text:
What are the odds of getting the following hands in euchre?
I really like euchre, but I stink at statistics.  Last time we were playing euchre, people would get some crazy hands, and we kept saying, "What are the odds?"
So now I ask you all, what are the odds?
Scenario 1:
What are the odds that you get dealt both Jacks of the same color, plus three other cards all of the same suit in the same color?  (i.e. Jack of Hearts, Jack of Diamonds, plus Ace, Queen and Ten of Diamonds).
Scenario 2:
What are the odds you get dealt four of the cards asked in Scenario 1, and the fifth card you need is the face-up card in the pile of 4 remaining cards?
Scenario 3:
What are the odds you get dealt all 4 Jacks?

Surprisingly, the odds for Scenarios 1 and 3 are the same! The chances for each:

Scenario 1: 0.0047%
Scenario 2: 0.0535%
Scenario 3: 0.0047%

Here are the calculations:
Note: The order you consider cards in these calculations doesn't matter, so I'll use the most convenient ordering possible.
Scenario 1:
First, there are 24 cards in a Euchre deck. To calculate the number of possible Euchre hands of five, you multiply the number of possibilities for each card: 24 × 23 × 22 × 21 × 20 = 5,100,480.
Then we need to figure out how many unique ways you can get Scenario 1. We'll calculate the draw this way: (I) number of Jacks (4), (II) number of Jacks matching the color of the first Jack (1), (III) number of remaining cards (A, K, Q, 10, 9) matching the color of the Jack (10; 5 in each suit of that color), (IV) number of remaining cards matching the suit of card III (4), (V) number of remaining cards matching the suit of cards II and III (3). This calculates as 4 × 1 × 10 × 4 × 3 = 480.
So the chance of getting Scenario 1 is 480 / 5,100,480 = 1 / 10,626.
Scenario 2:
This one is more complicated.
Number of ways to get a mismatch by missing a Jack: (I) number of Jacks (4), (II) number of remaining cards (A, K, Q, 10, 9) matching the color of the Jack (10; 5 in each suit of that color), (III) number of remaining cards matching the suit of card II (4), (IV) number of remaining cards matching the suit of cards II and III (3), (V) number of cards that don't complete the set (19). This calculates as 4 × 10 × 4 × 3 × 19 = 9,120.
Number of ways to get a mismatch by missing another card: (I) number of Jacks (4), (II) number of Jacks matching the color of the first Jack (1), (III) number of remaining cards (A, K, Q, 10, 9) matching the color of the Jack (10; 5 in each suit of that color), (IV) number of remaining cards matching the suit of card III (4), (V) number of cards that don't complete the set (17). This calculates as 4 × 10 × 4 × 3 × 17 = 8,160.
So, the number of ways to end up with four of the five cards in Scenario 1 is (9,120 + 8,160) / 5,100,480 = 6 / 1,771.
Then the number of cards that would match (3) out of the number of possibilities for the face-up card (19) is 3 / 19
So the chance of both occurring (and thus Scenario 2) is (6 / 1,771) × (3 / 19) = 18 / 33,649.
Scenario 3:
Number of ways of getting all four Jacks: (I) number of Jacks (4), (II) number of remaining Jacks (3), (III) number of remaining Jacks (2), (IV) number of remaining Jacks (1), (V) number of remaining cards (20). This calculates as 4 × 3 × 2 × 1 × 20 = 480.
So the chances of getting Scenario 3 is 480 / 5,100,480 = 1 / 10,626.

Hi Firefeather. I've been reading through some of the older posts recently, and I don't think the calculations above are quite right.  I think you're right that you can ignore ordering, but in that case the total number of hands isn't 2423222120, it's that divided by 54321 (or 24C5).  Also I didn't follow the logic in scenario 2 when it came to looking at the four remaining cards.

For scenario 1 you need to multiply by 2 to account for suit choices. If you're dealt both red jacks, it doesn't matter if the rest of your cards are Hearts or Diamonds, but in your calculation you require them to be the suit of the first jack.

And @tttppp is correct, 2423222120 is the number of ordered Euchre hands, so your denominators all must be changed.

This answer is so very, very wrong, getting virtually every single sub-calculation incorrect. It should NOT be the accepted answer. @LittleBobbyTables: If you return to the site, you could unaccept this answer and accept a correct one, as a small step towards rectifying your grievances on your sign-off statement.

Solutions (TLDR)
The probabilities of a single player getting one of the above hands, from a single deal, are:

Scenario 1: 0.09%
Scenario 2: 1.54%
Scenario 3: 0.05%

Below I have calculated the probabilities and also simulated them to check the calculations.
I have also simulated the scenarios to find the probabilities of any player getting one of the above hands, from a single deal.  I think this more closely matches the observations described in the question.  These probabilities are:

Scenario 1 for any player: 0.38%
Scenario 2 for any player: 5.98%
Scenario 3 for any player: 0.19%

Introduction
A standard euchre deck contains the 9, 10, J, Q, K and A in each of the four suits, making a total of 24 cards.  Each of the four players are dealt a hand of five cards, and there are four cards left over.
We will split the deck, D, into the four Jacks J, and the other twenty non-Jack cards N.  We refer to the player's hand as H, and the table cards as T.
We use the notation nCk to refer to the Combinations function; the number of ways of choosing k elements from a set of size n.
We refer to the probability of an event using the notation P(Event).  We will refer to the conditional probability of event 2 happening, given that event 1 happens, with the notation P(Event 2 | Event 1).  In several places we do not explicitly state that an event is conditional on another event, but hopefully the context makes it obvious when this is the case.
Scenario 1: What are the odds that you get dealt both Jacks of the same color, plus three other cards all of the same suit in the same color? (i.e. Jack of Hearts, Jack of Diamonds, plus Ace, Queen and Ten of Diamonds).
P(Scenario1) = P(H contains exactly 2 Jacks) * P(The Jacks are the same colour)
               * P(3 other cards are same suit) * P(The suit is the same colour as the Jacks)
P(H contains exactly 2 Jacks) = (Ways to pick 2 from J) * (Ways to pick 3 from N)
                                / (Ways to pick 5 from D)
                              = (4C2)(20C3) / (24C5)
                              = 6*1140 / 42504
P(The Jacks are the same colour) = 1/3
P(3 other cards are same suit) = P(The second card is the same suit as the first)
                                 * P(The third is the same suit as the other two)
                               = 4/19 * 3/18
P(The suit is the same colour as the Jacks) = 1/2
∴ P(Scenario1) = 6*1140 / 42504 * 1/3 * 4/19 * 3/18 * 1/2
               = 5/5313
               ~= 0.09%

Scenario 3: What are the odds you get dealt all 4 Jacks?
This calculation is slightly more straightforward than for Scenario 1, as we no longer need to worry about suits.
P(Scenario3) = P(H contains exactly 4 Jacks)
             = (4C4)(20C1)/(24C5)
             = 1*20 / 42504
             = 5/10626
             ~= 0.05%

Scenario 3 is exactly half as likely as scenario 1.
Scenario 2: What are the odds you get dealt four of the cards asked in Scenario 1, and the fifth card you need is the face-up card in the pile of 4 remaining cards?
This is harder to compute.  First we split the calculation down by distribution of H.  I've written d(H)=(3,2) to mean that H contains 3 Jacks and 2 non-Jacks. Note that it is only possible to get Scenario 2 if H contains 1,2 or 3 Jacks.  We refer to "Scenario 2" as S2 for brevity.
P(S2) = P(d(H)=(3,2))P(S2 | d(H)=(3,2)) + P(d(H)=(2,3))P(S2 | d(H)=(2,3))
        + P(d(H)=(1,4))P(S2 | d(H)=(1,4))
P(d(H)=(j,n)) = (4Cj)(20Cn) / (24C5)

Scenario 2, Part 1: Three Jacks in the hand
Consider the case where H contains 3 Jacks and 2 non-Jacks.
P(S2 | d(H)=(3,2)) = P(2 of the Jacks are same colour) * P(2 non-Jacks are same suit)
                     * P(Suit of non-Jacks is same colour as 2 Jacks)
                     * P(T contains at least one non-Jack of same suit)
P(2 of the Jacks are same colour) = 1
P(2 non-Jacks are same suit) = 4/19
P(Suit of non-Jacks is same colour as 2 Jacks) = 1/2
P(T contains at least one non-Jack of same suit) = 1 - P(T contains 0 non-Jacks of same suit)
                                                 = 1 - 16/19 * 15/18 * 14/17 * 13/16
                                                 = 514/969
∴ P(S2 | d(H)=(3,2)) = 1 * 4/19 * 1/2 * 514/969 = 1028/18411

Scenario 2, Part 2: Two Jacks in the hand
Next consider the case where H contains 2 Jacks and 3 non-Jacks.  We treat the cases where the Jacks are the same colour separately from the cases where they are different.
P(S2 | d(H)=(2,3)) = P(Jacks are same colour)P(S2 | d(H)=(2,3) and Jacks are same colour)
                     + P(Jacks not same colour)P(S2 | d(H)=(2,3) and Jacks not same colour)
                   = 1/3 P(S2 | d(H)=(2,3) and Jacks are same colour)
                     + 2/3 P(S2 | d(H)=(2,3) and Jacks not same colour)

When the Jacks are the same colour, we need exactly two of the three non-Jacks to share a suit, and that suit to be the same colour as the Jacks.
P(S2 | d(H)=(2,3) and Jacks are same colour) = P(Exactly 2 non-Jacks share a suit)
  * P(Suit is same colour as Jacks) * P(T contains at least one non-Jack of same suit)
  = (3*(4*15)/(18*19)) * (1/2) * (514/969)
  = 2570/18411

When the Jacks are different suits, we need all three non-Jacks to share a suit.
P(S2 | d(H)=(2,3) and Jacks not same colour)
 = P(All 3 non-Jacks share a suit) * P(T contains other Jack of same colour)
 = (4*3)/(19*18) * 4/19
 = 8/1083

