# For any given set, the probability that Jagdit wins the set is $2/3$. Find the probability that the player who wins the first set wins the match.

I am learning probability. There is a question I really don't understand:

Jagdit and Kamal are playing a badminton match. The first player to win $2$ sets wins the match.

For any given set, the probability that Jagdit wins the set is $2/3$. Find the probability that the player who wins the first set wins the match.

I tried to find the probability of $P(www), P(wwl), P(wlw)$ and my answer is $16/27$. Am I wrong? If so, what is the correct answer?

• You did fine on finding Jagdit's side, but missed out Kamal's side. Answers have given the complete working. – Nij Jul 4 '16 at 12:07

The probability Kamal wins the 1st and the 2nd is $$(\frac{1}{3})\times(\frac{1}{3})=1/9$$

The probability Kamal wins the 1st and the 3rd is $$(\frac{1}{3})\times(\frac{2}{3})\times(\frac{1}{3})=2/27$$

The probability Jagdit wins the 1st and 2nd is $$(\frac{2}{3})\times(\frac{2}{3})=4/9$$

The probability Jagdit wins the 1st and 3rd is $$(\frac{2}{3})\times(\frac{1}{3})\times(\frac{2}{3})=4/27$$

Therefore the probability that the player who won the first set wins the match, is the sum of these four:

$$(\frac{1}{9})+(\frac{4}{9})+(\frac{6}{27})=\frac{7}{9}$$

It's good practice to check your answer, so let's calculate the probability that the person losing the first set wins the match; it should equal 1- the answer.

The probability Kamal loses the first, but wins the 2nd and 3rd is $$(2/3)\times(1/3)\times(1/3)=\frac{2}{27}$$

The probability Jagdit loses the first but wins the 2nd and 3rd is $$(1/3)\times(2/3)\times(2/3)=\frac{4}{27}$$

The sum of these is $$\frac{6}{27}=\frac{2}{9}$$ which does indeed equal $1-\frac{7}{9}$, confirming our answer.

• "The first player to win two sets", not " the player to win the first two sets". – Nij Jul 4 '16 at 11:53
• Ah good point, I will correct. – samerivertwice Jul 4 '16 at 11:53
• Don't rush to downvote! – samerivertwice Jul 4 '16 at 11:54

You are looking for the sum of probabilities of Jagdit winning the match if Jagdit won the first set, and of Jagdit losing the match if Jagdit lost the first serve.

The probability of Jagdit ("J") winning the first set ("$S_1$") is $\frac{2}{3}$. Now there are two "good legs" further from here: either J wins $S_2$ or J loses $S_2$ but wins $S_3$. That is, one leg has $P=\frac{2}{3}$ and the other, $P=\frac{1}{3}×\frac{2}{3}$, for a total of $\frac{8}{9}$. As this relied on J initially winning $S_1$ the true probability of J winning the match without winning $S_2$ along the way, is $\frac{16}{27}$.

That's not all though. What if Kamal ("K") had won $S_1$?! There's some chance that K wins the match too! And in fact, for J to win the match instead, given that K won $S_1$, J must win both of $S_2, S_3$ - and this occurs with probability $(\frac{2}{3})^2=\frac{4}{9}$. Then the odds of K winning at least one of the next sets, and therefore the match, is $\frac{5}{9}$.

Again, this was conditional, so the true probability is $\frac{5}{27}$.

Stick them together and we get $\frac{21}{27}$ as the probability that whoever wins first, also wins overall!