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In a best of 5 tennis match, Alex and Bob are equally likely to be the better players. If Alex (Bob) is the better player, he wins a set with probability 0.75 (0.75) independently of the outcome of the other sets. Three sets are played, and Alex leads with two sets to one. What is the probability that Alex will go on to win the match?

Here is an attempt and I will try show you all the steps:

Let $A = \{\text{Alex is the better player}\}$, $B = \{\text{Bob is the better player}\}$, $T = \{\text{Alex leads two sets to one}\}$, $W = \{\text{Alex wins the match}\}$.

From the problem statement, $P(A) = P(B) = 1/2$, and $$P(T) = P(T|A)P(A)+P(T|B)P(B) = (0.75\times 0.75 \times0.25) \times 0.5 + (0.25\times 0.25 \times 0.75) \times 0.5 = 3/32$$

We want to calculate:

$$P(W|T) = P(W|T,A)P(A) + P(W|T,B)P(B)$$

Now to calculate $P(W|T,A)$, this is just the event that Alex goes on to win the match given that he is the better player and he leads two sets to one. The possible outcomes here is that he wins the next set, or loses a set and wins a set, and so $P(W|T,A) = 0.75 + 0.25\times0.75 = 15/16$. Similarly, we have $P(W|T,B) = 0.25 + 0.75\times0.25 = 7/16$.

The final answer, therefore, is $P(W|T) = 11/16$.

Any feedback is appreciated

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Outline: The fact that Alex is leading is evidence he is the better player. We need to take that into account by doing a conditional probability calculation.

You averaged the two numbers $7/16$ and $15/16$, giving equal weight to each. But for example the weight in front of the $15/16$ should be the probability that Alex is the better player, given that he is leading. This is greater than $1/2$.

The required probability $\Pr(B|T)$ can be calculated using $\Pr(B|T)=\frac{\Pr(B\cap T)}{\Pr(T)}$.

For the weight in front of the $7/16$, one can do a similar calculation, but it is easier to note that it is $1-\Pr(B|T)$.

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