of ways to fail provides an analytical line of attack.  As already noted there are clear ways that a single stack may fail.  The needed cards might also be unreachable within two stacks, three stacks or four stacks.  Once the conformations of cards for locking up the needed cards are known, their individual odds may be computed and combined to get an answer.

I've done quite a bit of work on this. We used a combination of Monte Carlo simulation and unsolvable layout detection to achieve the current best estimate of winnability (81.945 ± 0.084%). Unfortunately an analysis of the unreachable patterns of cards is not enough, as our method (current world leading) only detects ~71% of the unsolvable layouts.

Correction on the winnability estimate, actually 81.942 ± 0.081%.

However, if you started a list and enumerated the initial conditions -- I feel like I've seen this on a linux version of Solitare: the numbering of deck order, that is -- and you definitively decide a certain one is un-winnable, you then could compare notes across nodes (share with friends) and VOILA: a list of un-winnable starting deck stacks.
I've been starting to think the Windows 7 version has the un-winnable decks removed, ... I don't know, it's a little heavy-handed and smug about the statistics.

With 52! starting shuffles, you'll need an... inconveniently long... time before you have a good list.  Even after you solve the problem of determining unwinnable definitively.

52 factorial = roughly 8 followed by 67 zeroes. That's a lot of combinations. A 1TB hard drive would store about a trillion of these, and you'd need trillions of terabytes to store even a decent fraction. Not very practical unfortunately, just because of the astronomical number of probabilities involved. Probably easier to just store a certain number of demonstrably winnable games.

@JonathanHobbs Not all of them have to be stored to make the calculation. for 1 to 52! getdeck, try solving game, add to statistics at each point only one deck need be stored, and the statistics can be quite small.

@McKay You have to store quite a bit to develop a decent list, though. (I'm not sure which calculation you speak of.) As an aside also regarding the answer: the Windows 7 version actually just stores a few dozen thousand decks, and you're randomly given one each game. It might be they just picked a few dozen thousand decks known to be winnable.

@JonathanHobbs No, all you have to store is which deck you're looking at (which would need to go up to 52!, which means we'd need about 226 bits), and you'd need to store how many of them were solvable (another 226 bits, or less), and then one game of solitaire (which windows 3.1 was apparently able to store just fine), and the algorithm to actually solve the game. The data storage mechanisms do not need to be very much in order to do a full set of statistics on solvability. We're talking less than 1k of storage. Sure it would take a long time to do all these calculations. But not storage.

Oh, I was thinking about a different answer. I was thinking about calculating the percentage of shufflings that are solvable.

@McKay Right, yeah. This is discussing storing the individual games. 52! (a 68-digit long number) times 226 bits = a very, very large amount of bits (a 71-digit long number). Specifically: a 59-digit quantity of terabytes (the equation being: amount of numbers to store, divided by number of bits in a terabyte)

I've recently written a paper (accepted to ModRef 2024) on this exact topic. We used a programming technique called constraint progamming to find "blocked" layouts - a provably unsolvable layout requiring a card to move before itself (impossible). This can generate some quite complex layouts which are provably unsolvable:
In the above, assume that the four kings and two black sixes are available to be played in the stock. The seven piles in the top of the image represent the tableau. Relevant face-down cards have been shown as face-up for clarity (e.g. 7 of hearts).
We estimated the upper bound for the number of games of non-thoughtful (position of the face-down cards is unknown) deal-3 Klondike solitaire to have a 95% confidence interval of 81.942 ± 0.081%. This is the "most popular" variant of Solitaire.
This suggests 18.058 ± 0.081% cannot be solved, regardless of the moves made.

Great read.  How much of an improvement is 81.945 ± 0.084% from the results of Charlie Blake and Ian P. Gent?

Thanks for pointing this out @Cohensius: I read the winnability estimate from the wrong column in my slide pack. I originally quoted the Solvitaire winnability estimate, the correct new best estimate is now in the post.
Blake & Gent ran Solvitaire on 1,000,000 layouts, of which 157 didn't finish execution for deal-3 Klondike. We proved 63 of these layouts impossible. This only corresponds to a tiny improvement in the winnability estimate: reducing the 95% confidence interval by 0.003 percentage points.
We did find massive improvements in other solitaire games - currently in progress!

To add to the other great answers, this link has a nice explanation of how a deal is un-winnable.  Reasons for Getting Stuck in Klondike Solitaire

The two black 5's are blocking a red 6 and a lower card in their suit.  Thus they can not move to a different tableau pile and cannot move to the foundations.


text:
In 2 player, 10 card Gin, figuring out opponents cards
What tactics can you use to help determine what your opponent is looking for in two player 10 card Gin?

