# Finding out the probability of the first person to win

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.

• 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. Nov 21, 2018 at 10:48

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 $$\frac{1}{6}+(\frac{5}{6})^2\frac{1}{6}+...= \frac{1}{6}\sum_{k=0}^\infty{(\frac{5}{6}})^{2k}=\frac{1}{6}\frac{36}{11}=\frac{6}{11}$$.

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:$$p=P(W)=P(D=1)P(W\mid D=1)+P(D\neq1)P(W\mid D\neq1)=\frac16\cdot1+\frac56\cdot(1-p)$$

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 $$\frac{6}{11}$$.

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