Putting these two cases together we get:
P(S2 | d(H)=(2,3)) = 1/3 * 2570/18411 + 2/3 * 8/1083 = 2842/55233

Scenario 2, Part 3: One Jack in the hand
The final distribution of H to consider is 1 Jack and 4 non-Jacks.
P(S2 | d(H)=(1,4)) = P(At least 3 non-Jacks share a suit) * P(The Jack is the same colour)
                     * P(T contains other Jack of same colour)
P(At least 3 non-Jacks share a suit) = (4*4*3*15 + 4*3*2) / (19*18*17) = 124/969
P(The Jack is the same colour) = 1/2
P(T contains other Jack of same colour) = 4/19
∴ P(S2 | d(H)=(1,4)) = 124/969 * 1/2 * 4/19 = 248/18411

Scenario 2: Conclusion
So putting all of that together.
P(S2) = (4C3)(20C2)/(24C5) * 1028/18411 + (4C2)(20C3)/(24C5) * 2842/55233
        + (4C1)(20C4)/(24C5) * 248/18411
      = 4*190/42504 * 1028/18411 + 6*1140/42504 * 2842/55233 + 4*4845/42504 * 248/18411
      = 4670/302841
      ~= 1.54%

Simulation
To check my calculations I have written a Python script to evaluate ten million deals, to see how frequently each of the scenarios occurs.  The script is here.
I also used it to compute the probability of any of the four players satisfying each of the three scenarios for a given deal.  I thought this would be interesting, as this is what was observed in the question, although not actually the probability being asked for.
Probability for a single player   | Probability across all four players
Scenario1 | Scenario2 | Scenario3 | Scenario1 | Scenario2 | Scenario3
 0.0972%  |  1.5319%  |  0.05  %  |  0.3703%  |  5.9367%  |  0.1748%
 0.0927%  |  1.519 %  |  0.0442%  |  0.3716%  |  5.9698%  |  0.1914%
 0.0911%  |  1.5607%  |  0.0466%  |  0.3896%  |  5.9523%  |  0.187 %
 0.096 %  |  1.5207%  |  0.0442%  |  0.3753%  |  5.9583%  |  0.1853%
 0.0946%  |  1.5594%  |  0.0457%  |  0.3711%  |  5.9709%  |  0.1844%
 0.1004%  |  1.561 %  |  0.0466%  |  0.3824%  |  6.0065%  |  0.1916%
 0.0964%  |  1.5367%  |  0.0476%  |  0.3776%  |  5.9651%  |  0.1882%
 0.098 %  |  1.563 %  |  0.0452%  |  0.3775%  |  6.0131%  |  0.188 %
 0.0942%  |  1.5385%  |  0.0469%  |  0.3849%  |  6.0205%  |  0.1852%
 0.0983%  |  1.535 %  |  0.0418%  |  0.3641%  |  5.9913%  |  0.1931%
-----------------------------------------------------------------------
 0.0959%  |  1.5426%  |  0.0459%  |  0.3764%  |  5.9785%  |  0.1869%

Happily the simulation backs up the calculations.

Very nice and clear. Since you define all your other notation so well, you might want to mention that P(X) is the probability of X.

@shujaa Thanks!  I've made an edit to include that now.

@tttppp This is excellent work. Is the discrepancy between your Scenario2 simulation and calculation due to the fact that you checked for success in the simulation without subtracting the success of S1 and S3? ie. S1&S3 fulfills requirements of S2. It looks like the difference between sim and calculation is about the sum of S1+S3.

@ghoppe Thanks! I don't think that's the issue, but it's an interesting anomaly. The python script definitely excludes scenario 1 from scenario 2. My guess is that my arithmetic is wrong somewhere in the calculation, but I haven't spotted where yet!

I arrived at the same answers as tttppp, but I used more combinations and counting, so maybe it might help you understand the intuition behind it a little more. To reiterate, a combination C(n, x) tells you how many ways you can arrange n objects from a set (or group) of x objects, where the order doesn't matter. So if you have elements a, b, c, a combination treats the ordering (a, b, c) and (c, b, a) the same (it won't double count it).
Scenario 1:
What are the odds that you get dealt both Jacks of the same color, plus three other cards all of the same suit in the same color? (i.e. Jack of Hearts, Jack of Diamonds, plus Ace, Queen and Ten of Diamonds).
The total number of possible Euchre hands is C(24, 5) = 42504. There are 24 possible Euchre cards, and 5 in a hand. Order does not matter, so we use a combination.
We must now find the total possible ways of getting two jacks of the same color, and then 3 cards of the same color as the first jack. There are 2 possible color combos (Jack of Hearts and Jack of Diamonds, and Jack of Spades and Jack of Clubs). Order doesn't matter for either of them. For each of the two combos, there are two choices of colors for the remaining cards. (Jack Hearts/Jack Diamonds with 3 Diamonds; Jack Hearts/Jack Diamonds with 3 Hearts etc.). There are then 5 possible cards of each color to choose 3 from (6 of each color, and we already chose a Jack), so we have:
C(2, 1) * C(2, 1) * C(5, 3) = 2 * 2 * 10 = 40. There are 40 ways we could get two Jacks of the same color, and 3 other cards that are the same suit as one of the Jacks, when order doesn't matter.
Probability of scenario 1 = 40 / 42504 = 5 / 5313 ~= 0.09%
Scenario 3:
What are the odds you get dealt all 4 Jacks?
To answer this, we need to answer the question of how many hands are there that have 4 jacks? There are 4 jacks, and you need to choose 4 of them, so that is C(4, 4) = 1. There is only one way to have 4 jacks in a hand when order doesn't matter. There are 20 cards left, and you need to choose 1, so it's C(20, 4) = 20. There are 20 ways to have a non jack in a hand of 4 jacks. So there are 20 ways to form a hand of 4 jacks and one other card.
So this is 20 / 42504 = 5 / 10626 ~= 0.047%
I don't really know how to solve Scenario 2, so I would defer to previous responses.

You shifted the decimal too many times 5 / 10626 = 0.0004705 = 0.04705%


text:
Determining which areas to use in 2- or 6-player Power Grid
Pretty straight-forward question here: In the 2- and 6-player versions of Power Grid, you are instructed to use only 3 areas and 5 areas respectively, but it doesn't say how to pick the areas, leaving it up to the players.
What are good recommendations for determining which areas are in/out?

We use the system where player 1 picks an area to keep, player 2 picks an adjacent area, etc. until we have the right number of areas. I'm nearly positive this is how the designer, Friedemann Friese, taught us. If not then we picked it up from the earlier version of the game, the crayon-based Funkenschlag.
One tip: when the new (current) edition came out I had a chance to play with the designer explaining the rules. We used the pick-an-adjacent-area rule to pick our staring zones and were using the Germany map. With six players we picked five zones, and the one left out was the center zone, leaving a ring-shaped map. Friedemann immediately exclaimed, "We never playtested it that way!"
It turns out that's a pretty poor map to play on since you're restricted to expanding through other people's territory much earlier in the game than you normally would. The result was an extra-long game that frankly wasn't as fun.
So the tip: if you play the Germany map with 6 players, do not eliminate the center territory.

great answer, and great anecdotes!  Were you one of the original playtesters?

Unfortunately I was not. My first play was shortly after it was introduced at the Nuremberg International Toy Fair. I have had the opportunity to playtest some of Friedemann's games since then, though. Great fun to see how his brain works!

Generally, we decide as a group whether we want to use the more-dense or the less-dense areas of the map (at least on the U.S. map).  It's all about whether we feel like making everybody pay those high connection prices in the west.

I don't think it really matters so much.  But you could randomize it, or else agree to rotate through each of the possibilities on a game-by-game basis, to create as much variety as possible.
Many connoisseurs seem to think that Power Grid is much better for 3-5 players, mind you, so if the lack of hard-and-fast rules for how to set up the 2 or 6 player games is bugging you, it might not be a bad idea to try not to play with that many players, so far as possible!

That's funny, it's one of the games we always pull out when we have six players.  Guess I'll have to play it more with less to see if it's better that way.

@Lance Roberts -- agreed!  As soon as our gaming group hits six players, there's a good chance Power Grid comes out.

There are so many games that only take up to 5 players, that when your group is 6 large, you're delighted to find anything that will accommodate everyone!  But just because a game can take 6, doesn't mean it will be best with 6, alas...

the lack of hard-and-fast rules doesn't bug me, I was just curious what peoples' answers were.

We just let the guys who go first pick the areas one at a time.

I have always done it either a) randomly or b) let the newest player pick, which often leads to the same results as randomly.  :)  I have never had a case of only 2 players, but with 6 that seems to work pretty well.

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text:
Missions in Risk -- is Australia really an island?
For some reason, I agreed to play Risk with my wife and her friend the other night, and we decided to try the Mission cards in Risk because no one wanted to go for global domination.  I would soon be reminded why I don't play this game.
One of the mission cards states "Control two islands".  While there is nothing in the rules that states what exactly is an "island", to me, islands would be any of the following countries:

Great Britain
Iceland
Greenland
Indonesia
New Guinea
Madagascar
Japan

However, our friend also claimed Australia was an island, albeit a "very big island".  Pulling up the Wiki Answers article as well as the Wikipedia article state that Australia is a technically a continent and not an island, partly due to being on its own tectonic plate.  The Wikipedia article does mention that Australia has been dubbed the "island continent", which was enough to give our lawyer friend something to hold on to.
Barring common sense, are there any rulings as to which countries count for the mission card?

Worse: Arguing that GB should count as two islands.

Australia is not an island.
The rules for the 2003 edition define an island as follows:

A mission may require you to control an island. An island is a territory completely surrounded by water and only connected to other territories by sea-lines, for example Indonesia.

See page 18 of the rules here: http://www.hasbro.com/common/instruct/Risk_2003.pdf

that was the conclusion I had come to, but it's good to see it in print.  Once again, teeny-tiny italicized blue font is my nemesis.

