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Two people are playing a dice, they take turns to roll a dice and if one person rolls a "1", the game ends and the person who rolls a "1" wins. If he does not roll a "1", then the game continues until one person roll a "1", what is the probability of the first person to win?

I have no idea how to calculate this.

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  • $\begingroup$ Consider the sequence of $X_1, \dots$, where $X_i$ is the $i$-th roll, $X_i=1$ if the roll is 1 and $0$ otherwise. Then the first $i$ such that $X_i$ is following a geometric distribution $Y$. If $Y$ is odd, the player who started rolling wins; if $Y$ is even, the other player wins. $\endgroup$ – Stockfish Nov 21 '18 at 10:48
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Assume the dice is a fair 6 sided dice, and let's use the string with alphabet in $\{1,2,3,4,5,6\}$ to denote the outcome of the roll. The first person win exactly when the string representing the outcome is of the form $1$ or $ \_ ,\_1$ or $\_,\_,\_,\_1$ and so on where $\_$ are numbers in the set $\{2,3,4,5,6\}$. So now, that happens with probability $\frac{1}{6}+(\frac{5}{6})^2\frac{1}{6}+...= \frac{1}{6}\sum_{k=0}^\infty{(\frac{5}{6}})^{2k}=\frac{1}{6}\frac{36}{11}=\frac{6}{11}$.

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It is not really necessary to apply geometric distribution together with infinite sums.

If $p$ denotes the probability of the first player to win then we have the equality:$$p=\frac16+\frac56(1-p)=1-\frac56p$$leading to $$p=\frac6{11}$$

Concerning the first player observe that by not throwing a $1$ his opponent will have probability $p$ to win so that in that case the "new" chance to win will be $1-p$.


Formally if $W$ denotes the event that the first player wins, and $D$ denotes then the number rolled at the first throw then:$$p=P(W)=P(D=1)P(W\mid D=1)+P(D\neq1)P(W\mid D\neq1)=\frac16\cdot1+\frac56\cdot(1-p)$$

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Using $p = 1/6$, we have $P($the first player wins$)$ = $\sum_{i=0}^{\infty} P($the first 1 is in throw $ 2i+1) = \sum_{i=0}^{\infty} (1-p)^{2i}p = \frac{1}{6}\frac{36}{11}=\frac{6}{11}$.

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Notice that the game stops at the first instance of a $1$, which implies that all previous rolls must have given one of $\{2,3,4,5,6\}$'s. Also notice that the person who rolls first only rolls on odd instances. This means that what you need to find is the probability of the first $1$ being on an odd roll. So the first $1$ appears on one of the first, third, fifth... and so on. Note that the result is a geometric series with a ratio of $\frac{25}{36}$, and the first term as $\frac{1}{6}$. From here, you can just compute the infinite sum to be $\frac{6}{11}$.

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