In Gin, you don't get any hard information on what is in your opponent's hand until the hand is over.
This means you have to make use of whatever information you can get.  The primary source of information is the discards.  To be successful you need to keep track of both

What your opponent discarded
What you discarded

Pay attention to both suits and the number of each card.  For example, if your opponent discarded the 5 of clubs, he is most likely not interested in 5's or mid range clubs.
Observe your opponent, if he usually plays fast but pauses even briefly to consider the jack of diamonds you just tossed then he most likely has several cards around that.  A couple examples of what he might have.

10-K of diamonds
sets of 9's and 10's but he has diamonds of both and could make another run (like 7-8-9-10 of clubs)

Perhaps from previous discards and pickups you could narrow down the options as to what he might have.

I like your answer but I think that the first statement: "In Gin, you don't get any hard information..." is wrong.

In Gin, you do get some hard information on what is in your opponent's hand and constantly need to add some intelligent guesses on what is probably the opponent's hand.
Hard information:

A pickup from the discard by the opponent - surely that opponent hold this card. Probably this card is a part of a Meld or at least a combination.
Your hand & discard pile - surely opponent do not hold those cards.

Softer information:

Opponent's discards - almost surely are not part of a meld, that is to say, opponent probably does not holds two cards in the same suit with rank {+1,+2,-1,-2} or two cards in the same rank.
Cards that cannot make a meld since other cards are discarded.  For example, the K♠ is probably not in the opponent's hand if the Q♠ and two other Kings were discarded already.

The nice book: How to win at Gin-Rummy by Pramod Shankar suggests to keep a table of the opponent's likely hand and to mark any card that is for sure in or for sure not in the opponent's hand, for example:

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/2.5/


text:
What reasons would a designer want to employ squares over hexes in a battle simulation game?
In many games I have played, initial editions (eg the 1960s version of Midway) use squares for their boards.
Often, later editions (eg the 1990s edition of Midway) change to a hex pattern.
Hexes get around awkward movement rules (like being able to move 14 squares, or 10 on the diagonal (10 / sqrt of 2 is very close to 14)), but I wonder when squares would give a better feel for the environment?
In Clue it makes sense because they represent tiles in the hallways (and while you can be whisked around the house by being involved in an accusation or via tunnels, the hallways analogy makes sense). However, in a simulation game, I cannot see how squares could be useful.
When are they a good mechanic in simulation game design?

It is possible to use squares that are offset every row/column (like a brick-wall pattern), but still have 6 directions of movement like a hex-board. Hex movement has advantages of being equa-distant from all adjacent squares/hexes, but works better when the points are oriented perpendicular to most movement (I.e. if you have two armies squaring off against each other on the top and bottom of a map, you want the hex points left and right

@user1873 that's a  fascinating thought I hadn't considered

@user1873 Cubitos uses such an offset square grid, for example. Functionally, it's equivalent to a (slightly stretched) hex grid.

Well one point towards squares is simplicity. It is a much easier system to grok (outside of weird rules for diagonals)
The most important thing to consider is what your playing field is. If you are talking about a mostly open plain with very few obstacles, then a hex field is great as it avoids the diagonal rule cleanly.
However if you were going with, say, an urban layout, then a hex system would have a harder time with all the straight edges (not impossible, but difficult) while also not helping too much, since diagonals aren't a huge deal with hallways.

Indoor areas or city terrain can be easier to represent using squares because they tend to follow 90 degree angles anyway. Most games with movement in a dungeon or similar places use squares.
There are board and miniature games that combine hex or free movement outside of cities with square movement in cities. I have seen tabletop RPGs played with a hex map for overland terrain movement, but switching to square grids for encounters and dungeon exploration.