For those curious, Australia split into Eastern and Western Australia. Western Australia has always wanted to do that

Within this context, I would define an island as a territory which is completely surrounded by water. By this definition Australia would not be an island, since in the game it's divided into two territories. That said, from a mechanical standpoint there's no difference between an island and a landlocked territory.

Consulting Wikipedia, I see that the Secret Missions have been changed.
Unfortunately, I cannot find a full list of the newer mission cards.
Looking at the old cards, there are several of the form:

capture Europe, Australia and one other continent
capture Europe, South America and one other continent
capture North America and Africa
capture Asia and South America
capture North America and Australia

As long as your set of mission cards contain missions worded similarly then I think most reasonable people would grant you that Australia is a continent and therefore not an island.

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text:
What are the rules for who wins when the gold is reached in Saboteur 2?
In Saboteur 2, I am having trouble understanding precisely which dwarves win in certain situations. Could someone clarify what's already in the rulebook?
In particular, I was wondering: if a dwarf from the green team reaches the gold, does the blue team also win (and vice-versa)?
The way that made the most sense in my head was: if any dwarf of ANY kind reaches the gold, then winners include: the boss, the profiteer, and dwarves on teams for which a path to the gold is NOT blocked by a door. However, the rules seemed to say that WHICH dwarf reached the gold actually mattered, which doesn't really seem sensible.

I agree the the rules are poorly phrased in the "A team wins if:" section.  It provides an outcome tree, but clearly fails to provide a branch for what happens if a dwarf from the other team creates a connection to the treasure, and the way is not blocked for that team by a door of the wrong colour.
Later in the rules, under "Splitting The Treasure" it states:

Depending on whether there is an uninterrupted path from the start to
  the treasure or not, the Gold-Diggers or the Saboteurs win the round.
  If the path to the treasure is blocked by a door, only the dwarves
  from the team of the same color win.

I think from that paragraph it becomes pretty clear that an uninterrupted path unblocked by any door would be a victory for all Gold-Diggers.  I think your initial instincts were right and you shouldn't worry too much about the sloppily-phrased earlier paragraph.  These things do happen, much to the distress of OCD rules lawyers like us!

FWIW, while this answer could be right, I don't really find the reasoning presented in it particularly convincing. I've asked a separate question about whether there are any errata or other official sources that would explicitly state which interpretation of the rules is the intended one.

The green team wins because they reached the end, played the last connection. The blue team does not win in your scenario.
Though I still don't understand how many gold to distribute myself. I do know the above.

This seems to have been responding to the other answer, but answers need to address the question itself - we're a Q&A site, and we don't work like a forum or do post-to-post discussion. See our [tour] for more information. I recommend if you want to back this up with a citation (e.g. the description you mentioned), you quote what you're citing.

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text:
Do geologists ever count as "winners" in Saboteur 2?
In Saboteur 2, the number of dwarves who win a round matters quite a bit: the more dwarves who win, the fewer points are awarded to each.
Geologists play by funny rules, though, and get points in another way. My question is: do geologists count toward the number of "winners" for a round, thus reducing all the other winners' points? Or are they left out of the winner count?

No.  The geologist does not count as a winner.
As noted in the rules, they are indifferent as to who wins and loses, caring only about the crystals.


text:
Do you get points if you are broken/trapped in Saboteur 2
In Saboteur, if the good dwarves reach the gold, they distribute points to themselves but skip over any good dwarves with broken equipment.
Does this rule still apply for point distribution in Saboteur 2, or does your broken status not matter for collecting points after a round?

It does not. Note that this rule is optional in the original rule set.
Answered by the Designer, Frederic Moyersoen on BGG:

The broken tools variant was introduced to have some competition between the gold-diggers.
The extension provides more chaos and competition. Therefore, this variant rule has been skipped during the design of the game. However, if you wish, you can still use it.


text:
Tactic cards in combat in Warhammer: Invasion
Look at the following combat scenario:

Both players had counted their forces strength and assigned the damage to opponents forces (assigned but not yet applied)
Now according to the rules players can play Tactics cards (as in between every other combat phases).

The question is: can one player play now the tactics card that will give him extra strength? The damage has been assigned already and having this ability would move the game back to previous phase (damage assignment) but the rule book doesn't clarify this subject.

The rules don't state anywhere that you can roll back to a previous phase, and the core rules and FAQ don't explicitly cover this scenario.
I would say that while you can play a card that gives him extra strength, there will be no net effect since damage has already been applied, and once damage is applied, that's it.


text:
In Mahjong , what is the chance to get a hand that allows a player to finish on their first turn?
Made hands are very rare in mahjong. Is there a table that lists the possibilities of starting with a complete hand?

@Closers: Please let a comment. I will glad to follow your votes if I know and agree your arguments.

@Bigown - The question isn't clear what a "made hand" is.  I'm also concerned about a flood like we had for the poker questions before... these seem like shotgun questions

@bigown I didn't vote to close, but I think this is the concern; I have posted some further discussion as well: http://meta.boardgames.stackexchange.com/questions/212/multiple-questions-same-author-similar-topics

All SE sites have to deal with vague or ambiguous questions.  You ask the poster to clarify. If they do, great, if not, answer as best you can.  I'm not sure 'close all vague' questions is the right response.

I didn't know that the term "made hand" was unclear, I'm not a native speaker and know (and misused, on second thought) the term from poker. This question is asking for the chances to get a hand that allows a player to finish on their first turn.

I did my best to define a "made hand" in my answer below.

I wrote a program to simulate mahjong hand dealings.
It's not verified yet, but so far, the ratio appears to be around 1 in 100,000 games. (The output was the chance that the case happened to at least 1 out of 4 players)
I had the fixed program run for a bit (1.7 billion draws):
3004 mahjong in 1057000000 random draws: 0.00028%, 1 in 351864
4392 mahjong in 1526000000 random draws: 0.00029%, 1 in 347449
4930 mahjong in 1695000000 random draws: 0.00029%, 1 in 343813

At this point, the state space of the random number generator is likely rather close to exhausted, so further research requires a different implementation (this is easy, I just did not bother yet).
If the math of @Jan is correct, it would be pretty easy to replace the RNG with a permutator and just perform a whole-space search to calculate a precise result. The above program calculates around 2m hands per second on my box, so it would take 40-ish hours.

Update: I implemented a proper RNG, optimized the program slightly and let it search through 25 billion random draws:
 737s: 15433 mahjong in  5000m random draws: 0.00031  %, 1 in 323981. 6.78m checks/s
1557s: 31080 mahjong in 10140m random draws: 0.00031  %, 1 in 326254. 6.51m checks/s
2281s: 46008 mahjong in 15000m random draws: 0.0003067%, 1 in 326030. 6.58m checks/s
3116s: 60971 mahjong in 20000m random draws: 0.0003049%, 1 in 328024. 6.42m checks/s
4020s: 75921 mahjong in 25000m random draws: 0.0003037%, 1 in 329289. 6.22m checks/s

The last line is the current state (previous lines for picturing variance).

Not sure I know what a 'made hand' is.  Are you asking what are the odds of being dealt the Nine Gates?  If yes, the odds are about 5000:1 in a 136 tile set. This page has a combinatorial treatment of the 13 orphans vs the Nine Gates.

Just to clarify: The 5000:1 ratio described in your interesting link is not the chance to get Nine Gates but the ration between Nine Gates and 13 orphans. Both of them are, as far as I can tell, a lot rarer than 0.02%. I heard that even regular players often get them only once in their life.

I wrote a program that draws a random 14-tile set from a reservoir of all possibles tiles (expect flowers and seasons, sorry):
then the program checks if the hand is a mahjong. So far here are my results:

This means that for 71 millions random hands, I got 151 mahjong hands. We can say it has converged (almost).
Conclusion: You will get a "tinwu", "tenhou", "heavenly hand" aka a wining hand by chance, once every ~471 000 draws, on average.
Here are example of drawn mahjong hands:
r: dots
d: dragons (b: white, v: green, r: red)
v: winds
b: bamboos
c: characters
hand number 433035
mahjong!
['b1', 'b1', 'b3', 'b4', 'b5', 'c6', 'c7', 'c8', 'r5', 'r5', 'r5', 'r7', 'r8', 'r9']
hand number 443206
mahjong!
['b1', 'b2', 'b3', 'c3', 'c4', 'c5', 'r2', 'r3', 'r4', 'r6', 'r7', 'r8', 'v4', 'v4']
hand number 833989
mahjong!
['b2', 'b2', 'b6', 'b7', 'b8', 'r1', 'r2', 'r3', 'r4', 'r5', 'r6', 'r7', 'r8', 'r9']
hand number 2194755
mahjong!
['b1', 'b2', 'b3', 'b4', 'b5', 'b6', 'c6', 'c7', 'c8', 'r1', 'r1', 'r7', 'r8', 'r9']
hand number 2338547
mahjong!
['b6', 'b7', 'b8', 'c7', 'c8', 'c9', 'dr', 'dr', 'r1', 'r2', 'r3', 'r6', 'r7', 'r8']
hand number 3803349
mahjong!
['b2', 'b3', 'b4', 'b9', 'b9', 'c6', 'c7', 'c8', 'r3', 'r4', 'r5', 'r8', 'r8', 'r8']
EDIT: My simulation continues to converge. I have now drawn 2186 mahjong hands from 966 millions random draws. I did fit a distribution on my dataset. I got an exponential distribution of parameter mu= 442501 +/- 9466.47 (that a 2% error) According to wikipedia, the exponential distribution is the probability distribution of the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate. This is coherent I think, at least qualitatively.
It could be awesome if someone could deduce this theoretically (that's too difficult for me). Then we could write a paper!

Could you have your program create a (large) list of examples of mahjong and non-mahjong draws, so I can compare with my program? It seems they converge to different values.

Good idea but I've try to generate such a file. It quickly gets really big and I have a limited data plan :( How large is enough?
I use the algorithm proposed by Victor Nicollet here.
I can send you my code (in python) but a lot of comments and variable are in French and it's a bit messy.
I suspect my program fails to see some Mahjong if it is wrong.

