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 approximate probabilities

And here is the data presented as a pie chart

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

^{6}C

_{1}ways to choose which number will be a pair and

^{5}C

_{2}ways to choose which of five dice will be a pair, then there are

^{5}C

_{3}× 3! ways to choose the remaining three dice

^{6}C_{1} × ^{5}C_{2} × ^{5}C_{3} × 3! = 3600 outcomes

** 2 pairs** – There are

^{6}C

_{2}ways to choose which two numbers will be pairs,

^{5}C

_{2}ways to choose which of five dice will be the first pair and

^{3}C

_{2}ways to choose which of three dice will be the second pair. Then there are

^{4}C

_{1}ways to choose the last die

^{6}C_{2} × ^{5}C_{2} × ^{3}C_{2} × ^{4}C_{1} = 1800 outcomes

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.

**– – – –**

**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 computed by calculating ^{6}C_{n}. Now multiply this by the number of distinguishable 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 combinations for the hand is ^{6}C_{n} × n!/s!

Since the dice are independent variables, each combination is subject to permutation taking into account the number of indistinguishable dice in each of the n groups according to the general formula n!/n_{1}! n_{2}! . . n_{r}!

The total number of outcomes for any poker dice hand is therefore

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 combinations (^{6}C_{n} × n!/s!) is the same in both cases but there are twice as many permutations for 1 pair. Similarly with Full house and 4 of a kind, the number of combinations is the same but there are twice as many permutations for Full house.

**– – – –**

Note that

is reducible to

**– – – –**

P Mander April 2018