In the game of Yahtzee, five dice are tossed simultaneously. Find the probability of getting

a. full house

b. 4 of a kind

Bases on wikipedia

Full House = A three-of-a-kind and a pair

Four-Of-A-Kind = At least four dice showing the same face

And total number of cases is 6^5=7776

I really don't understand this game,


1 Answer 1


The answer depends on whether you want to count only the throws that satisfy only the requirements for a full-house/four-of-a-kind entry, or also the throws with all five dice the same, which you could use either in the full-house/four-of-a-kind row or in the Yahtzee (five-of-a-kind) row. The difference is the probability of a Yahtzee, which is $1$ in $6^4=1296$; I'll calculate the probabilities excluding five of a kind.

In both cases, full house and four of a kind, there are $6\cdot5$ different choices for the numbers. For full house, there are $\binom52$ choices for the positions and for four of a kind there are $\binom51$. Thus the probability for a full house (excluding five of a kind) is


and for four of a kind


Note that these are just the probabilities for the first throw, which I think is what you asked for; the more interesting part of the game is deciding which dice to keep to optimize the chances on subsequent throws.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .