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In a dice game, the player rolls three dice simultaneously, and then he may roll two more times any number of his dice(0, 1, 2 or 3). The player wins the game if all three dice have the same number on top after the last roll. What is the best strategy , and with that strategy what is the probability of winning?

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Do the two additional rolls have to include the same dice or the same number of dice, or can the player choose independently for the first and second additional roll which of the dice to reroll? –  joriki Mar 23 '13 at 7:13
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Assuming that the dice to reroll can be independently chosen in the two additional rolls, there is no single optimal strategy. An optimal strategy is one that leaves $2$ or $3$ equal dice unchanged and rerolls the remaining die in case of $2$ equal dice, and rerolls either all dice or all but one die if there are no equal dice.

After the first roll, the probabilities that $2$ or $3$ numbers are equal are $3\cdot6\cdot5/6^3=5/12$ and $6/6^3=1/36$, respectively, so the probability that none are equal is $1-5/12-1/36=5/9$.

If $3$ numbers are equal, we win. If $2$ numbers are equal, we have a $1-(5/6)^2=11/36$ chance of winning by rerolling the remaining die. If no numbers are equal, we might as well reroll completely, and then the probabilities are the same as on the first roll, but now we only have a $1/6$ chance of winning if two dice are equal on the second roll.

Thus the overall chance of winning is

$$ \frac1{36}+\frac5{12}\cdot\frac{11}{36}+\frac59\left(\frac1{36}+\frac5{12}\cdot\frac16+\frac59\cdot\frac1{36}\right)=\frac{2539}{11664}\approx0.2177\;. $$

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