Then, to get the probs of each of the maxes (1-5) after your second roll is given by matrix-multiplying the row vector consisting of the first row (initial) probs by the 5x5 transition matrix. Obviously, the odds of each of the maxes (1-5) on one roll is just given by the first row of the matrix. (You can think of the initial "state 1" as rolling all the dice.) The prob of getting the max matching dice = 2 when rolling 5 dice (900/1296) goes in the second column of the first row, etc. So for the 5-dice example, the prob of getting no matching dice when rolling 5 dice (120/1296) goes in the first column of the first row. The first row in the transition matrix contains the probs of the outcome(s) of the first roll. I believe I could work out the odds given the amount of dice rolled the same on the first roll, then work out the chances of transitioning to the next state using the Markov grid, but I am missing the logic step in the middle to work it out without the first roll being defined. So, specifically, I am looking to determine the chances of me getting beat by someone rolling the Yahtzee in 2 rolls, and what the odds where for me to roll it in 3 rolls. I rolled 5 5's on the first roll, no 5's on the second roll, and 5 5's on the third roll. Right now the record in this game is 3 rolls. So, basically, what would be the odds of rolling a Yahtzee in 2, 3, 4 etc rolls at the outset of the turn. Is there a way to determine the odds of getting a yahtzee within a specific number of rolls without taking the initial roll into consideration? So if I roll 4, 5, or 6 of the same value on the first roll, it can be used to calculate the odds of getting the rest of the dice the same on the second roll, and then on the 3rd, 4th etc roll. I was working on the chart that you provided, but I was definitely doing it the long way your math skills make me a proverbial caveman.įrom what I read of the Markov analysis, it will show what the odds are of completing the Yahtzee considering what the first roll is. Cool that you verified heehaw, appreciate that. Thanks whosnext for the info and link, that is exactly what I needed to get started. If anyone wants to confirm/deny the expression or the results given above, it would be appreciated. Hoping I did it right, here is what I get for N=10. Of course, for the smaller values of N this can be done readily by hand. Divide Permutation(6,E) by the product of the factorials of the Uj, and multiply by N! divided by the product of the factorials of the Ai.Įssentially, this is a product of the number of ways to select specific dice outcomes times the number of ways, given the combination of dice outcomes, of ordering all such permutations.Īlthough the formula may seem strange or complicated, it is rather straightforward to code up in a spreadsheet or any other programming application. Then add up the following terms for each K. Only consider combinations for which E<=6. Let Uj be the count in Ai for each of the unique elements of Ai. Let Ai be the elements of the combination in question. Let E be the number of elements of the combination in question. To find out how many ways there are to make K-of-a-kind from N 6-sided dice (where K is the maximum number for the roll - so for a roll of 10 dice with a 5 of a kind and 3 of a kind and 2 singletons, K would be 5), you need to consider every combination of positive integers that add up to N having a maximum of K. I had to dust off some cobwebs in the permutation part of my brain, but I think I derived (that is, rediscovered) the formula for the distribution of the possible outcomes of the initial roll of N dice. I looked at this when I got home tonight. ( I did look for info on Google, and understand some of it, but I find myself thinking about the problem in different ways, and getting different answers. My main question now is how to work out the odds of getting all 10 dice the same on the second, third, fourth, fifth, roll and so on.Īlso, I am wondering how to calculate the odds of rolling 5, 6, 7, 8 or 9 of the dice the same on the first roll. Or actually, since there are 6 ways to get a natural, I assume it would then be 6/60,466,176 or 1/10,077,696.įeel free to correct me if any of this is wrong. I have been looking online at odds and probability websites and have figured out a few of the basic questions I had concerning the odds of doing this.įrom what I believe so far, the odds of a natural Yahtzee would be 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6 = 1/60,466,176. On the first roll, we will set aside whichever number on the dice is most prevalent, and then keep rolling until they are all the same. A Yahtzee is when all 10 dice are the same number. We roll 10 dice simultaneously, trying to get a Yahtzee with all ten dice. My Brother and I have been playing a variation of Yahtzee.
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