A and B are playing a game by alternately rolling a die, with A starting first. Each player’s score is the number obtained on his last roll. The game ends when the sum of scores of the two players is 7, and the last player to roll the die wins. What is the probability that A wins the game


On the first roll it is impossible for A to win. on the other hand B can win as long as the number on A's die and the number on B's die adds up to seven. The combinations which allow B to win are: $$ S=\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\} $$ which is 6 out of a total of 36 combinations so the probability of B winning in the first round is $ \frac{1}{6} $ and the probability of A winning is 0.

In the second roll of A a has a winning probability of $ \frac{1}{6} $ using the same logic as for B in roll1 and in the next roll of B again B has a winning probability of $ \frac{1}{6} $. I can continue to do the above on and on but it will only cycle back to the first scenario so i stop here.

Now if B wins in round 1 then A loses and can not proceed to the second roll. Therefore the probability of a winning is probability of B not winning in the first round into the probability of A winning in the next, which is $$ (1- \frac{1}{6}) (\frac{1}{6}) $$ which is $\frac{5}{36}$

Unfortunately this is not the correct answer as per the answer key. Can someone please point out where I have gone wrong. Any help is appreciated.Thanks :)

EDIT: $$ $$ on any next attempt the probability of A winning can be written as $(\frac{5}{6})^2$ of the previous since there is a $(\frac{5}{6})$ probability that A will not win and a $(\frac{5}{6})$ chance that B will not win hence leading to the next attempt.

thus the answere is $(\frac{1}{6}) ((\frac{5}{6})^2+(\frac{5}{6})^4+...)$ which is equal to $\frac{1}{6} (\frac{1}{1-(\frac{5}{6})^2)})$ which is 6/11 and the right answer. EDIT $$ $$ thus the answere is $(\frac{5}{36}) ((\frac{5}{6})^2+(\frac{5}{6})^4+...)$ which is equal to $\frac{1}{6} (\frac{1}{1-(\frac{5}{6})^2)})$ which is 5/11


Your value of $\frac{5}{36}$ gives the probability of A winning on A's second roll. But of course there may be later chances for A to win.

  • $\begingroup$ If i try to consider all possible scenarios i will become very complicated. Is there any way to simplify it down? $\endgroup$ – Shash Jun 8 '18 at 15:26
  • $\begingroup$ One possibility would be to show that the probabilities of A winning on various rolls form a (convergent) geometric series. $\endgroup$ – paw88789 Jun 8 '18 at 15:28
  • $\begingroup$ Thank you! that helped $\endgroup$ – Shash Jun 8 '18 at 15:40

In your last equation, the factor $1/6$ should be the $5/36$ that you found above. This gives $5/11$ not $6/11$. You can get the answer without using infinite series, as follows: Let $p$ be the probability that A wins. Consider the situation after B's first roll. B wins $1/6$ of the time. If this doesn't happen then B is in the situation that A was in at the start. So B wins the game with probability $1/6 + (5/6)p$. But B's probability is also $1-p$. This gives $p = 5/11$.

  • $\begingroup$ I have changed the solution,thank you for your help! $\endgroup$ – Shash Jun 9 '18 at 0:52

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