# a sequential game of dice

consider the following game: 10 dice are tossed and those showing 3 are more are retained. [those showing 2 or less are discarded.] the remaining dice are tossed again and those showing 4 or more are retained. finally, the remaining dice are tossed once more and those showing 5 or more are retained. if $Z$ denotes the number of dice retained after the third round of play, what is the probability that $Z = 0$?

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The probability that a given die survives all three rounds is $(4/6)(3/6)(2/6) = 1/9$. Thus the probability that no dice survive is $(8/9)^{10} \approx 31\%$.

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yuval's answer shows that what is involved here is a triply-compounded binomial distribution. the usual setting is two stages: generate X ~ bin(n, $p_1$) and then Y ~ bin(X, $p_2$). so marginally, Y ~ bin(n, $p_1p_2$). – ronaf Sep 8 '10 at 3:02

For a much lengthier approach, the problem can be decomposed into three smaller problems: the chance of discarding all the dice in one round, two rounds, or three rounds.

The chance of discarding all the dice in the first round is

$$(2/6)^{10}$$

Your chance of having $n$ dice remaining after the first round is $(2/6)^{10-n} (4/6)^n$, and your chance of discarding all the dice at this stage are

$$\sum_{i=1}^9 (10)choose(n)(2/6)^i (4/6)^{10-i} (3/6)^{10-i}$$

where $i$ is the number of dice eliminated in the first pass.

Finally, your chance of having $m$ dice remaining after the second round is $(10)choose(n)(2/6)^{10-n} (4/6)^n (n)choose(m)(3/6)^{n-m} (3/6)^m=(2/6)^{10-n} (4/6)^n (3/6)^n$, and your chance of discarding the remaining dice at this stage is

$$\sum_{i=1}^9 \sum_{j=1}^{9-i} (10)choose(i)(2/6)^i (4/6)^{10-i} (10-i)choose(10-i-j)(3/6)^i (4/6)^{10-i-j}$$

where $i$ is the number of dice eliminated in the first pass and $j$ is the number of dice eliminated in the second pass.

The sum of those three probabilities gives you the answer.

For me, it works out to

Removing all dice in first: 0.00001694

Removing all dice in second: 0.01732459

Removing all dice in third: 0.29060462

Removing all dice in any round: 0.30794615

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unkz's method is of course correct. and if one needed that final number .307..., this gets it. but [alas], like many computationally obtained numerical results, the method gives no insight as to what the number actually represents. – ronaf Sep 8 '10 at 3:12