What is the nature of your game?  Combat, commerce, both?  This might shape the role of movement in your game as well.  Hex-based maps makes movement slight less expensive, especially when trying to around obstacles.
For example, if the player is trying to move their token from South to North, but another player's token is in their way.  On a square-based map, it would take four moves to get around the obstacle (W, N, N, E).  On a hex-based map, it would take only three (NW, N, NE).  This is assuming you're not using a facing/turning mechanic, such is used in Battletech.
The Civilization series on the PC also made the transition from squares to hexes.  My friends who play the game were all pleased by the change because it made movement smoother and less frustrating.
Unfortunately you can't foresee where your game will go in the future.  I'm sure hexes where the right solution for Battletech when the scenarios were all played on an open battlefield.  But hexes make things really weird when you try to take the battle into a city.


text:
Probability of i event_A and j event_B on n dice
I am trying to figure out probability for a game using 6-sided dice. The basic idea is that the player rolls n dice and gets a success for each 6 rolled. So far, so simple; I'm using Python's scipy.binom.pmf function which gives the probability of rolling i successes on n dice with a given chance of success, 1/6 in this case.
But the game offers a "bonus". With the bonus, in addition to 6s being successful, one 5 in the roll is also a success. (So a roll of 2, 3, 5, 5, 6, 6 would be two successes normally, but three with the bonus.)
How can I compute the probability of i successes on n dice with this bonus condition?

It is unclear to me when the "bonus" applies.  Is it decided before the roll (like "roll with advantage" in D&D), or is it something that applies based on the outcome of the roll (like "if there's a natural 6, then one 5 can count as a success as bonus")?

Do yu mean that m**exactly-- one 5 is a bonus success, ort at least  one 5? How many successes is a roll of 6 6 5 5 4 3?

I know your using python but there is a useful website here https://anydice.com.   You can work out many dice probability there and with a bit of coding create some complicated combinations and rules.

Are you looking for exactly i successes or at least i successes?

@DavidSiegel OP already provides an example which is equivalent to yours and answers your question.

@Mycroft, the bonus is decided before the roll, as in "6 dice, with a bonus" or "5 dice, no bonus."

@DavidSiegel At least one 5 is one bonus. In your example, that would be 3 successes.

@L.ScottJohnson, exactly i successes.

P(i successes) = ((# of ways to get i 6s and no 5s) + (# of ways to get i-1 6s and some 5s)) / 6^n
(# of ways to get i 6s and no 5s) = (n choose i) * 1^i * 4^(n-i) [pick which dice must be 6s, 1 way for each of those to be a 6, 4 ways for each of the others to be neither 6 nor 5]
(# of ways to get i-1 6s and some 5s) = (n choose i-i) * 1^(i-1) * (5^(n-(i-1)) -
4^(n-(i-1))) [pick which dice must be 6s, 1 way for each of those to be a 6, 5 ways for each of the others to be non-6 except then we're overcounting by the 4^(n-(i-1)) ways that they are ALL neither 6 nor 5]
Dropping the unnecessary powers of 1, overall your probability can be computed exactly as:

[Note: I am using the convention that (n choose -1)=0, unlike Python's math.comb which does not allow negative inputs.]

If you haven't seen "choose" before, see https://en.wikipedia.org/wiki/Combination. It's math.comb in Python

@Benjjamin Cosman, I think I see what you're  getting at, and I think you are on the right track, but there's still something wrong or I'm not interpreting you right. Here's the Python code I've got:

pSixes = comb(n, i)   * pow(1/6, i)   * (    pow(4/6, n - i))
pBonus = comb(n, i-1) * pow(1/6, i-1) * (1 - pow(4/6, n - (i - 1)))

result[i] = pSixes + pBonus

For 3 dice, that produces P(1 success) = 0.925925925925926, P(2 successes) = 0.33333333333333337, and P(3 successes) = 0.032407407407407406, ...

Which doesn't add up to 1.0 and doesn't match @StartPlayer's results (or my manual 3 dice attempt). I think something is wrong with the P(i-1 6s and some 5s) expression.

Oops; fixed. I've now checked that this one gives the same answer as @StartPlayer (and adds up to 1 :) )

That is great! I never would have come up with that.  Thanks!

I used [Any Dice] website to work this out.
The code below is probably not the cleanest but I only learned this site today. The function below will count 1 success if at least one 5 and add to that how many sixes rolled.  This is getting results of to 5d6.

function: contains VALUES:s in SEQUENCE:s {
 FIVES: 0
 SIXES: 0
 loop P over {1..#VALUES} {
 if (5 = SEQUENCE) {
  if FIVES = 0 {
   FIVES: 1
  }
 }
  SIXES: SIXES + (P@VALUES = SEQUENCE)
 }
 result: FIVES + SIXES
}

output [contains { 6 } in d6]
output [contains { 6 } in 2d6]
output [contains { 6 } in 3d6]
output [contains { 6 } in 4d6]
output [contains { 6 } in 5d6]

These are the results that I've formatted to be a bit more readable than the anydice website gives me. If you roll 1d6 the chance of 1 success (A  6 or 5) is 33.3%
If you roll 3d6. the chance of 2 successes is 18.05% (that's either 2 6s and no 5s or one 6 and at least one 5)

