You roll ten dice and record the maximum number $x$ of matchings. What is the probability that that $X=x\in[1,10]$?

Question

This question is inspired by a dice game called "Tenzi". You throw ten dice and try to achieve the maximum number of matching face values of the dice. What is the probability that a maximum matching of size $x$ occurs?

For $x\in[6,10]$, this is easy. The number of ways this can occur is $6{10\choose x}5^{10-x}$, i.e., selecting the face that gets matched (6 ways), choosing which of the $x$ dice get matched, and finally selecting all possible ways the remaining dice get matched.

When we get to the case where $x=5$, this method overcounts when we have two matchings of five dice. So, the number of ways this can occur is $6{10\choose 5}(5^5 - 5)$. The subtraction of the five at the end is because of the five possible cases where we have another matching of five dice resulting in the same value of $x$.

This suddenly gets very tricky when $x=4$. I'm thinking the inclusion/exclusion but I don't really know how to implement that. We must guarantee that no other matchings of size 4 or larger occur.

Obviously, $P(X=1)=0$, but for $x=2,3,4$, I don't know what to do.

Any ideas?

Answer 1

For $x=2$, you need to split it into disjoint events, and then add up their probabilities:

• The probability of $[2,2,2,2,2 ]$, which is $\dfrac{\frac{(2+2+2+2+2 )!}{2!\times2!\times2!\times2!\times2! }\times\frac{(5 )!}{5! }\times\binom{6}{5}}{6^{10}}$
• The probability of $[2,2,2,2,1,1]$, which is $\dfrac{\frac{(2+2+2+2+1+1)!}{2!\times2!\times2!\times2!\times1!\times1!}\times\frac{(4+2)!}{4!\times2!}\times\binom{6}{6}}{6^{10}}$

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For $x=3$, you need to split it into disjoint events, and then add up their probabilities:

• The probability of $[3,3,3,1 ]$, which is $\dfrac{\frac{(3+3+3+1 )!}{3!\times3!\times3!\times1 }\times\frac{(3+1 )!}{3!\times1 }\times\binom{6}{4}}{6^{10}}$
• The probability of $[3,3,2,2 ]$, which is $\dfrac{\frac{(3+3+2+2 )!}{3!\times3!\times2!\times2 }\times\frac{(2+2 )!}{2!\times2 }\times\binom{6}{4}}{6^{10}}$
• The probability of $[3,2,2,2,1 ]$, which is $\dfrac{\frac{(3+2+2+2+1 )!}{3!\times2!\times2!\times2!\times1 }\times\frac{(1+3+1)!}{1!\times3!\times1}\times\binom{6}{5}}{6^{10}}$
• The probability of $[3,3,2,1,1 ]$, which is $\dfrac{\frac{(3+3+2+1+1 )!}{3!\times3!\times2!\times1!\times1 }\times\frac{(2+1+2)!}{2!\times1!\times2}\times\binom{6}{5}}{6^{10}}$
• The probability of $[3,3,1,1,1,1]$, which is $\dfrac{\frac{(3+3+1+1+1+1)!}{3!\times3!\times1!\times1!\times1!\times1}\times\frac{(2+4 )!}{2!\times4 }\times\binom{6}{6}}{6^{10}}$
• The probability of $[3,2,2,1,1,1]$, which is $\dfrac{\frac{(3+2+2+1+1+1)!}{3!\times2!\times2!\times1!\times1!\times1}\times\frac{(1+2+3)!}{1!\times2!\times3}\times\binom{6}{6}}{6^{10}}$

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For $x=4$, you need to split it into disjoint events, and then add up their probabilities:

• The probability of $[4,4,2 ]$, which is $\dfrac{\frac{(4+4+2 )!}{4!\times4!\times2! }\times\frac{(2+1 )!}{2!\times1! }\times\binom{6}{3}}{6^{10}}$
• The probability of $[4,3,3 ]$, which is $\dfrac{\frac{(4+3+3 )!}{4!\times3!\times3! }\times\frac{(1+2 )!}{1!\times2! }\times\binom{6}{3}}{6^{10}}$
• The probability of $[4,4,1,1 ]$, which is $\dfrac{\frac{(4+4+1+1 )!}{4!\times4!\times1!\times1! }\times\frac{(2+2 )!}{2!\times2! }\times\binom{6}{4}}{6^{10}}$
• The probability of $[4,3,2,1 ]$, which is $\dfrac{\frac{(4+3+2+1 )!}{4!\times3!\times2!\times1! }\times\frac{(1+1+1+1)!}{1!\times1!\times1!\times1!}\times\binom{6}{4}}{6^{10}}$
• The probability of $[4,2,2,2 ]$, which is $\dfrac{\frac{(4+2+2+2 )!}{4!\times2!\times2!\times2! }\times\frac{(1+3 )!}{1!\times3! }\times\binom{6}{4}}{6^{10}}$
• The probability of $[4,3,1,1,1 ]$, which is $\dfrac{\frac{(4+3+1+1+1 )!}{4!\times3!\times1!\times1!\times1! }\times\frac{(1+1+3 )!}{1!\times1!\times3! }\times\binom{6}{5}}{6^{10}}$
• The probability of $[4,2,2,1,1 ]$, which is $\dfrac{\frac{(4+2+2+1+1 )!}{4!\times2!\times2!\times1!\times1! }\times\frac{(1+2+2 )!}{1!\times2!\times2! }\times\binom{6}{5}}{6^{10}}$
• The probability of $[4,2,1,1,1,1]$, which is $\dfrac{\frac{(4+2+1+1+1+1)!}{4!\times2!\times1!\times1!\times1!\times1!}\times\frac{(1+1+4 )!}{1!\times1!\times4! }\times\binom{6}{6}}{6^{10}}$

Answer 2

Rolling $10$ dice at once produces a string of length $10$ over the alphabet $[6]$. There are $6^{10}$ such strings, all of them equiprobable. Each string induces a partition of $10$ into $\leq6$ parts. We are interested in the probability distribution $(p_k)_{1\leq k\leq10}$ of the size of the largest part. In order to determine the $p_k$ one has to count the frequencies of all $35$ such partitions. The OP and barak manos have indicated how to do this, but have not provided final numerical results. Here they are:

A common strategy for this game is to keep the number showing up the most on the first roll. Assume this is the three. Denote by $E_n$ the expected number of additional rolls if we need $n$ more threes. Then $E_0=0$. Furthermore the number of threes in a simultaneous roll of $n$ dice is binomially distributed. This implies that the $E_n$ satisfy the following recursion: $$E_n=1+\sum_{k=0}^n{n\choose k}\left({5\over6}\right)^{n-k}\left({1\over6}\right)^k\>E_{n-k}\ .$$ Solving for $E_n$ gives $$E_n={1\over6^n-5^n}\left(6^n+\sum_{k=1}^n{n\choose k}5^{n-k}\>E_{n-k}\right)\qquad(n\geq1)\ .$$ Doing the numerical calculations results in the following table for the $E_n$:

From the above two tables we now can compute the overall expected number $R$ of rolls using this strategy: $$R=1+\sum_{k=2}^{10} p_k\>E_{10-k}=15.3485\ .$$We have not proven that this is the best strategy. In fact one might think of repeating the first roll of all ten dice until the largest resulting part is $>2$. This modified strategy produces a marginally better result: $$R'={1\over 1-p_2}+\sum_{k=3}^{10}{p_k\over 1-p_2}\>E_{10-k}=15.3442\ .$$