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08-16-2002, 12:39 PM
I know that the probability of any one die hitting one number (assuming 'perfect dice') is 1/6. But how do you determine the odds of hitting a certain 'hand' in a game such as Yahtzee? The part that baffles me is the fact that there is a maxium of 15 rolls, a miniumum of 5, you may hit the Yahtzee on any of the rolls, you may not hit it at all, you may hit 3 the first time and toss 2 the second time, etc. Could someone help me out or direct me to a reference? Thank you.

08-16-2002, 03:50 PM
It depends exactly how you're choosing which dice to discard at each stage. If you take the strategy of simply going straight for yahtzee and nothing else (i.e. take the rank represented most and discard the rest), then it's doable. I'll just assume that's what you meant. The problem is kind of undoable without assuming some fixed strategy for discarding dice.

What I'm going to do is let Y(n,r) = probability of getting r occurences of a particular (fixed) die value, on a single roll of n dice. Once we calculate all the Y(n,r), the solution is much easier to see. I just chose 'Y' for 'yahtzee', no special reason.

First of all, how many ways are there to roll n dice a single time? We're considering n ordered rolls of a single die, with replacement. There are 6^n ways to do this. Out of these 6^n ways to roll n dice, how many ways can we get, say, exactly r 1's? (or 2's, 3's, etc.) We have to choose where those r 1's are going to go among the n. There are C(n,r) (n choose r) ways of doing this. Now, for these r dice, we have no choice...they're already 1. But for the others, we have a choice of 5 die values (everything BUT 1.) Thus, there is a total of C(n,r)*(5^(n-r)) ways of getting exactly r 1's. Thus,

Y(n,r) = C(n,r)*(5^(n-r))/(6^n)

We need to enumerate all the possible ways of getting ourselves up to a yahtzee in the 3 rolls, i.e. all increasing sequences from 0 to 5.

5

4,5

4,4,5

3,5

3,4,5

3,3,5

2,5

2,4,5

2,3,5

2,2,5

1,5

1,4,5

1,3,5

1,2,5

1,1,5

Suppose we're just interested in getting a yahtzee with 1's (forget about 2's, 3's for the moment). If we can calculate the probability of all the different scenarios above, then the sum would give us the probability of ending up with yahtzee 1's in the end. Then, we would just multiply this result by 6 to get the probability of a yahtzee of ANY die value. Let's go!

For '5', this is straightforward to compute, it's just

Y(5,5)

For '4,5', we have to get 4 of a kind, then 1 on the 2nd roll given the 4 of a kind on the first. The probability is therefore

Y(5,4)*Y(1,1)

For '4,4,5', we get

Y(5,4)*Y(1,0)*Y(1,1)

and so on, and so on,...

It's just a matter of carrying through the computations, now. Although, there may be a really good 'intuitive' way of seeing the solution without doing much computation, that I've just missed completely.

08-17-2002, 01:39 AM
Sorry, I goofed. This isn't quite right. It will give a rough number which should be pretty accurate, but it's not the exact probability. The reason is that I didn't account for the case of getting 2 of a kind on roll 1, and then the other 3 being a different 3 of a kind on roll 2, for example. Also, I counted twice when you get two pair, when each of these should only be counted once (22445, you can only count it as trying toward a 2-yahtzee or a 4-yahtzee, not both). The idea is the same, but I don't see any other way to find the exact probability other than enumerating all the possible cases in a decision tree and assigning probabilities in the way I did above. That's just a lot of number-crunching.

08-17-2002, 02:08 AM
I was going to mention that. Also, if you want to compute the probability of just getting a yahtzee in 1's assuming you are only going for 1s as you did, then this result can be computed much more easily as:

[1-(5/6)^3]^5

That is, each die has up to 3 rolls to get a 1, so the chance of not getting it is (5/6)^3, the chance of getting it is 1-(5/6)^3, and getting it on all 5 dice is this raised to the 5th power. This should make the other computations easier as well.

08-17-2002, 09:06 AM
There are 7776 possible rolls for first roll

Five numbers the same = 6 combinations

Four = 6*5*5 = 150

Three = 6*25*6 = 900

Two = 6*100*10 = 6000

All different 6.5.4.3.2= 720

If two or more the same we have 11/36 chance of each dice pairing the ones we keep in the next two rolls.

Chances of winning in this scenario are:

6/7776 + (11/36)(150/7776) + (11/36)^2(900/7776)

+(11/36)^3(6000/7776)

=3.95%

If all different, we throw again, same combinations, but now we only have 1/6 for last roll, so

720/7776 of the time we get:

6/7776 +(1/6)(150/7776) + (1/6)^2(900/7776)+

(1/6)^3(6000/7776)=1.08% but we need to multiply

through by 720/7776

which is approx 0.01% (i lost figures in rounding)

and the remaining chance is

720/7776 * 720/7776 *6 /7776 = not much

total therefore is 3.96% +/- 0.01% due to operators lack of calculator skills

08-17-2002, 12:48 PM
Ok everyone, thanks for the responses.

08-19-2002, 08:23 AM
I don't think it will matter to 2dp but I have failed to consider the rolls when you start with two of a kind and roll three of a different kind.

just so the problem can be completed you would need to add these in.

Sorry