Probability of a dice game Here is my question... n players playing a dice game which any of them who trows a 6 the game is over and he wins. They do it 1 by 1 .So what is the probabilty of winning for any 1 of them ?
My calculation about it is like
 $$ P =\left(\dfrac1{6} \right)^{n} $$ 
Is that right ?
 A: Here is another way: let $p_k$ denote the probability of the $k$th player winning. Then $p_k=\frac{5}{6}p_{k-1}$ for $1<k\le n$. Now use $\sum_{k=1}^np_k=1$.
A: Hint:
First round: player 1 wins with probability $\frac{1}{6}$, player 2 with $\frac{5}{6}\cdot\frac{1}{6}$, $\cdots$, player n with $\left(\frac{5}{6}\right)^{n-1}\cdot\frac{1}{6}$.
Second round: player 1 with $\left(\frac{5}{6}\right)^n\cdot\frac{1}{6}$, player 2 with $\left(\frac{5}{6}\right)^{n+1}\cdot\frac{1}{6}$, and so on...
The winning probability of each given player equals the sum of his winning probabilities in each round.
A: There are $n$ players each with $p=\frac{1}{6}$ chance of winning on their roll.
After 1 round there is a $(\frac{5}{6})^n$ chance that the game has not ended and the situation is exactly the same as it was at the start of the game. Therefore we only need to normalise the probabilities over that first round by dividing by $1-(\frac{5}{6})^n$.
Player 1 has a $\frac{1}{6}\frac{1}{1-(\frac{5}{6})^n}$ chance.
Player 2 has a $\frac{5}{6}\frac{1}{6}\frac{1}{1-(\frac{5}{6})^n}$ chance.
In general, Player $k$ has a $(\frac{5}{6})^{k-1}\frac{1}{6}\frac{1}{1-(\frac{5}{6})^n}$ chance.
So for a 2 player game, Player 1 has $\frac{6}{11}$ and Player 2 has $\frac{5}{11}$.