Simple check: did your program include the 7pairs? It is a valid hand, but not in the form of 3*4+2. My program did not include the 7 pairs (nor the 13 orphans but that's negligible). It now does. I don't think it will help the probability to go to ~1/350000 but I can check.
Other simple check: my program seems to be more strict than yours. So I suspect it does not count some mahjongs as valid (excluded the 7 pairs). Why don't you send me your list of valid hands so I can check if my program agrees? If you can format them as in my comment, it will be easier for me.

Mine included 7 pairs and 13 orphans. Sure, I'll prepare a list and upload it.

Here's a list of 500 valid hands: https://gist.github.com/nerai/7f524d2b0a08f57a89534bde3eb21c62 or
https://gist.github.com/nerai/8a7ab081c7687acce6b93c2fee01fe32 (your format)

For reference, here's a list of 100k invalid hands: https://gist.github.com/nerai/6f8de4b4205dbfc64b1b4556ceddbf03

Thank you for the effort of formatting!
Just to be sure:
'r'=G=green dragon
'v7'=R=red dragon
'v'=B=white dragon
'b'=N=north wind
This is correct?

I've found those hands which my program finds invalid:
['c5', 'c6', 'c7', 'c7', 'c7', 'b4', 'b4', 'b5', 'b5', 'b6', 'b6', 'v2', 'v2', 'v2']=56777m445566sSSS
['c2', 'c3', 'c4', 'c5', 'c6', 'c6', 'c7', 'c7', 'c8', 'b2', 'b3', 'b4', 'v1', 'v1']=234566778m234sEE
['c3', 'c4', 'c4', 'c5', 'c5', 'c6', 'r2', 'r3', 'r4', 'b1', 'b1', 'b2', 'b3', 'b4']=344556m 234p11234s
['c3', 'c3', 'r6', 'r7', 'r8', 'b5', 'b6', 'b6', 'b7', 'b7', 'b8', 'b9', 'b9', 'b9']=33m 678p 566778999s
Are those valid in Japanese MJ? My program checks for HK Mahjong. I assumed the valid hands were the same. Apparently not.

Sorry, I got the translation of the honors completely wrong. Your mapping is correct. I should have checked this more carefully

The hands you showed seem to be valid to me. The first one is 567+77+456+456+SSS, the second is 234+456+567+234+EE, the third is 345+456+234+11+234, the fourth is 33+678+567+678+999. I should add that I'm not knowledgeable about any of the rulesets, I'm not an active player.

You are absolutely right they are correct. I and my program got confused because of the ordering of the tiles... They all have in common that they have a double chi=chow... That's my mistake! I guess, that's why my program is too strict. I'll try to fix that.

The https://tenhou.net/sc/prof.html page contains statistics on yaku. As I understand it, Tenhou (天和) is found in 0.00046% of the hands that ended in someone victory, and Chiihou (地和) in 0.00091%. Thus, one case is obtained for approximately 100,000 or less.

Interestingly, Chiihou is currently listed as 0.00015. If my Japanese is on point, these statistics were drawn from a pool of 650k games (which is to say, relatively few).

Note: I am basing this answer on Japanese Mahjong rules; Chinese rules will probably lead to a different outcome.
Tenhou, Chihou or Renhou hands must be valid hands, i.e. they must satisfy the requirement of having four groups of three and one pair; except for Chiitoi (seven pairs) and Kokushi Musou (13 orphans) which we must treat separately.
The four groups of three can technically be kan, pon or chi. For a kan-containing hand to qualify, this would imply Rinshan Kaihou which complicates the picture a little. However, combinatorically it does not matter whether we draw the winning tile first and the kan-completing tile second or vice-versa (there is no confounding discard) so even if Saki is playing the outcome should be the same and the discussion can be limited to pons instead of including kans.
This leads us to the following cases:
Kokushi Musou: there are 13 possible hands
Chiitoi: the pairs can be taken from 1–9 in three colours, four winds and three dragons (34 tile faces). As most rules specify that no pair can be repeated, this gives us 34C7 = 5,379,616 possibilities
Four chi: there are 21 possible starting tiles for the chis (1 through 7).

If each starts on a different tile that gives 21C4 = 5985 possibilities.
If two chi are identical (Iipeikou), that gives us an additional 21C3 × 3 = 3990 possibilities. It is not possible to have exactly three chi identical as that would be a three pon + chi hand. Further, it is not possible to have all four identical as that would be a special case of a three pon + chi hand.
However, there are an additional 21C2 = 210 hands of Ryanpeikou (two pairs of identical chis that are distinct from each other).
The final pair makes it slightly more complicated. Naively, one might assume that there are 34 possible pairs but the real number may be smaller for a given hand due to chi using three or four of a number tile.

If all chis start on different tiles: a given number 4 through 6 will be the only one used three times in exactly 16 hands; 3 and 7 add another 17 hands each for a total of 82 per suit. In addition, there are 6 hands per suit in which two consecutive numbers are used three times (e.g. 1-2-2-3-3-3-4-4-4-5-5-6 composed of the chi 1-2-3, 2-3-4, 3-4-5, 4-5-6)
If the hand has Iipeikou, each number 3 through 7 will be unavailable in exactly 10 × 16 = 160 hands per suit. Two consecutive numbers are unavailable for pairs in 76 (2/3 and 7/8), 72 (3/4 and 6/7) and 68 (4/5 and 5/6) hands per suit or 216 overall. There are 3 possible combinations per suit which render two non-consecutive numbers unavailable (e.g. 1-2-3-3-3-4-4-5-5-5-6-7), meaning there are 219 hands per suit with two pairs unavailable. Finally, three consecutive numbers can be unavailable; for 3/4/5 through 5/6/7, three hands each, for 2/3/4 and 6/7/8 two hands each.
If the hand has Ryanpeikou, each number 3 through 7 is not available for pairing in exactly 1 hand per suit and two consecutive numbers beginning on 2 through 7 are not available in exactly 1 hand per suit.
(It is not possible to remove the 1’s or 9’s from the list of possible pairs using only chis)

therefore, including the pair there are:
3 × 6 × 32 + 3 × 5 × 33 + (210 – 3 × 6 – 3 × 5) × 34 = 7089 Ryanpeikou hands
3 × 13 × 31 + 3 × 219 × 32 + 3 × 160 × 33 + (3990 – 3 × 13 – 3 × 219 – 3 × 160) × 34 = 133,749 Iipeikou hands
3 × 6 × 32 + 3 × 82 × 33 + (5985 – 3 × 6 – 3 × 82) × 34 = 203,208 hand that have neither
7089 + 133,749 + 203,208 = 344,046 all-chi hands.

One pon: again, assuming three different starting numbers there are 21C3 or 1330 ways to choose the chis plus an additional 21 × 20 = 420 Iipeikou options.

With Iipeikou, the numbers 3 through 7 can be unavailable for pairing in two cases per suit. Two consecutive numbers from 2/3 to 7/8 are also unavailable in two cases per suit. As only the choice of Iipeikou matters for which tiles can be pon, there are 3 × 10 × 31 = 930 hands where one can choose from 32 possible pairs and 3 × 12 × 31 = 1116 hands where one can choose from 31 possible pairs.
Without Iipeikou, the only way to make a digit unpairable is for the three chis to start on three consecutive digits; this gives 5 options per suit and leaves 31 choices for the pons.
The more difficult case is when the pons are restricted but not the pair.

If one wishes to remove exactly one possible pon, this restricts the possible placement of the third chi:
When removing 5, the third chi must have a different suit (14 options)
When removing 4 or 6, the third chi can be one of the same suit or a different suit (15 options)
When removing 3 or 5, there are two possibilities within that suit for the third chi (16 options)
Overall 76 options
If one wishes to remove two consecutive pons, the following restrictions are in place:
When removing 4/5 or 5/6, one option remains in that suit for the third chi (15 possibilities)
When removing 3/4 or 6/7, two remain within that suit (16 overall)
When removing 1/2 or 7/8, three remain (17 overall)
Overall 96 options
Two non-consecutive pons can be removed; there are three options to do this per suit and it requires all three chis
There are eight options per suit to remove three consecutive numbers from being ponable and this uses all three chis; e.g. 1-2-2-3-3-4-4-5-6 (individual chis: 1-2-3, 2-3-4, 4-5-6).

Therefore, there are:
930 × 32 + 1116 × 31 + (420 – 30 – 36) × 31 × 33 = 422,406 Iipeikou options
3 × 5 × 31 × 32 + 3 × 76 × 33 × 33 + 3 × 99 × 32 × 33 + 3 × 8 × 31 × 33 + (1330 – 3 × 5 – 3 × 76 – 3 × 99 – 3 × 8) × 34 × 33 = 1,460,808 options without Iipeikou
Altogether 422,406 + 1,460,808 = 1,883,214 one-pon options.

Two pons: it’s best to decide where to put the two chis first.

Assuming Iipeikou, there are 21 possible chis. Furthermore, there are 31 remaining tiles that can be the pons (pons and Iipeikou cannot overlap), so 31C2 × 21 = 9765 main hands. The pair is only restricted by the pons (Iipeikou will leave two of each number) so there are 32 possible pairs per hand giving 312,480 hands.
Assuming all independent chis, there are 21C2 = 210 ways to choose the chis. We now have to decide how we can choose the pons. As above, the pons will be the only thing restricting the pair so there will always be 32 possible pairs.

For each number 3 through 7, there is exactly one combination of chis that will make this and only this number unavailable for pons. This means 15 × 33C2 = 7920 main hands.
For each number 2 through 7, there is exactly one combination of chis that will make this and the consecutive number unavailable for pons. This means 18 × 32C2 = 8928 main hands.

Thus, for chis, pons and pair there are 7920 × 32 + 8928 × 32 + (210 – 15 – 18) × 34C2 × 32 = 3,716,640
Overall, that is 3,716,640 + 312,480 = 4,029,120 hands with two pons.

