Liar's poker

For the book by Michael Lewis, see Liar's Poker.

For the card game sometimes known as Liar's Poker, see Commune (card game).

Liar's poker is an American 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.

Liar's Poker probabilities

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 ⁿC_r X x 0.1^X x 0.9^Y-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) = 8ⁿC_r0 x 0.1⁰ x 0.9⁸ = 0.4305
P(1 six) = 8ⁿC_r1 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.

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.

Example game

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.

In popular culture

• In his 1989 book Liar's Poker, Michael Lewis details how Salomon Brothers traders would play liar's poker.
• A version of liar's dice is played in Pirates of the Caribbean: Dead Man's Chest between Will Turner, Bootstrap Bill, and Davy Jones.
• A game of liar's poker was played in an episode of the TV series Hustle (Season 3, Episode 3) where one of the main characters plays and loses against two merchant bankers.
• Characters on the show Quincy M.E. were often seen playing Liar's poker.
• Anne O Faulk's "Holding Out" uses the game as a plot point.
• In the 1977 movie Semi-Tough, Burt Reynolds' and Jill Clayburg's characters play an ongoing game of liar's poker periodically throughout the movie.
• In the 2011 movie Hall Pass, the group of characters play a game.
• Elliott Gould's and Jim Bouton's characters play a round as friends in the beginning of the 1973 neo noir film, "The Long Goodbye".
• In The Wire episode Dead Soldiers, Tommy Carcetti and Anthony Gray play a game.
• In the 1965 film " Cat Ballou " the sheriff is confronted playing Liar Poker at the barn dance
• In the WKRP in Cincinnati episode "Herb's Dad," Herb's father, and later Herb himself, play Liar's poker with Johnny and Venus.