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We have a two player game with this rules:

First player throws two dice and the bigger number will be his number.

Second player throws only one die.

If the number of first player was bigger than that of the second player, the first player wins; if not, the first player loses.

For example if first one have 2 and 4 and second have 4,first one loses and second one wins.

Please help me finding probability of winning of first player.

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closed as off-topic by Jonas Meyer, Micah, Claude Leibovici, Normal Human, Zachary Selk Jul 18 '15 at 4:34

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up vote 6 down vote accepted

You care most about the maximum of the first roll. This will be 1 with probability 1/36. It will be 2 with probability 3/36 etc. (it will be n with probability (2n-1)/36)

The first player will win if the second player throws less than his max.

$\large P=\frac{1}{36}\frac{0}{6}+\frac{3}{36}\frac{1}{6}+\frac{5}{36}\frac{2}{6}+\frac{7}{36}\frac{3}{6}+\frac{9}{36}\frac{4}{6}+\frac{11}{36}\frac{5}{6}=\frac{125}{216}$

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We find the probability that the second player B wins.

If B tosses a $6$, she wins with probability $1$.

If B tosses a $5$, she wins with probability $\left(\frac{5}{6}\right)^2$. For then B wins precisely if the first player A gets $\le 5$ twice in a row.

If B tosses a $4$, she wins with probability $\left(\frac{4}{6}\right)^2$.

And so on.

So the probability B wins is $$\frac{1}{6^3}\left(6^2+5^2+4^2+3^2+2^2+1^2\right)=\frac{91}{216}.$$

Thus our required probability is $1-\frac{91}{216}$.

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