# Poker dice probability

I am working on a maths excersice and got stuck on this question where I need to calculate the probability of poker dice.

The game poker dice is played with 5 dice. It's possible to get one of the following hands:

Poker: All dice have the same value (ie 3,3,3,3,3). Four of a kind: 4 of the 5 dice have the same value (ie 3,3,3,3,1). Three of a kind Two of a kind Street: (1,2,3,4,5 or 2,3,4,5,6) Full House: (333,22) Two pair: (1,1,2,2,3)

Now I have to find the probability of these hands.

I know there are 6^5 = 7776 different throws.

For the poker there are 6 different values possible (111111,222222,333333 etc) so the probability is 6/7776

For the four of a kind theres 6*5*5 = 150 150/7776

But at the three of a kind is where I get stuck (and the other hands), wikipedia tells me there is an probability of 1200/7776. I don't know how they got the 1200.

If there is someone who could help me I would be very thankful.

Thanks,

Rico (Ps English isn't my first language)

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Three of a kind:

6 ways to choose which kind it is that appears three times.

5-choose-3 (which is 10) ways to choose the three times the chosen kind appears.

5 ways for the five other kinds that could appear on one of the other two rolls, and 4 ways for the four remaining kinds to appear on the remaining roll.

6 times 10 times 5 times 4 gives 1200.

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1. 3 of a kind is [wwwxy]. There are 3 ranks, wxy.
2. Left hand side is ranks. Right hand side is number of dice (6C3)(3C1).(5C3)(2C1)(1C1) / 6^5
• 6 ranks choose 3, there are six dice faces.
• 3 ranks choose 1 .
• 5 dice choose 3, you only need 3 dice to get 3 of a kind
• 2 dice choose 1.
• 1 dice choose 1.
3. Solve above and you get 1200/7776.
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