Liar's poker is a bar game that combines statistical reasoning with bluffing, and is played with the eight-digit serial number on a U.S. dollar bill. Normally the game is played with a stack of random bills obtained from the cash register. The objective is to make the highest bid of a number that does not exceed the combined total held by all the players. The numbers are usually ranked in the following order: 2,3,4,5,6,7,8,9,0 (10) and 1 (Ace). If the first player bids three 6s, he is predicting there are at least three 6s among all the players, including himself. The next player can bid a higher number at that level (three 7s), any number at a higher level (four 5s) or challenge. The end of the game is reached when a player makes a bid that is challenged all around. If the bid is successful, he wins a dollar from each of the other players, but if the bid is unsuccessful, he loses a dollar to each of the other players.
Liar's dice is a similar game played with dice, often as a drinking game.
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The chances that the other players have at least the amount of a number you need to be able to call your bid when challenged, can be determined by the following two formulae:
Formula 1 P(at least X times C) = 1 - binomcdf (Y , 0.1 , X-1)
With:
X = amount of the needed number
C = the needed number, which has a probability of 1/10 = 0.1
Y = the amount of unknown numbers, which is equal to 8 x amount of extra players
Example 1: you are playing a 2-player game and you want to determine whether the other player has at least 2 more sixes.
P(at least 2 times six) = 1 - binomcdf (8 , 0.1 , 1) = 0.18670...
So you have a chance of 18.69% that the other player has at least 2 sixes
Example 2: you are playing a 5-player game and you want to determine whether the other players have at least 4 more sevens.
P(at least 4 times seven) = 1 - binomcdf (32 , 0.1 , 3) = 0.3997...
So you have a chance of 39.97% that the other 4 players have at least 4 sevens.
Formula 2. In order to calculate the probability of at least X times C, you have to subtract each probability from X=1 till X=X-1 from 1.
P(X times C) = Y nCr X x 0.1X x 0.9Y-X
With:
X = amount of the needed number
C = the needed number, which has a probability of 1/10 = 0.1
Y = the amount of unknown numbers, which is equal to 8 x amount of extra players
Example: you are playing a 2-player game and you want to determine whether the other player has at least 2 more sixes.
P(at least 2 times six) = 1 - P(no six) - P(1 six)
P(no six) = 8nCr0 x 0.10 x 0.98 = 0.4305
P(1 six) = 8nCr1 x 0.11 x 0.97 = 0.3826
P(at least 2 times six) = 1 - 0.4305 - 0.3826 = 0.18670...
So you have a chance of 18.69% that the other player has at least 2 sixes
Overview probabilities of the at least needed amount of a specific number for a 2-player game to a 6 player game.
| Number Needed |
Extra players | ||||
|---|---|---|---|---|---|
| 1 player | 2 players | 3 players | 4 players | 5 players | |
| 1 | 0.56 | 0.81 | 0.92 | 0.97 | 0.99 |
| 2 | 0.19 | 0.49 | 0.71 | 0.84 | 0.92 |
| 3 | 0.04 | 0.21 | 0.44 | 0.63 | 0.78 |
| 4 | 0.01 | 0.07 | 0.21 | 0.40 | 0.58 |
| 5 | 0.00 | 0.05 | 0.09 | 0.21 | 0.37 |
| 6 | 0.00 | 0.00 | 0.03 | 0.09 | 0.21 |
| 7 | 0.00 | 0.00 | 0.01 | 0.04 | 0.10 |
| 8 | 0.00 | 0.00 | 0.00 | 0.01 | 0.04 |
So for example if you need 3 more of a specific number, the chances in a 2 player game are 4%, in a 3 player game 21%, in a 4 player game 44%, et cetera.
There are some tactics which are mathematically based and should be fully understood by the players in order to make it a full bluffing game.
Above are the odds that the other players have at least a specific amount of a needed number. It is possible that a player comes in a so-called 'damned if I do, damned if I don't' situation. Assuming that by challenging you will definitely lose, and by raising you will definitely be challenged while not being able to call your bid, you should always raise in a 2-player game, raise in a 3-player game if your odds are above 25%, raise in a 4-player game if your odds are above 33.33% or, in other words, raise in a n-player game if you odds are above (n-2)/(2n-2).
Example: You're in a 5-player game. Your serial number is 53653158. The last bid was 7 threes, which you deem is highly possible, since you already hold 2 threes. You can outbid by bidding 7 fives. You need 4 more fives to be able to call your bid, which is a chance of 40%. The tactic above states that you should raise if your odds (40%) are above (n-2)/(2n-2), with n being the number of players. (5-2) / (2x5 -2) =0.375x100% = 37.5%<40%, so statistically you should raise.
Overview of probabilities which need to be surpassed to raise in a "Damned if I do, damned if I don't" - situation:
| 2-player game | 3-player game | 4-player game | 5-player game | 6-player game | |
|---|---|---|---|---|---|
| (n-2)/(2n-2) | always raise | 0.25 | 0.33 | 0.38 | 0.40 |
| max. needed numbers | always raise | 2 or less needed | 3 or less needed | 4 or less needed | 4 or less needed |
As is stated before, Liar's Poker is all about bluffing, so you shouldn't stick closely to these statistics and tactics.
If every player follows the exact mathematical formulae, a possible game is the following. Keep in mind that the order of least to most valuable number is 2-3-4-5-6-7-8-9-0-1.
Player 1: 21068274
Player 2: 44789800
Player 3: 27706500
Player 4: 63523655
Player 1 begins
Player 1: 3 twos (has 2 twos - 92% chance others have another two)
Player 2: 4 fours (has 2 fours - 71% chance others have another two fours)
Player 3: 4 zeros (has 3 zeros - 92% chance others have another zero)
Player 4: 5 fives (has 3 fives - 71% chance others have another two fives)
Player 1: Challenge (can only outbid if others have at least 4 more of two, six, seven or eight, which is a chance of 21%, and 21%<33%)
Player 2: 5 zeros (has 2 zeros - 44% chance others have another three zeros)
Player 3: 6 zeros (has 3 zeros - 44% chance others have another three zeros)
Player 4: Challenge (can only outbid if others have at least 4 more fives, which is a chance of 21%, and 21%<33%)
Player 1: Challenge (can only outbid if others have at least 5 more twos, which is a chance of 9%, and 9%<33%)
Player 2: Challenge (can only outbid if others have at least 5 more fours, eights or zeros, which is a chance of 9%, and 9%<33%)
Player 3 has been challenged by all the other players. Each player tells his amounts of zeros. For Player 3 to win, together they have to have at least 6 zeros. They have exactly 6, so Player 3 wins and the other Players have to pay him the agreed amount.
This game was played with four players who fully understood and applied the mathematical formulae, but in Liar's Poker it's about bluffing and trying to influence other players' decisions to your benefit, while keeping these statistics in the back of your mind.