Probabilities in Liar's Poker

Last Update: June 30, 2003

Rules

1. Liar's Poker is played using randomly picked currency from your wallet. The denomination does not matter. Hoarding ringers is strictly not allowed.
2. All players must agree on the stakes, for example $1 per person per round. You do not have to use the exact bill you are playing for, for example you can use a $20 bill although playing for only $1.
3. A rule should be set for who goes first, for example whose letter in the serial number is lowest, or who won the last time. Who goes first is not very important, in my opinion.
4. A hierarchy of numerals should be established. I prefer zeros are low and nines are high.
5. Players in turn bid on the combined numbers in all serial numbers, your own and those of the other players.
6. Each player must in turn either declare a higher hand than the player player or challenge.
7. In a 3+ player game all players must challenge to end the game.
8. Eventually a player will be challenged. Then the combined serial numbers will be used to determine if the last hand called exists. For example if the challenged hand is four eights then there must be at least four eights on all serial numbers. If players trust each other than can simply declare how many of the given number they have, of course the challenged player reserves the right to see the bills if he so requests.
9. If the serial numbers support the challenged player then the player will win the agreed upon stakes from each other player. Otherwise the challenged player must pay each other player the agreed upon stakes.

Let's look an example. Suppose there are three players playing for $1 stakes with the following serial numbers.

Player 1: 06742088
Player 2: 92859819
Player 3: 07202503

Here is the play of the game, player 1 goes first.

Player 1: 2 zeros
Player 2: 2 fives
Player 3: 3 zeros
Player 1: 3 eights
Player 2: 3 nines
Player 3: 4 zeros
Player 1: 5 zeros
Player 2: challenge
Player 3: 6 zeros
Player 1: challenge
Player 2: challenge

At this point there must be 6 zeros for player 3 to win. There are only 5 so player 3 must play player 1 and 2 $1 each. Had player 2 had a zero then player 3 would have won.

Strategy

In 3+ player games it often happens that a player is in a damned if you do damned if you don't situation. Assuming that by challenging you will definitely lose, and by raising you definitely will be challenged, you should always raise in a 2-player game, raise if your probably of winning by doing is 25% or greater in a 3-player game, 33.33% in a 4-player game, and (n-2)/(2n-2) for n players. Of course nothing is ever certain, so this scenario is admitedly unrealistic.

It often happens that you need at least one other player to have at least one of a certain number for you to win. Assuming nothing about the other player's numbers (again an admitedly unrealistic assumption) the following table shows the probability of the total number of any given number according to the number of other players.

So if you are playing with two other players and you have 3 fives and call four fives the probability of winning if you are challenged is 1-0.185302 = 0.814698. However if you need two fives the probability drops to 1-0.185302-0.329426 = 0.485272.