## Probability and Poker Dice

Posted: August 1, 2019 in probability
Tags: , , CarnotCycle is a thermodynamics blog but occasionally it takes five for recreational purposes

Poker dice is played with five dice each with playing card images – A K Q J 10 9 – on the six faces. There are a total of 6 × 6 × 6 × 6 × 6 = 7776 outcomes from throwing these dice, of which 94% are scoring hands and only 6% are busts.

Here are the number of outcomes for each hand and their probabilities And here is the data presented as a pie chart The percentage share of 5 of a kind (0.08%) is omitted due to its small size

A noticeable feature of the data is that the number of outcomes for 1 pair is exactly twice that for 2 pairs, and likewise the number of outcomes for Full house is exactly twice that for 4 of a kind. But when outcomes are calculated in the conventional way, it is not obvious why this is so.

Taking the first case, the conventional calculation runs as follows:

1 pair – There are 6C1 ways to choose which number will be a pair and 5C2 ways to choose which of five dice will be a pair, then there are 5C3 × 3! ways to choose the remaining three dice 2 pairs – There are 6C2 ways to choose which two numbers will be pairs, 5C2 ways to choose which of five dice will be the first pair and 3C2 ways to choose which of three dice will be the second pair. Then there are 4C1 ways to choose the last die Conventional calculation gives no obvious indication of why there are twice as many outcomes for a 1-pair hand than a 2-pair hand.

But there is an alternative method of calculation which does make the difference clear.

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A different approach

Instead of starting with component parts, consider the hand as a whole and count the number (n) of different faces on view. The number of ways to choose n from six faces is 6Cn. Now multiply this by the number of ways of grouping the faces, which is given by n!/s! where s is the number of face groups sharing the same size. The number of face combinations for the hand is 6Cn × n!/s!
Since the dice are independent variables, each face combination is subject to permutation taking repetition of faces into account with the permutation-repetition formula 5!/n1! n2! ..  where n1, n2 etc are the number of repetitions of a given face.

Thus the total number of outcomes for any poker dice hand can be calculated with a single formula:  It is easy to see from the table why there are twice as many outcomes for a 1-pair hand than a 2-pair hand. The number of face combinations (6Cn × n!/s!) is the same in both cases but there are twice as many dice permutations for 1 pair. Similarly with Full house and 4 of a kind, the number of face combinations is the same but there are twice as many dice permutations for Full house.

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P Mander April 2018

1. Peter Mander says:
2. Jack says: