Probability and Poker Dice

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 ⁶C_n. 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 ⁶C_n × 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 (⁶C_n × 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