1d6
0 success, 66.6666666667 1 success,33.3333333333
2d6
0 success, 44.4444444444 1 success, 47.2222222222 2 success,8.33333333333
3d6
0 success, 29.6296296296 1 success, 50.462962963 2 success, 18.0555555556 3 success, 1.85185185185
4d6
0 success, 19.7530864198 1 success, 48.225308642 2 success, 26.2345679012 3 success, 5.4012345679 4 success, 0.385802469136
5d6
0 success, 13.1687242798 1 success, 43.4799382716 2 success, 31.9573045267 3 success, 9.90226337449 4 success, 1.4146090535 5 success, 0.0771604938272

Thanks for that! I wrote the 3 dice case manually, and it matches with your results. The problem is that once you get up to 8 or 9 dice, it's too slow (and I've forgotten how to get multiple dice loops :-)).

The 1 dice case was clearly correct and I drew about a 6x6 grid to check 2d6 results were correct as well.  An advantage of this site is you can write code to check and kind of odds.  if you have a bag or 7 tokens and two are "success" you can work that out as well by saying d7 as a 'dice'.    It takes a bit of coding and the site docs seem clear enough.


text:
Suburbia - what is the rule of Reputation?
Introduction
In Suburbia you track your population, reputation, and income. The goal is to have the most population when the cards run out. Income is needed to purchase tiles to increase your population.
Question
As I understand from what the rules say about reputation is that in the event of a tie with population at the end of the game,  the one with the highest reputation is the winner. My question is, does reputation do anything else in the game other than that? If that's the only reason for it, and ties are pretty unlikely, I have no reason to keep it up.

please verify that the link I added is the game you referring (just Suburbia, not "CIties Suburbia").  If that the case, then reputation determine the amount of population that is added each turn.

In the subject line, do you mean "role of Reputation"?

Reputation = number of people that added to the total, at the end of the turn.
The relevant part in the rules:

ADJUST POPULATION (BASED ON REPUTATION)
Increase or decrease your Population on the Population Board by the amount where your Reputation Cube is located. If Population gets to 0, any additional movements backwards cost $1 per Population. If the player cannot pay this fee, nothing happens to the player...he just stays on the 0 spot.
As your borough grows, it will cost more to maintain and its “small town” quality will decrease --
this is represented by Red Lines on the population track. When a player’s Population crosses a Red
Line, then that player’s Income ◯ and Reputation █ are both reduced by 1 (moved to the left one
space) for every Red Line the player moves past. If a player’s Population dips below a Red Line (this
can happen if he has a negative Reputation) he moves both his Income Cylinder and his Reputation
Cube forward (moved to the right one space).

It's probably worth copying the relevant text of the rules to the answer as links to images may not last forever.


text:
Identify this Solitaire with 4 piles of 4 cards
I am looking for a Solitaire game, the rules are:

Setup: deal 4 piles of 4 cards. Only the top card is face up. Then you flip over the top card in your hand. That becomes the starter card for the first row.
Play: you can play any card from that suit on the row. You may not start another row until you get a starter card from another suit. Once two or more rows are started you can only play a card on a row if the same card in a different suit is on the row above it.
If you take a card from one of the four piles, you turn over the next card.

You win the game when all of the cards have been played.

maybe you can find your game in this index of ~540 solitaire games:  https://www.bvssolitaire.com/solitaire-game-list.htm

I'm not sure what the best cite for it is, but I believe this game is called Osmosis.

https://www.bvssolitaire.com/rules/osmosis.htm


text:
Is there a reliable free cell solver?
I am trying to find a solution to game 9371310 and the online solver at
https://fc-solve.shlomifish.org/
instructions try to use 6 free cells when only 4 exist. Here is where the instructions go terribly wrong.
Is there a more reliable solver out there?

I don't have that problem when I try to solve that board on that online solver.  What process did you follow?

Did you possibly select 6 from the drop down for number of freecells?

I get the OP's result when and only when switching "Freecells Number" from "Default" to "6". Just switch it back to "Default".

