# Yahtzee Fun

 PF Gold P: 117 If you’ve ever played the game of Yahtzee, you’ll know that often times, the last line to be filled in is YAHTZEE. Let's say I want to calculate the probability of rolling Yahtzee on a single turn. To this end, suppose that after you roll five dice, you are allowed to select any of the five dice and roll them again. At which point, you may select any of the five dice and roll them for a third time. This would help with calculating the probability of rolling YAHTZEE after any of the three rolls. First, consider the situation where you have exactly i dice of the same value and you re-roll the other 5 − i dice. Assuming that you have i dice of the same value that you are not going to re–roll, let pi,j denote the probability that you end up with exactly j dice of the same value after you re–roll the other 5 − i dice. (please let me know what you think of this) Note that if j < i then pi,j = 0 since you certainly have at least i dice all of the same value after you roll the other 5 − i dice. I want to use a transitional matrix to calculate this which is easy with the binomial distribution, but let's say we wanted to calculate each value in the matrix using basic counting knowledge and probability theory.
 P: 256 It sounds like a reasonable way to do it to me. Let me see if I understand it right: after the first roll you are in one of five states (you have 1, 2, 3, 4, or 5 of some number, which you will try to match from then on). And then you get two more rolls, each time rolling the ones that don't match your number. How else would you calculate p_ij other than with the binomial distribution? You've got 5-i dice to roll, and you need to pick up j-i more of the established number on that roll.
 PF Gold P: 117 lol That's what I'm trying to figure out. I was told that there's another way, but don't know how to figure it out.
P: 256
Yahtzee Fun

 Quote by silvermane lol That's what I'm trying to figure out. I was told that there's another way, but don't know how to figure it out.

Hmm. Is it possible they meant another way to do it besides using transition probability matrices? For instance, you could do it recursively. But I think it's easier to do it your way, with matrices.

Do you get to assume that you start out with a pair, or do you have to consider all the possible situations after the first roll? If so, you have to find the probabilities for that roll separately (since a particular number has not been established yet).

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