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Two players are rolling two dices, if they get 6 Player A wins, if they get 7, player B wins, else they rolling the two dices again...

What is the probability that A will win?

I'd like to get any ides how to solve it...

Thank you!!

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up vote 3 down vote accepted

Here is a more complicated way to get to the result. This will broaden your knowledge:

Assume X is the possibility that player A wins. His favorable rolls are 5 (15,24,33,42,51) out of the 36 possible, 6 rolls will make him lose, while the rest 25 will make the game a tie so far and they will roll again. On the next roll the probability will be the same. So,





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I really like this solution – Cruncher Jul 14 '14 at 13:12

There are $11$ possible cases: $\{ (1;5), (2;4), (3;3), (4;2), (5;1), (1;6), (2;5), (4;3), (3;4), (2;5), (1;6) \}$. The first $5$ make $A$ win, the other $6$ make $B$ win.

Hence the probability that $A$ wins is $\frac{5}{11}$.

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I arrived at 5/11 a little bit differently. I know the probability of 6 is 5/36. Probability of 7 is 6/36. Together these only make 11/36 possible cases. So we need to scale each probability up by 36/11 which simplifies to 5/11. – Cruncher Jul 14 '14 at 13:10
@Cruncher,@Crostul: This method is incorrect. Consider the following game. On each round $n$ starting with $n=1$, a $2(n+1)^2$-sided dice is thrown. If it shows $1$, A wins. If it shows $2$, B wins. Otherwise the game continues. According to the above method, since in each round each of A and B are equally likely to win, thus for the whole game each will win with probability $\frac{1}{2}$. But that is wrong as the game will continue on forever with probability $\frac{1}{2}$, so each player will only win with probability $\frac{1}{4}$. The other answers don't have the same problem. – user21820 Jul 15 '14 at 13:08
@user21820 That is an entirely different game. The fact that it works with the other answers is irrelevant. In the current game, enumeration works. Period. The game goes on forever with probability 0 as long as you use the same dice on each iteration. And this works for any number of sides on the die. As long as its fixed. – Cruncher Jul 15 '14 at 15:52
@Cruncher: I didn't say whether the other answers work with the game I described. But we must show that the probability of the game ending is $1$, which many people do not realize is actually important for the reason I give. So this answer as written is not valid. – user21820 Jul 15 '14 at 16:21
@user21820 it's generally accepted that if each trial has a fixed, non-0 chance of ending, then the probability of ending is indeed 1. This "reroll" or "redo" logic, is intuitive, and widely used. In ANY case. The method is not wrong. At best you can say the justification is incomplete. But saying: "The probability of eventually rolling one of these cases is 1 as $\prod_0^\infty{25/36}=0$" – Cruncher Jul 15 '14 at 16:54

Here is one idea...

  • There are $5$ out of $36$ options to get $6$:
    • $1+5$
    • $2+4$
    • $3+3$
    • $4+2$
    • $5+1$
  • There are $6$ out of $36$ options to get $7$:
    • $1+6$
    • $2+5$
    • $3+4$
    • $4+3$
    • $5+2$
    • $6+1$
  • All the remaining $25$ options are neither $6$ nor $7$

So the probability that $A$ wins on attempt #$N$ is $\displaystyle{(\frac{25}{36})}^{(N-1)}\cdot\frac{5}{36}$

In order to complete the answer, as suggested by @VishwaIyer:

Calculate the probability that $A$ wins at some point in time, by adding up the probabilities:


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I would say that this answer be correct. – georg Jul 14 '14 at 10:44
@georg: Thank you :) – barak manos Jul 14 '14 at 11:12
@georg This answers a different question. And doesn't actually give a concrete answer to this question. – Cruncher Jul 14 '14 at 13:08
@Cruncher, his solution is correct in the fact that it's an infinite geometric series with first term $\frac{5}{36}$ and common ratio $\frac{25}{36}$. So the infinite series sum is $$S = \frac{\frac{5}{36}}{1-\frac{25}{36}} = \frac{5}{11}$$ Which in fact is the probability that $A$ will win. – Vishwa Iyer Jul 14 '14 at 15:34
@VishwaIyer: Thank you. I left this answer as "an idea", because I haven't been really able to explain why the sum of the infinite series is the answer, since each term in the series gives the probability that the event will occur "at that specific point" (i.e., only one term in the entire series "can actually occur"). – barak manos Jul 14 '14 at 15:41

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