Sanankou: It is impossible for the pons to interfere with the chi, so it is safe to say that the main hand has 34C3 × 21 = 125,664 possibilities. The pair might seem to pose a problem but actually it does not as there is no way a single chi would remove any number from the pair pool in addition to the pons; thus there are 125,664 × 31 = 3,895,584 possible Sanankou hands.
Toitoi/Suuankou: finally something easy! There are 34C4 = 46,376 options for the four triplets and each leaves 30 possible pairs giving 1,391,280 Toitoi/Suuankou hands.
Assuming there is no error in my maths, there should thus be 16,922,873 hands that can win Tenhou, Chihou or Renhou.
To determine how likely it is to win, consider that there are 136 tiles in the game of which you choose 14. This is the same for Tenhou and Chihou and it doesn’t matter that your tile picks are non-consecutive. 136C14 = 4,250,305,029,168,220,000 = 4.250 × 1018 so there are 4 quintillion 250 quadrillion possible starting hands. A quick division of the two values shows that the probability of Tenhou or Chihou is around 3.982 × 10–12 or somewhere around 1 in two trillion.
The above chain of thought is incorrect. It assumes that all tiles are individual but there are four identical copies of each tile.
I was dreading having to calculate how many combinations with limited repetitions (drawing 14 from 34 distinct tiles but allowing each tile to be repeated a maximum of 4 times) because it seemed a daunting task but thankfully JMoravitz over at Maths.SE already did that work for me and showed me a shortcut. Using their method, there are in fact only:
47C14 – 42C9 × 34 – 37C4 × 34C2 = 326,446,402,010 or 326 billion possible hands.
Thus, the odds of getting Chihou or Tenhou are much better than my initial estimate and come in at around 5.18 × 10–5 or 0.005183967 % or a 1 in 20,000 chance.
As part of an old running joke from over at Chemistry.SE, I will add: or 5 m%.
Renhou is essentially like Chihou except you get one, two or three tries at the 14th tile depending on whether you are South, West or North wind. Thus, naively I want to assume that the probability of Renhou is somewhere between Chihou and 3 × Chihou but I would have to confirm that with a mathematician.

No, I don’t know why I did this; no, I did not really have the time; yes, I had to rewrite a couple of times because I noticed errors in my maths; no, I am not certain that my maths is error-free now; yes, this XKCD is relevant.

Thank you for this detailed explanation. Do you know why it contradicts the other answers?

@mafu I don’t, I suspect I have an error in my maths somewhere. However, it’s interesting to see that this question now has attracted to large-number approximations whose results differ by a factor of 4 and where my answer comes in in the middle …?

To make sure, do you account for the different probabilities of the various unique hand compositions? You've reduced the total number of unique hands to 326 billion, but some of them are more or less likely (winds)

@mafu Whether a hand contains x North Winds or x 7-sous is irrelevant for how likely the hand is because there are four North Wind tiles and four 7-sou tiles. But now you’ve got me unsure whether all hands are really equally probable; i.e. if it matters whether a hand has one, two, three or four of the same tile.

I've tested (not sure if this is what you mean) what happens if the probablity of drawing the same tile twice (or more) is the same as drawing it just once: the probability of a complete hand doubled (1 in ~170k instead of 1 in ~340k).

Here is yet another answer from this page. 以上により、天和の確率は12859078207674÷4250305029168216000 ＝約0.000003025 (約1/330530)となります。
I think it means that the probability to get a tenhou=tenwu is 1/330530. Unfortunately i don't follow precisely the derivation because it is written in Japanese.
So far the answers are:

1/350000 according to @mafu's program ==> Is the program counting everything?
1/19290 according to @Jan's maths ==> Is the maths correct?
1/442501 according to @Pierebean's program ==> Is the program counting everything?
1/217391 according to tenhou.net ==> Is N large enough? As Mafu says: 650k is not much.
1/330530 according to rascalhp ==> Can someone who speaks Japanese check the maths and compare it to Jan's ?

Come on people, we can crack this!

I don't remember the further results I got from the program I wrote back then. I'll dig it out and let it run again.

(edited my answer with the results)

Notably, the linked page has 4250305029168216000 as the divisor, which @Jan said to be not correct. Maybe the calculation of the dividend is different to account for this.

I’m off to a bad start as Rascalhp’s stats already have a different number of total possible dealer hands ^^'

(I should also add that I can’t be pinged if I am not yet in the comment thread so I didn’t get your ping above but I should get all subsequent pings in this comment thread. Also, posts don’t ping.

"Made" hands are those that satisfy certain requirements, specifically four "sets" of either triplets or "three in a row," plus a pair. There are also a handful of "special cases" of "made hands" that don't meet these requirements. If, after taking a tile, you don't have a "made hand" (in either regard), you need to discard a tile and try again.
The "special cases" are easily countable. The more common cases can be counted using a branch of math called "combinatorics" (the math of combinations). Increasingly powerful software programs can do this easily.
The older "authorities" probably lacked "combinatoric" capabilities, hence no tables from them. The answer (in the form of a table, or a descriptive function), could now well exist somewhere in the world. That may or may not include the western world, depending on how popular the topic of mahjong is among western problem solvers. But the answer appears attainable, and may soon be, if it hasn't been already.

The probability for the Chinese Classical Game has been fully calculated as follows:
4 pungs  0 chows plus pair     2137006080
3 pungs, 1 chow  plus pair    67441343634
2 pungs, 2 chows plus pair   774405495000
1 pung,  3 chows plus pair  3809325606336
0 pungs, 4 chows plus pair  6715895624442
The Chinese Classical Game includes the special '13 orphans' hand but does not accept 7 pairs as it is considered relatively easy compared to other limit hands.
13 Orphans  1308622848
7 pairs  1569298171584
Comparing these totals with the combinatorial 136 C 14 gives these results
Regular hands + 13 Orphans      =         1 in 373800 *
Regular hands + 13 Orphans + 7 pairs = 1 in 328467 *

There is a small element of possible double counting in that 3 identical chows can be rearranged to produce 3 separate pungs. There are 279,401,280 of these and the effect is approximately +8

Are you able to add a reference to where this work was done?

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.


text:
Infiltrating in new Caprica
Suppose you are playing a game with a Cylon leader. He is not infiltrating. The humans reach New Caprica. Is there now no mechanic by which a Cylon Leader can infiltrate? So the only way for a Cylon leader to infiltrate is for him to already have been infiltrating when we arrived?

That is correct; there is no way for a Cylon leader to start infiltrating while on New Caprica. (See the Unofficial FAQ at BGG)
If a Cylon leader needs the humans to win and wants to be able to help them out on New Caprica, s/he needs to be infiltrating when they get there. Otherwise, s/he won't be able to use the Human actions, which is the main way to help the humans.

Makes perfect in-universe sense: once arrived on New Caprica, the humans get to see many multiple copies of each model... kinda hard for a random Cylon to start infiltrating at that point.  A Cylon who wants humans to win should, you know, join the humans.  ;)


text:
What do the abbreviations BB, SB and D mean in Texas Hold 'Em?
http://www.flopturnriver.com/Holdem-Expected-Value-10-Players.php
Maybe D is dealer and B is blind, but what is BB?

BB = Big Blind
SB = Small Blind
D = Dealer


text:
Building a shipyard during canal phase
In brass, there are only a small number of shipyards available, but it seems in the canal phase this is reduced further. There is a specific rule that says a shipyard cannot be built at the places connected via a rail link, but i don't understand why.
If i have a city card, meaning i can build within that city, what is stopping me build a level 1 shipyard? It has no materials needed to be shipped, so why is this place any different to any other city on the map? Am i missing a specific rule somewhere?
This has been a rule that initially confused out group, until we played online and realised the correct usage (which the rules do backup).
My confusion is that the specific rule says something like

Hence the shipyard in the canal phase
  cannot be built at this location

And i don't understand the rational.

[a level 1 shipyard] has no materials needed to be shipped

I believe you've got that part wrong: A shipyard does require coal (look at the brown square at the bottom-right corner of the tile), so a city must be connected to a source of coal through canals or rails before you can build a shipyard there.
http://www.boardgamegeek.com/image/307212/brass
Barrow-in-Furness and Birkenhead may only have rail links (notice that there's no blue line) to the rest of the map, from where coal may come.
Therefore during the canal phase, the only possible shipyard location is Liverpool as you can build a canal link to Wiggan or other potential coal sources; but there is no such potential canal link out of Barrow-in-Furness or Birkenhead.

That was my error. I was looking at the level 0 tile, and saw no materials needed. Thanks!


text:
What board games have good quality versions available for play online for free?
What are some worthy online versions of board games?
They should be:

Complete working versions
On a site that is easily accessible (so no spam, lots of pop-ups or requires handing over excessive information to sign up)
Is a true version of the original (same rules, look and feel, game play) so no poor quality clones
Reasonably active community so there are enough players to have a game any time of day

EDIT:
To clarify they should be online games that replicate an existing board game. Not just any online board game.


text:
How to eliminate advantages gained from the seating position of inferior players?
In many games, such as Liar's Dice or On The Underground, there is an advantage to be had by sitting immediately after (in the turn order) the player who is least experienced or just plain bad at the game. Are there any widely applicable ideas for resolving this problem? Or, is there some house rule or variant that improves this aspect of a particular game?

The same advantage happens in Puerto Rico, you always want the new guy on your right (and you want the 'best' player to your left).

As long as you take the time to explain as much as possible beforehand, that will go a long way in my opinion.  One of my groups will explain the game in almost painful detail before starting, and the other will just throw me into the game and constantly say, "Oh, by the way, we forgot to tell you this" after I've done something stupid.  Guess which group I prefer?

In Dominion, there are really only two cards that depend on the person directly next to you.  Those cards are Possession (person to your left being good is favorable) and Contraband (person on your left being bad is favorable).  Of course, one thing to explain to bad players if THEY have Contraband is that it needs to be the very first treasure they play, so that the other person doesn't know how many coins you have.

