We have 3 players, one of them wins if he gets a $1$ throwing one die, what is the probability that the first player wins?

Question

K is the "rounds of turns" (sorry I'm translating from another language).For example, if k=2 the player 1 throw the die, then the player does it, k=3 the same etc. The dies aren't thrown simultaneously but by turns. The first player to throw a 1 wins. So they will play until one player throw a o1 then they stop.

The probability that at any turn, the first player wins, is

$$\left(\dfrac{5}6\right)^{3k-3}*\left(\dfrac{1}6\right)\\$$

For two players, the probability that the first player wins is

$$\left(\dfrac{5}6\right)^{2k-2}*\left(\dfrac{1}6\right)\\$$

Keeping two players, the probability that the second player wins is

$$\left(\dfrac{5}6\right)^{2k-1}*\left(\dfrac{1}6\right)\\$$

Am I right or am I missing something?

Thanks.

Answer 1

You seem to be stating the probability of winning at the $k$th turn. If you want the probability $P_n$ that the nth player (out of 3) wins, you need to sum over all $k$.

$P_1=\frac{1}{6}\sum_{k=0}^\infty (\frac{5}{6})^{3k}=\frac{36}{91}$. Similarly $P_2=\frac{30}{91}$ and $P_3=\frac{25}{91}$

Answer 2

I like to approach these problems as follows:

Let $p_i$ be the probability that Player $i$ wins. If Player $1$ doesn't win on the first roll, then the probability that Player $2$ wins will now be $p_1$ (because Player $2$ is now in the position Player $1$ occupied). Similarly, if neither of the first two players win on their first rolls, then the probability that Player $3$ wins will be $p_1$.

Thus, $p_2= \frac 56 p_1$ and $p_3 = \frac 56 p_2 = \frac {25}{36} p_1$. We also know that $p_1+p_2+p_3=1$. These equations yield the solution $p_1= \frac {36}{91}$.

In general, this reasoning shows that if there are $k$ players and each player's probability of winning in a single turn is $x$, then $p_1= \frac{x}{1-(1-x)^k}$ (and $p_n= (1-x)^{n-1}p_1$).