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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 ?

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That's the probability when $n$ players throw $n$ dice and everyone wins. Each one plays in turns, if he wins the game's over(Fun and difficult part). – Inceptio Mar 24 '13 at 13:49
Is there a limited number of turns, or may the game go on forever (so long as noone rolls a $6$)? – Cameron Buie Mar 24 '13 at 13:52
The game keeps goin on untill 1 of the n players get 6 in their turn. – serkan Mar 24 '13 at 13:53
up vote 1 down vote accepted


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.

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If player one wins, game's up! How do you know about the $n$th player's probability of winning? – Inceptio Mar 24 '13 at 13:57
ahh so you say that i have to count the rounts too if n players cant get 6 in their turns. Aha got the message thx. – serkan Mar 24 '13 at 13:59
Given the information that player one wins, then his winning probability is 1 and others' 0. However we do not have this kind of information a priori. – Chen Yang Mar 24 '13 at 14:01
Consider a general case rather. If player $k$ wins. The probability of others winning is $0$ – Inceptio Mar 24 '13 at 14:02
Yeah, I agree with your conclusion. However this says that the conditional probability of others winning given the information that player k wins is 0. I think here we need unconditional probability. – Chen Yang Mar 24 '13 at 14:05

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 $\sum_{k=1}^np_k=1$.

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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 $\frac{5}{11}$.

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