Assign seats randomly.  Card draw, dice roll, etc.
A part of strategy in games is knowing the situation you are in and adjusting your game play.  If you get the unlucky draw of being to the right of this person, you'll need to adjust your strategy knowing the person to your left may not make the play an experienced player may make.  Many times I make a play I normally wouldn't make when I sit the right of an inexperienced player if I cannot afford for the person to the left of the player to make specific play.
So in my opinion situational awareness should dictate your strategy.
After every game we play, we re-assign seats randomly.

I think you can't have a serious, cutthroat game with a wide range of skills, so relax. Gently coach the new players, but no more than they're willing to receive. Play a couple of dummy rounds for the new players and then restart if needed.
If you're really serious, you can investigate ways to handicap the game, but that's really hard to do without ruining the balance of most games.

Board game skill definitely transfers from game to game. If you really don't want to play Puerto Rico with an unskilled player because the person to their left will have an advantage, then play different games with them that do not have this problem.
For example. Power Grid has turn order that changes based on how many cities a player has. Of course good players will purposefully hold back on purchasing houses to get better position in the next round, but it is still a great mechanism to make sure that inferior players have limited ability to affect the game in a negative way.
Other games that solve this problem are Caylus, Agricola, Citadels, and pretty much any other game where turn order can be bid for, and is not set in stone. Also, many game sthat have limited player interactivity, such as pure racing games, do not suffer because turns are effectively, if not actually, simultaneous. Also, this problem obviously does not occur in two-player games or co-op games.
I generally highly discourage any modification to games. Players are not game designers, and without incredibly thorough play-testing by professionals, any so-called house rule you come up with will probably just ruin the game, even if you don't realize it.
If you absolutely feel compelled to modify a game, here is a general method to follow. First, determine if it is better to go first or last in any given round. For the sake of example, let us say it is advantageous to go first. In the first round of the game proceed as normal. Then using some quantifiable measure, rank the players as to who is in the lead. Since going first is advantageous, have the person in last place go first, second to last place go second, and so on. Redo the turn order at the beginning of every round by this same measure.
This modification will in all likelihood ruin your game, as skilled players will take advantage of the system to go first/last when they need to. It also allows for increased chance of error, when someone no doubt accidentally takes their turn out of order. Yet, it will fix the problem of sitting to the left of the weakest player under most circumstances.

This is a great answer.  I don't feel the last 2 paragraphs are needed though.

If you don't realize it, was the game really ruined? And how are you supposed to TRULY understand a game if you've never modified it? I personally modify my games all the time, and I see no reason to discourage people from doing so if it makes the game more fun for them. :D

We long-ago made a set of 'random position' counters for games where player order matters; one with the name of each player in our group.  Then we deal out the seating for the game at random.  This doesn't solve the problem in any individual game, but on average everyone comes out even.  Getting the worst player as in a good place for your or the best as your most-likely attacker becomes just another random factor of the game - in the long run, luck evens out.
I don't know any games offhand that are absolutely ruined by this kind of position difference, unless the gap in skills is so great as to be a problem no matter what - in which case you might as well admit that and make it a relaxed teaching game.

"I don't know any games offhand that are absolutely ruined by this kind of position difference..."
FWIW: 7 Wonders would be such a game.  If there's one player who is slightly worse at drafting games than the rest of the players, the player to their left has a notable advantage.

@invisiblejon: Thanks for that; I haven't played 7 Wonders, and it's interesting to know that there's a case where it really matters.

Glory To Rome is another such game.  In my gaming group, Craftsmen are so highly valued as patrons that we (almost) always use Jacks to lead/follow as Craftsmen.  When the new player is using his actual Craftsmen to lead/follow the Craftsman role, this gives the next player an easy way to pick up a Craftsman as a client, which makes all of their future Craftsman actions much more powerful.  I used to be this weak player that handed out Craftsmen to the veterans, until I learned better.

When it's clear that one player has an advantage over all the others, my group often has fun by conspiring against that player, and if people are playing the game for enjoyment it can end up being pretty fun. Once my family noticed that I tended to win more than any of them at Settlers of Catan, they started to be very careful about what trades the offered me and, if I was winning, they'd all combine forces to make sure that robber was on me at all times. The same thing happens in Puerto Rico.
Experienced player to new player before me:

"No don't pick craftsman! That means he'll get to ship his goods first and get more victory points than us! Pick Builder so he doesn't get that stack of $$$!"

New player thinks about this fact and reconsiders. Picks the Builder, thwarting my plan. The other players had fun conspiring against me, I had fun being conspired against, and the unfair advantage I would've had was gone. What's not to like?
Now I don't win as much any more, and when I do it's far more of a treat. My opponents have fun trying to thwart my plans and I have fun scheming to win and laughing when I don't. If you have a new player at the table you probably won't have the super-cutthroat experience you might have if all the players were of equal skill level, but that doesn't mean you can't have fun. :D

I think it really depends on the game and the situation.
If I am new to a game, it doesn't really matter where I sit, until I become experienced enough to understand the positives and negatives of a seat position.
If I'm just a bad player, well, then I am not in any better position no matter where I am.
This being said, we try to shuffle seating regularly when playing several rounds of a single game.  Depending on the game, we will also sit according to final score/placement, winner getting the first seat, then on down to the left.  This helps with games like Puerto Rico and some other games that tend be a bit harder for those going first.

Winner first then down to the left means the best player has the worst player to their right, which is a huge advantage. Why not loser first then up to the left, so that each player except the worst has a slightly worse player to the right?

@Sparr - Again, it depends on the game.  For example, When playing Puerto Rico, the 3 & 4th place players have a distinct starting advantage over the first (and second).  I can see where having the least experienced on your right would be helpful also.
My suggestion would be to roll dice for seating position, or have opposite players move 1 space right, just mix it up.

Given honest players, some of whom have a grasp of the game at hand and this problem, I have often found that simply seating players in order of skill/experience mitigates this problem somewhat.

The problem with that is that the worst player sits next to the best player.

If you seat them worst to best (in order of play), then the best player goes after the second-best player, and the person to benefit from the worst player's play is the second-worst player.

If there is too much of a difference between players such that winning is not an option for the "lesser" players, try to focus on having fun. The players won't return if they always lose.
Another option is to team up. Make a team with a couple of an experienced players and a new player, so the new player can learn from the pro without the need to be beaten.

My recommendation would be to switch the seating every few turns/times around the table.  Not so often that it slows down the game, but just enough that nobody sits next to the less experienced player for a time that would give them a major advantage.  Just my opinion on the topic.

This question makes me wonder:
How many games where turn order is the same from round to round would be "ruined" if you selected turn order randomly every round, with the caveat that no player could go twice in a row?
In other words, what if you re-organized turn order at the end of every round?
Note that I don't think this is a very practical solution, since the act of re-ordering turns at the end of each round would add at least another two minutes per round, but it's conceptually interesting.

That's not an answer, but a comment. An interesting comment, at that.


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Game where you construct a line through the board blocking the other player?
This is a two-players game, where each players construct a line.
The first player has to join the top and the bottom of the board whereas the second has to join the left and the right. The two lines cannot intersect.
Do this game has a name?
Where I could find other informations about it?
PS:
I’m currently writing an HTML5/Canvas implementation here (only in French for now, and the “AI“ plays at random, the smart AI is not finished yet), so you can see more precisely what I’m talking about.
I’ve called it “Percolation” because this looks a little like percolation, but this is probably not the “official” name.

My first thought was "Blockbusters" :)

Do you remember where you saw the game first? It sounds like a game I invented and made a website for awhile back I called "Oct." Anyway, here's the website: hexandoct.net16.net

The game is called Bridg-It and it is a specific implementation of the "Shannon switching game".  There's a very elegant win for the first player which is based on pairing the edges together.  I think it's in a book by Martin Gardner, but I can't seem to find it online.  There's another solution here, but it's not as easy to understand.
Edit: The solution is given in New Mathematical Diversions by Martin Gardner, and was found by Oliver Gross.  "Make the first play as indicated by the black line at lower left in the diagram. Thereafter whenever your opponent's play  crosses the end of a  dotted line, play by crossing the other end of the same line."

Thank you very much for your answer! I have found a strategy for this game which seems to always win, but I never managed to prove that this is indeed an always winning strategy (when playing first). Thanks to your links, I will perhaps be able to prove or disprove it :-)  By the way, the other solution (with the electrical circuit) does not seem optimal, I managed to beat it twice while playing second.

TwixT "... was popular in the 60s and 70s as part of the 3M Bookshelf Series.
TwixT is a "connection-type" game, very similar to Hex . Other related games are Havannah... "Connection type" means the object is to form a path connecting something to something else. This requires very different thinking from the standard battle-type game such as chess or checkers.

On a personal note, I loved Twixt when I was about ten or eleven years old.  It was the first thinking game that let me excel enough to feel smug.  Plus the futuristic people on the cover looked ludicrously intelligent (and smug).

Quoridor is similar but not identical - two to four players try to reach the opposite end of the board with their pawn. Each turn, instead of moving, they can choose to place a wooden wall piece as an obstacle to block the opponent's path.


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In Settlers of Catan, how do you overcome bad initial settlement placement?
In Settlers of Catan, how do you overcome bad initial settlement placement?  In a lot of games of Catan I've played, initial settlement placement seems to really make or break the game for a player.  Are there any good strategies for overcoming a weak starting placement?

I think the main thing you need to do is correctly identify your strengths and weaknesses, and don't try to force a strategy you're not set up for.  If you don't have abundant brick and wood, don't even bother going for Longest Road.
Even the worst placement will have some resources they are rich in.  Try to get a port that capitalizes on your resources, and make the necessary trades with other players to get the resources you're poor in.
If you're rich in...
Brick or Wood -- try for longest road and build plenty of settlements
Grain or Ore -- upgrade your starting settlements to cities and buy development cards
Sheep -- go for Largest Army via development cards, or depending on your next best resource, you'll probably follow one of the other two basic strategies
Of course even with all of this, sometimes the dice just won't drop your way :)

I won a game recently using this. I didn't have Brick or wood, I didn't have a 6 or 8 terrains, but I was close to a port, that gave me a lot of advantages. I did a lot of trades, and with a port, my trades were always good to me. Catan is that game very balanced that we could overcome initial difficulties.

