# Where I am going wrong in this probability question?

1,2,3 are three players, two players are selected at random, they play and then the winner plays with the other player, and if he wins then he is tournament victor. Given, the probability with which i wins over j is i/i+j.Find the probability that 1 wins the tournament.

My approach: There are three possibilities with with 2 players out of 3 are selected. (1,2), (1,3), and (2,3). If (1,2) are selected first (with probability 1/3), then 1 wins with 1/1+2=1/3 probability. Now, 1 plays with 3 and wins with 1/1+3=1/4 probability. P(1) = 1/3*1/3*1/4 When (1,3) is selected, applying similar analogy, P(1) = 1/3*1/4*1/3 Adding both gives, P(1 wins) = 1/18.

## 1 Answer

You're missing the case where $$(2,3)$$ is chosen first and $$1$$ wins against whoever wins here.

• Do I have to consider two cases (3,2) and (2,3) for separate winners in this first group? – Roopesh Singh Jun 24 at 10:16
• @Mathenatic For the entire question: you need the cases $(1,2)$ where 1 wins against 2 and later against 3, $(1,3)$ where 1 wins against 3 and later against 2, $(2,3)$ where 3 wins against 2 and later loses to 1, and $(2,3)$ where 2 wins against 3 and later loses to 1. The final answer should be $\frac{1}{18}+\frac13 (\frac{2}{5}\frac{1}{3} + \frac{3}{5}\frac{1}{4})$ – Amit Rajaraman Jun 24 at 10:21
• What if it asks, 1 wins given that 1 didn't play first round. Then, P(1 wins | 1 did not play) = P(1 wins and didn't play first round)/ P( 1 did not play) = 17/60 / 1/3 = 17/20? – Roopesh Singh Jun 24 at 10:53
• @Mathenatic The probability that $1$ wins and didn't play the first round is $\frac{1}{3}(\frac25\frac13+\frac35\frac14)$ and the probability that $1$ didn't play the first round is $\frac13$. So, the conditional probability should just be $\frac{17}{60}$. – Amit Rajaraman Jun 24 at 10:57
• @Mathenatic I am reasonably sure that whatever I did is correct. – Amit Rajaraman Jun 24 at 11:10