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Suppose we are playing a game where we start with six dice and we roll all the dice simultaneously. After the roll, if any of the dice is a duplicate of another we remove it entirely from the game. For example, if we roll 1 1 1 2 2 5, we would remove three dice: 1 1 and 2. After removing the applicable dice (or not removing any at all), we grab the remaining dice and roll again.

What is the expected number of rolls before we arrive at only one dice?

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  • $\begingroup$ This is a markov chain problem, but I'm not sure how best to compute the transition probabilities. $\endgroup$ – Thomas Andrews Sep 28 '16 at 20:51
  • $\begingroup$ (Not a solution.) Consider the $6\times6$ matrix $A$ defined by $A(i,i)=i/6$ for every $1\leqslant i\leqslant6$, $A(i,i+1)=1-i/6$ for every $1\leqslant i\leqslant5$, and $A(i,j)=0$ for every other $(i,j)$, then the number of remaining dice performs a Markov chain starting from $6$ with transition probabilities $q(i,j)=A^{i-1}(1,j)$ and the number $N$ of rolls to arrive at one unique die is such that $P(N\leqslant n)=q^n(6,1)$ hence $E(N)=\sum\limits_{n\geqslant1}(1-q^n(6,1))$. $\endgroup$ – Did Sep 28 '16 at 22:15

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