If your initial placement is bad, hopefully you can still build a little.
The first couple road placements are critical.  By then you should be able to figure out which commodities are in short supply this game.  Either because they haven't been rolled yet, or they're sitting on a 3 and an 11.
Use your first two roads to build for that scare resource, claim it, and then see who will overtrade you for it!

If you find you're often badly placed, initially, it may be that you're either making errors there, or misjudging where other players are going to place their initial settlements. It's easy to pick your optimal placement when in a four player game you place fourth-fifth, and you can usually figure where the fourth player to play will go when you place third-sixth. If you go first you usually grab one of the two best producing junctions, and should aim for brick and wood since if you have those two with your first settlement, if your options are limited when it gets back to you at least you're able to move, and other players are often hungry for brick and wood early in the game, meaning you're the trading king. Going second-seventh is often the toughest slot, since the first player to go typically grabs the best brick and wood combination (assuming there is one), and there often isn't another good brick and wood combo on the table. Then, of course, by the time it gets back to you pickings tend to be slim.
Still and all, given the good advice people have offered, above, you really shouldn't run into setup situations too often at the beginning of the game that seriously handicap you. One of the three main lines of play should be open to you. It's unusual to be stuck with setups where all you get are ore, sheep, and brick; or wheat, sheep, and wood; those triumvirates that leave you unable to move, unable to build, and unable to pick up development cards. Avoid those scenarios if you possibly can, since you're absolutely dependent on the tender mercies of other traders to bail you out and let you progress. Even so, if that's the best situation available to you, you may need to try to build to the nearest 2:1 port of a resource you're rich in, just to get the flexibility that'll get you back in the game.

Diplomacy.  During the mid game, constantly point out how badly you are doing, encouraging others to trade with you in preference to other players - even offering you better deals.  This can allow you to develop a couple of settlements and get back in the game while the others are playing stop-the-leader.

Initial placement is critical, but once you've seen the layout, you can devise a strategy that will get you something.  One of the big goals you should be going for is number diversity, so that you're always getting something.

When there is a valuable resource(the average amount of pips for each resource is 11.6) that has less than 10 pips, and there is no red hex corresponding to that resource, and there is only a 4 pip for that resource(ie there  is only a 9,3, and 12 hex for wheat, then try to claim the 9 and block other people from it) and don't trade with people for that resource unless it's 3:1(and usually if you want to trade 3:1 they have a harbor, else usually at least 4:1). So if you don't have that resource as initial placement, you should try to get a port and trade what you have(for example you have 2 cities and they correspond to a 8 and a 9 for sheep, get a sheep port)  with the bank.

If you have a bad settlement set up, you need to hope for good dice. For instance, if you have key resource hexes that are twos and threes or 11s and 12s, you should hope that those numbers are rolled a disproportionate number of times. You might play as if they were good numbers, and build your strategy of getting "more than your share." If that works, great. If not, better luck next time. Know that even players with resources on "6" or "8" numbered hexes will have the dice pass them over in a single game from time to time.
Another number that tends to help you is 7s. Those numbers hurt "everybody," but they hinder others more than they hinder you. More to the point, if you really have bad settlements, you will be seen as less of a threat and have the robber placed on you less than usual. The 7s are particularly good if you roll them, because you can move the robber to a hex owned by players that have access to the goods that you lack, and steal from them.
if you are using a "come from behind" strategy, head for a port, particularly a 2 to 1 port in things that you are "long" in. Also, if you are behind, people will be more willing to trade you the goods that you need.

I made up something I haven't really tested thoroughly, but if you have enough tiles you could build a complete hidden map(tiles upsidedown) underneath your visible map and and have a player pay a wood and ore to make a shovel to dig for new ground beneath the "bad" tile and wait one round with a blank tile.)
It's an addition, you may or may not like, but hey.
(We usually remove the 2's and 12's and you're able to revise one of the numbers drawn so the gamble is more to the strategy than lady fortune smiling upon you.)

Welcome to the site! This is a Q&A site, not a discussion forum. Your response will be removed by an admin because it does not answer the question being asked here. Feel free to stick around and ask and answer questions about board games!

"how do you overcome bad initial settlement placement?" is the question and I gave an anwer about how I overcame and the questioner can overcome bad initial settlement. Q&A.

Sorry, it wasn't clear that your variant was intended to address poor initial placement. And while variants can be all well and good, the OP was asking about strategies to overcome the issue; changing the rules of the game so that the issue doesn't happen so often doesn't really address that.

Resign.
In my experience, it's almost impossible to work yourself out of last place in this game. There's often a struggle between the top two players, but if you're a few turns in and somebody has badly blocked your expansion, I think you're doomed.

As a chess player, I recognize the right time for resigning, but in a group game like Catan, it seems quite poor form, as you are significantly altering the game for others. (Should they remove your settlements? Or just deal with the fact that you have made some places unsettleable?) Be a man, play it out.


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Monopoly Alternatives
A recent Board Games with Scott episode [Breaking Up the Monopoly] has brought up an interesting question:  When someone that you know asks to play Monopoly with you, what game do you suggest playing with them?  I assume that you don't end up playing Monopoly with them.  If so, that's cool, just tell me why you do that instead (alternative rules, versions, etc.).
Try to limit your answer to one game, and explain the mechanic or "hook" that you use to entice them away from the dreaded Boardwalk, Park Place, and Free Parking.

It would be nice if there were more boardgames out there that were "a lot like Monopoly" (buying properties, making money off them) while being, you know, actually good.  There's a game by respected Eurogame designer Wolfgang Kramer called "The Market of Alturien" which is a lot like a Euro version of Monopoly.  Only one problem: it's an absolute stinker that doesn't really work on any level (the best I can say about it is it might be quite good to play with children).  Looks like even Kramer couldn't make a good game out of Monopoly!

Settlers of Catan
For adults, I would suggest Settlers of Catan, since it's a much better game to me and a lot more fun to play.  If your willing to play a full game of Monopoly, then Catan is a prefect fit since it's actually shorter.  There's also a much more nuanced trade mechanic, and a board to stack our your share of.  I see it as a great chance to introduce them to Euro games if they haven't played one before.
For young kids, if they want to play Monopoly, I wouldn't dare be a spoil sport and suggest something else.

What do you classify as "young kids?" If they're old enough to play regular Monopoly, as opposed to some simplified kids' version, I would still suggest a decent boardgame instead. There are a couple of reasons for this: 1) If you enjoy yourself, you'll be more likely to play with them again next time instead of resenting the hours of your life wasted on Monopoly. 2) You might be responsible for introducing them to more interesting, challenging, and entertaining games than Monopoly.

Monopoly by the rules
I'd suggest playing Monopoly with no house rules to keep the game competitive and short.

No bonus given for landing on Free Parking
No restriction on when properties can be bought, e.g. not until one trip around the board.
No building houses or hotels if the bank has none to sell. They're meant to be a limited resource.
Auction properties. If someone doesn't want to buy a property, then it will be auctioned to the highest bidder.

Playing by the actual rules makes Monopoly a much shorter and more enjoyable game.

-1 I'm sorry, but I really think it would still totally suck (even if less than with random custom rules)

I've suggested this in the past.  It works surprisingly well.

Yeah, when I've done it with friends, they're always surprised that that Monopoly is supposed to be that quick, and they end up enjoying it a lot more.

Another person in favor of Monopoly by the rules.  "Tricked out," it's tedious.  By the book, it's actually pretty fast.

You can play the "speed" monopoly from the original rulebook. This is where you deal all the properties randomly to players before you begin the game, essentially saving you the first ~30mins of the game and start with less starting cash.

Acquire
I am surprised no one has mentioned this classic (back in print as of 2008) which is quite thematically similar (paper money, real estate).  Except, it is shorter, more strategic, and has no dice rolling (but still some randomness from tile drawing).  It is also pretty easy to learn.

Power Grid
For me, it has to be Power Grid. It has many of the attributes of Monopoly, such as bartering of which power station to buy and is economically driven, plus the rules are pretty simple. Furthermore, it is such a great game, it is a brilliant introduction to the world of strategic board games, and I have known a few players hooked in this way.

Monopoly Deal
A card game based on the board game, everyone wins here -- they get to play something in the spirit of Monopoly, and you don't have to deal with a long, drawn-out game of landing in Jail and haggling for properties, or worse, reaching a stalemate where no one will budge.
The card game is fun, very inexpensive to purchase, and plays quickly.  It works better with multiple players, but is just as fine playing with two players.

I have it; I never played it, but quickly reading the rules just once, it seemed promising.

Easy Money / Money Money
Faster Pace - Everyone plays the same pawn. Everything that happens concerns all players.
Shorter Play - No stalemates. The game ends when the bank runs out of money, which takes a more or less predictable amount of time.
Moaaar Money - In the middle of the board are stacks of bundles worth one million, half a million, 200k and 100k a piece!
Granted, there is no buying or building properties, auctions or event cards. Basically it trades some of the complexity and duration for fun and more action.

Top Shop
So sue me, it's technically a videogame (for the original Playstation), but it's basically monopoly set in a mall, translated to Japanese and then back into English. It's a heck of a lot of fun, and plays pretty much just like a boardgame. Up to six can play at a time.
If you land on an empty square you can open a store. If you land on an opponent's store you have to buy at least one item, but can buy two. If there are no items in a store you've landed on you can take over it. If you land on your own store you can restock ANY one store you own. There are random events, and points you get that will let you do various game changing events.

Pandemic
The cooperative nature, quick play, and ease of play (especially without the expansion) makes it an instant shoo-in for me.

I like Pandemic, but is it in any way related to Monopoly?

Not particularly.  I am a big fan of it for just that reason - often people who are asking to play monopoly (in my experience) don't have a understanding of what other options there are.  Giving them a cooperative board game experience that is fast and fun is often eye opening.  Although if you are looking for a similar experience, then I second Catan.  You could also look at Reiner Knizia's Palazzo, or other games by him.  They tend to focus around market-style systems.

