My Brother and I have been playing a variation of Yahtzee. We roll 10 dice simultaneously, trying to get a Yahtzee with all ten dice. A Yahtzee is when all 10 dice are the same number. 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.

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.

From 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. 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.

Feel free to correct me if any of this is wrong.

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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.

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Also, 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.

( 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. )

The initial case is straightforward combinatorics. Subsequent cases can be derived via a Markov chain analysis.

See the link below.

http://www.datagenetics.com/blog/january42012/

I looked at this when I got home tonight. 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.

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. Let E be the number of elements of the combination in question. Let Ai be the elements of the combination in question. Let Uj be the count in Ai for each of the unique elements of Ai. Only consider combinations for which E<=6.

Then add up the following terms for each K. 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.

Essentially, 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.

Although the formula may seem strange or complicated, it is rather straightforward to code up in a spreadsheet or any other programming application. Of course, for the smaller values of N this can be done readily by hand.

Hoping I did it right, here is what I get for N=10.

1 0
2 4,082,400
3 32,004,000
4 18,774,000
5 4,721,220
6 787,500
7 90,000
8 6,750
9 300
10 6
----------
Total 60,466,176

Hope this helps.

P.S. If anyone wants to confirm/deny the expression or the results given above, it would be appreciated.

I get the same thing for that one.

There are only 2 possible compositions: we can have 5 doubles, or 4 doubles & 2 singletons.

N(5 doubles) = 6! * 9!! = 680400
N(4 doubles, two singles) = C(6,2) * 45 * 28 * 15 * 6 * 2 = 3402000
Total = 4082400

Thanks whosnext for the info and link, that is exactly what I needed to get started. Cool that you verified heehaw, appreciate that.

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.

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From what I read of the Markov analysis, it will show what the odds are of completing the Yahtzee considering what the first roll is. 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.

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, basically, what would be the odds of rolling a Yahtzee in 2, 3, 4 etc rolls at the outset of the turn.

Right now the record in this game is 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.

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.

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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.

Hopefully that makes some sense.

The first row in the transition matrix contains the probs of the outcome(s) of the first roll. 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 prob of getting the max matching dice = 2 when rolling 5 dice (900/1296) goes in the second column of the first row, etc. (You can think of the initial "state 1" as rolling all the dice.)

Obviously, the odds of each of the maxes (1-5) on one roll is just given by the first row of the matrix.

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. That gives a new row 5-vector.

Then use that output 5-vector as the input into the next go-around to give the probs of the maxes after three turns. And so on.

For the 10-dice case, the probs given in my earlier post are the elements that go in the first row of the transition matrix (and form the first "input" to get the probs after two rolls). But you'll need all the other elements of the 10x10 upper-triangular transition matrix.

I have not given any thought as to how easy or hard it will be to derive all the elements of the transition matrix in the 10-dice case.

Maybe I'll have time to look into it this evening.

Assuming whosnext's post #3 is correct (I didn't double check the math but the logic all looks good), you have about a 1/25256 chance of rolling a Yahtzee in the first two throws (~1/25319 of exactly 2).

Higher numbers than that get really hard. The simplest idea would be to just apply binomial distribution for the later rolls, but that would be undercounting because of cases where for example on the first roll you roll 3 3s, and then when you reroll the 7 other dice you hit 5 5s, I assume you keep the 5s there? If you are "locked in" to the number you keep the first time the math is probably pretty doable.

I could not think of an easier and more reliable way to derive the 10-dice transition probabilities than a brute force approach so I coded up a simple routine. Here is what I get for the numerators for the transition probabilities of moving from P matches to Q matches (denominators listed at bottom).

T(10,10) = 1

T(9,10) = 1
T(9,9) = 5
---------- 6

T(8,10) = 1
T(8,9) = 10
T(8,8) = 25
---------- 36

T(7,10) = 1
T(7,9) = 15
T(7,8) = 75
T(7,7) = 125
---------- 216

T(6,10) = 1
T(6,9) = 20
T(6,8) = 150
T(6,7) = 500
T(6,6) = 625
---------- 1,296

T(5,10) = 1
T(5,9) = 25
T(5,8) = 250
T(5,7) = 1,250
T(5,6) = 3,125
T(5,5) = 3,125
---------- 7,776

T(4,10) = 1
T(4,9) = 30
T(4,8) = 375
T(4,7) = 2,500
T(4,6) = 9,380
T(4,5) = 18,870
T(4,4) = 15,500
---------- 46,656

T(3,10) = 1
T(3,9) = 35
T(3,8) = 525
T(3,7) = 4,380
T(3,6) = 22,050
T(3,5) = 68,145
T(3,4) = 119,700
T(3,3) = 65,100
---------- 279,936

T(2,10) = 1
T(2,9) = 40
T(2,8) = 705
T(2,7) = 7,200
T(2,6) = 47,250
T(2,5) = 209,720
T(2,4) = 612,500
T(2,3) = 739,200
T(2,2) = 63,000
---------- 1,679,616

As mentioned previously, the T(1,Q) probabilities are simply the distribution of outcomes on the initial roll of 10 dice given in a previous post above.

Based upon this transition matrix, I turned the crank several times to derive the probability (given in percent below) of a 10-dice yahtzee after rolls from 1-25.

1 0.000010% (1 in 10,077,696)
2 0.003959 (1 in 25,256.25)
3 0.076727 (1 in 1,303.3)
4 0.466870 (1 in 214.2)
5 1.589086 (1 in 62.9)
6 3.839748 (1 in 26.0)
7 7.428515 (1 in 13.5)
8 12.326330 (1 in 8.1)
9 18.310969 (1 in 5.5)
10 25.052697 (1 in 4.0)
11 32.195696 (1 in 3.1)
12 39.415298
13 46.448523
14 53.103631
15 59.256349
16 64.839305
17 69.829223
18 74.234560
19 78.084903
20 81.422558
21 84.296258
22 86.756725
23 88.853697
24 90.634077
25 92.140902

Oh wow, I did not consider that scenario, or the effect that it would have on the math.

The way we have been playing lately though takes that scenario out of the equation. Basically, since the scores have gotten better, and harder to beat, we roll all the dice over and over until we hit a roll with five matches. We then separate the five (or more) dice and continue with the second roll with the remaining dice.

So, we never get locked in. I am not sure what a binomial distribution is, but I appreciate the help. Thanks for verifying the chance of me losing the current record.

I didn't want to quote the whole post with the Markov transition matrix, but that helps immensely whosnext.

It makes it very easy to see the odds of moving from one outcome to the next. That will be very handy next time we play the game.

I am surprised how seemingly easy it is to get the Yahtzee in three rolls (I now fear more for my record being tied.) Lol.
I thought it would be more difficult to attain.

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I will undoubtedly have further questions, but right now I am going to take a little time to go over the links and answers provided. Right now I feel a little like I did in high school, looking in the back of the book for the answers, lol.

I appreciate the help, and will say that the questions I asked were answered completely. Thanks for the tutorial.

add my 2 cents.
I always late
found my Excel sheet for this and updated it.

Now in Google here for those that care

Google Worksheet Link

here is a transition matrix snapshot
(I use a UDF in Excel so it cannot raise the matrix in Google)
left it in for the probabilities.


seems to agree with other post.
enjoy
Happy New Years
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