Imagine this very simple bet;

Every player rolls 5 dice, the player with the highest ranking 'hand' wins.


Hand rankings;
1 One pair
2 Two pair
3 Three of a kind
4 Full house
5 Four of a kind
6 Five of a kind
7 Straight (1-5 or 2-6)


Are these hand rankings in the correct order with their probability?
(Five of a kind more likely than a straight?)

What would be the probability of rolling the different hand rankings?

Good place to start; https://en.m.wikipedia.org/wiki/Poker_dice

Thanks! It's all in there

To illustrate one way to derive these results, consider the case of three of a kind, e.g., 222 4 6.

Using combinations, the occurrence probability is the number of combos (ways) trips can occur throwing 5 dice divided by the number of possible combos.

Some basics.
Number of combos of selecting r things out of n different things: C(n,r) = n!/(r!*(n-r)!)

Number of combos to select k things out of n different things, with r1 alike, r2 alike, etc.: n!/(r1!*r2! *...), SUM(ri)=k

The denominator: We have for each die 6 ways it can fall.
With 5 throws, the total number of ways is 6^5 = 7776

The numerator. We are selecting 3 numbers out of 6, e.g., (2,4,6). Number of ways is C(6,3). With 5 throws, 3 are alike and two are different. Number of ways is 5!/(3!*1!*1!). There are 3 ways to select the number that is tripped (e.g., the 2).

Therefore

Prob(trips) =[3*C(6,3)*(5!/3!)]/6^5 = 1200/7776

There are other ways to get this result, some perhaps easier. I didn't derive this immediately but the more you do this the "less harder" it gets.

NOTE: I didn't say "simpler".

Because you can just say that the first three are the same, the fourth is different, and the fifth is different still. 6/6 x 1/6 x 1/6 x 5/6 x 4/6, and then multiply by ten since there are ten ways the two unmatched dice can be put into this sequence.

Wow thanks!