Vegas Showdown
I like the limited resources and bidding to get different rooms.  It can get pretty competitive, depending on the group, but I enjoy the mechanics and think of it as Monopoly only a little more fun and with a definite endgame.

I came across a board game called Anti-Monopoly. The concept's a little different from Monopoly , its just the opposite :)

Yeah That game sucks. Big time.

Perhaps Monopoly Express?  It's got the Monopoly theme, but it's a dice game that takes about 25 minutes to play.
In that vein, Risk Express is actually quite fun (but hard to find in the USA).  It's another dice game version of a Hasbro board game that keeps the flavor, but plays quickly.


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In Secret Misson Risk which player is the winner when 2 missions are simultaneously achieved?
When playing the North American 1993 version of Secret Mission Risk, which player is the winner when 2 missions are simultaneously achieved?
Example: If your mission is, “Conquer Asia & Africa” and you have just one more territory to conquer to complete your mission, let’s say Madagascar which also has the last red troop on the map while another player has the mission “Destroy all Red Troops”, who wins when the last red troop dies and Madagascar is captured?
Does the other player holding the mission “Destroy all Red Troops” win as the red troop died before Madagascar was occupied? Does the player holding the mission “Conquer Asia & Africa” win as that player was the contributor to the final action accomplishing both missions? Would this game be considered a tie as both missions where accomplished simultaneously? Would the winner be the first to reveal their mission?

I've always assumed that it had to be your turn in order to have completed the mission.

In the 2003 version of Mission Risk, where you must accomplish 4 different missions on 4 different turns, it is indicated right on the mission card when you must reveal the mission, at start of turn, end of turn or just after gaing reenforcement of troops.  My question was related to the 1993 version of North American Risk.  Thoses missions can be acomplished at any time.

From these rules:

The player who completes his or her mission first-and reveals the Mission
  card to prove it-wins.

The rules make no mention of ordering, or of only being able to win at a specific time. If both players achieve a mission at exactly the same time, and both reveal their completed missions, then they have both fulfilled the win condition. You could assign a single winner in any arbitrary manner, but personally I'd say they both win; after all, does victory have to be exclusive?
Extra note: regarding the timing of defeating the final army and capturing a territory, the rules say:

As soon as you defeat the last opposing army on a territory, you capture that territory and must occupy it immediately.

While the rules don't specifically set out timing, I'd say that the "as soon as" and "immediately" mean that capturing a territory happens at the same moment as defeating the last army.


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Good board games for single player
I love playing board games but often there's not many people around that share my passion. For that reason I'm looking for board games that can be played with just one player. I've discovered Arkham Horror few months ago and I think it's a fabulous game but I was wondering for some time now if there are other good board games that are as good or even better for single person?

Are you interested in games specifically designed for one person, or are games that can be played by 1 player with small modifications acceptable?

Both types will be fine.

Flagged for wiki conversion. See: http://meta.boardgames.stackexchange.com/questions/66/recommend-a-game-questions

Agricola has a supported one player option.  I find it doesn't have the same feel.  It's more of a puzzle, but it is still very enjoyable.

You can look for games with single-player mode. There are few realy good games with this.

One of the best game - Agricola
has single player game.
Race for the Galaxy with an expansion "The
Gathering Storm" has also single
player mode.
Pandemic - designed for 2-5 players, but you still can play alone (because it its cooperative game).

While not necessarily "board games" there are a few card games that have a nice solo option:

The Isle Of Dr. Necreaux has a great pulp sci-fi theme
Space Hulk: Death Angel is a good single or two player game that recreates the feel of the board game well.
Chainsaw Warrior — you, a chainsaw, and stuff that needs chainsawing... nuff said.

Death Angel looks pretty good and it's cheap. I'll probably get this one soon. Thanks.

Death Angel can actually be played with up to six players.  It's really quite good.  I've never come close to winning the single-player game.

There are a couple fun ways to play Carcassonne with one player:

Build an outline of a 10x10 grid with randomly chosen tiles.  Attempt to fill it in completely.
Just play solo (possibly with a limited number of tiles and/or meeples) and try to maximize your score. Two examples.
The version in the Carcassonne iPhone game.  Draw+place tiles with no meeples.  Complete roads and cities of successively larger sizes starting at size 2 (features completed out of order do not count).  There are a few ways you could score this.
The iPhone game uses a method that may be too heavy for use by hand: tiles to compete scoring features are free, other tiles cost based on the size of the rectangle that encloses the area of play (cost=length+width after the tile is placed); you lose if the total cost exceeds 1000 before your size 6 road and city are built.
For less bookkeeping, you could try to maximize the road+city size you build while minimizing the size of the area of play.

The recent Castle Ravenloft boardgame has a few scenarios for a single player that are quite fun to play.

Le Havre has an enjoyable one player experience.
The game involves managing many goods while buying up valuable businesses with an eventual goal of having a shipping empire.

Roll Through the Ages. This is a terrific civ game disguised as Yahtzee! It's a fun game in general, but I play it as a solitaire all the time on my iPhone. The app is a great deal at 5 or 6 dollars. In fact, I own only the digital version! But it's a great game in either form.

Lord of the Rings is a cooperative game, but a friend of mine highly recommends playing it solitaire. I would guess that other cooperative games could be played this way, but it would depend on how much information was intended to be hidden from the other players.

Arkham Horror supports single player play. However, it would probably be more fun to have someone else to co-miserate with!

I play many games (Settlers, Puerto Rico, Ticket to Ride) against the computer.  Their strategies get old after a while, but it gives me some practice.  Lots of the popular games are also online.

Board game against the computer?

@RaYell, yep, there's ton's of board Games on the computer now.

For me the best thing about board games is that you touch and see game elements. If I want to play on the computer I'll probably pick Modern Warfare :) But that's just me.

I think the initial idea was what games could be played solo without a computer.

A lot of people don't know that there are good board / turn based games online, I think it's a fair answer.

I've had lots of fun with a few free print-and-play dice games.  The rules and game boards can be printed out for free, though you'll need to provide your own dice (obviously).  Here are a few good ones that I keep coming back to:
D-Day Dice

This dice game depicts the attack on Normandy Beach on June 6, 1944.  You start with a few soldiers and a printout of the beach divided into sectors.  Roll dice to recruit more soldiers, recruit specialists, gain courage (allowing you to advance), and so on.  It's a lot of fun, and actually quite difficult to beat.  I won the first game I played, and I haven't won since!
The d6 Shooters

In this game, you play the role of the leader of a posse escorting an outlaw to his trial in Reno.  By rolling dice, you gather food, mine for gold, move toward Reno, and fight any outlaws you encounter on the way.  There's a lot to like about this game, and it plays in about 20 minutes.  It's a game that falls into that "Just one more time!" category.
One nice thing about both of these games is that there are several different scenarios available, and new ones come out periodically.  Just follow the links above to their respective pages on BoardGameGeek.com, and check out the "Files" sections for the rules and downloadable material.
P.S. - If you enjoy these games, it might be worth your while to laminate the player sheets.  That way, you can use dry-erase markers and simply erase when you're done.  No need to keep printing the game boards over and over again!

Word games would be great for this, especially Bananagrams and Boggle.
You can play against yourself to improve your fastest time in Bananagrams or your highest score in Boggle.

I think some great games have been named, but we should not forget (Peg) Solitaire, one of the best known games made specifically for just one player.

+1 It's good to remember that there are just games that are designed for one player from the ground up!

The Age of Steam expansion Barbados is a one-player map. The random cube distribution at the beginning makes it interesting to replay. Note that this map is included in the 3rd edition of Age of Steam.

I've played Scrabble often as just a single player.
All rules stay the same, but you're just looking to optimize every play for the most number of points (and I personally disallow myself to turn-in my tiles).
You could also optimize for always playing the longest possible word, or finishing your personal game in, say, 25 minutes.

Al Cabohne (a Bohnanza-spin-off) has a single player option. In that case you draw random cards for your opponents (the bean mafia). Works quite well in my opinion.
Other than that I know some riddle-games, there you solve crimes or similar things. You given some hints and place markers on the grid to represent encounters. That's more like a puzzle than a game though.

Blokus
Blokus has several interesting single player puzzles listed in the rules if that's what you're looking for.
Finding a way for four players to all get all their pieces on the board is a pretty cool challenge of analysis and spatial calculations.

Labyrinth

Labyrinth takes 1 or 2 players inside the Islamist jihad and the global war on terror. With broad scope, ease of play, and a never-ending variety of event combinations similar to GMT's highly popular Twilight Struggle, Labyrinth portrays not only the US efforts to counter extremists' use of terrorist tactics but the wider ideological struggle—guerrilla warfare, regime change, democratization, and much more.

Wilderness War is a fantastic game.  I'm looking forward to seeing what Volko Ruhnke does with this.

I played my first game of this yesterday using the learning game.  Its very good.

Ricochet Robot is very cool and funny, if you like puzzles. It's "the game with which you shout numbers in your garden". You must find the shortest way from a place to another, using very dumb straight-going robots.
If you prefer strategy games, you have Vinci, which is the first version of Small World and has a single player version which is very good! I tried it, the player doesn't win every time! The idea is: you play with "automatic players", that can be compared to AI in video games. They obey very basic rules and the player doesn't decide for them what to play, it's impressive how it works well! Maybe the game can't be found nowadays, though, but you can still get a Small World and play with this special single player rule from its ancestor, the two games are exactly the same.
It's very easy to play alone at Once Upon a Time, in which you must create a fairy tale. If you do, use a microphone to keep a track of your masterpiece! Maybe Disney will make a movie out of it someday.
As other mentionned, cooperative games are good as well! My preference goes to Forbidden Island, playable in less than an hour.

Elder Sign is closely related, obviously, and can be played normally with one player. It would be better to play as least two investigators (three if you feel like it), as one investigator tends to end in either certain doom or boring steamrolling. (It's also quite cheap and very, very compact compared to Arkham Horror.)