The solver you mention do solves this deal with 4 free cells.
For MS Freecell Deal #9371310
: TD 2S 8H 4S JC QC 6S
: 2H 4C 7S 8S 9D KD 3D
: 5S 3C 9S 9H 7D AC KC
: 5D 7H 9C 5C AD 6D JD
: TC 4D 3H AS QD KS
: 3S 8C QS AH KH TS
: 2C QH 8D 4H JH JS
: 5H 2D 6C 6H TH 7C

Used the default settings, it solved it in few seconds:


text:
Card Game Name: 5 points and you’re out
I recently learned a new card game and the person didn’t know the name. Maybe someone else does. Goes like this...
A round begins with each player having been dealt 4 cards except the starting player is dealt 5. Cards are scored as face value except all face/ace/joker cards are 10 points. Each turn, you lay your highest card (ideally) and then draw from the deck or the top card off the discard. You can also lay multiples if they are the same number/face. Once someone reaches a card score of 14 points or lower, they can show their hand and play stops (or continue to attempt to get a lower hand). If they show and aren’t the lowest, they get a point against them. Otherwise, the person with the highest card score gets the point against them. A new round then starts with shuffled cards. Once anyone hits 5 points, they are out and rounds continue until only one person is left, the winner.

It definitely sounds like a variant of "Golf." I've played several (including Skyjo and Play Nine), but not this version. Most allow the player to at least see the top of the draw pile before deciding if they want to keep it (by discarding). It also sounds like "Silver" by Bezier games, but of course that doesn't use a standard deck of cards.

Your game's type is Draw and discard.  Knowing the nationality of your teacher could narrow your search.  Your description sounds a lot like a variant of Yaniv / Jhyap.
Yaniv description in Pagat:

Deal: Five cards are dealt to each player

The objective of the game is to have a lower total value of cards in your hand than the other players. When you think you have achieved this, you call "Jhyap" in Nepal or "Yaniv" in Israel to stop the play and compare hands.

Each game is composed of rounds, and each player's total score is updated after each round. The goal of the game is to keep your score to a minimum. Players with more than an agreed maximum, for example 200 points, drop out of the game, which is then continued until there is only one player left.

That card game is called 14.   ..

can you add a reference to the rules of the game 14?

In 14, the initial deal is 14 to each player with an extra card dealt to the player to the dealer's right. It differs in other ways as well. https://en.wikipedia.org/wiki/Quatorze


text:
Grandma's Patience
I've been given some cards numbered 1 to 16 (6 cards with each number). Apparently you can play a patience style game called Grandma's patience, trying to build up 6 columns of numbers, but I can't quite work out how to play it. Does anyone know this game or where I can find the rules?

Searching for "solitaire with six suits" turns up a game called "Klondike Six," played exactly like ordinary Klondike but with (wait for it) six suits.
https://thepolariscollective.wordpress.com/2014/12/27/hoyles-six-klondike-six/
The version described on that web page apparently uses a deck of 90 cards, with fifteen values in each suit. So, not quite what you've got, but pretty close.
Or, six suits of thirteen cards each makes "Jumbo Klondike."
https://politaire.com/help/jumboklondike
Of course, this assumes that the cards you have are suited (e.g. the 2 of clubs is distinguishable from the 2 of horseshoes, or whatever). The rules of Klondike also assume that within the 6 suits, 3 suits are black and 3 are red. What do your cards look like?
Also, obviously, if the cards were given to you in a box with writing on it, it would help to know what that writing said.

Thank you. The cards aren't suits, just numbers 1-16. I don't have the box. Maybe it can be played in a similar way to Klondike though. I'll have a look.


text:
Is there such a thing a 3 possible in Spades?
We were playing Spades.  My new girlfriend said she had 2 and 3 possible.  I laughed and asked what she meant by the 3 possible.  We stopped the game to figure this out.

Spades: 4,6
Hearts: K,2,3,5
Clubs: K,7
Diamonds: K,9,8

How she should have bid?
In the end she only took 2. After knowing that, how she should have bid?
is a 3 a legit bid with this hand?

The possible trick winners of this hand are the three Kings and the short ♣'s which allows a 3rd trick ruff/cut.
Having only two spades is a weakness. It means that there is a higher probability an opponent has a long ♠ suit.
The game of Spades encourage conservative bids since an overtrick (bag) results in a smaller loss compared to an undertrick.
I would bid 2 with this hand almost always.  If I am at the 4th sit and the sum of bidding is 7 before my bid, I would bid 3. (probably when someone bids nil)
To answer the question of the headline, in the bidding phase each player must say a single number between 0 and 13, nothing more.  She can't say: "2 and maybe 3". Of course that in a game among friends she can say whatever.

There are a lot of "in between" bids where I come from!

@TheChaz2.0, What do you mean? can you bid: 2.5?  how the scoring goes if you get the lower/higher bid?

Table talk, not actual bids.