The Riddler Express

Question

If a baseball team is truly .500, meaning it has a 50 percent chance of winning each game, what’s the probability that it has won two of its last four games and four of its last eight games?

Answer 1

One way to answer this question is by recognizing that the number of games the team has won can be represented by a random variable with a binomial distribution. A random variable X with a binomial distribution represents the number of successes in n independent trials where each trial has probability of success p. The probability of k successes out of the n independent trials for k = 1,....,n is given by:

P(X = k) = _nC_k p^n-k(1-p)^n-k where _nC_k = n!/k!(n-k)!

This is also known as a probability mass function. To say a random variable X has a binomial distribution is to say it has this probability mass function. Sometimes we write X ~ Bin(n, p) to indicate that X has a binomial distribution.

For this problem, we can assume the baseball team's chances of winning each game are independent, and then the games they play can be considered independent trials. Winning a game can be considered a success, with probability p = 0.5. So, if X represents the number of games the team won out of the last n, then X ~ Bin(n, 0.5).

Now, we're ready to calculate the probability we're interested in! That is, we want to find the probability of the event that the team won two out of its last four games, and four out of its last eight. With a bit of thinking, we can see that this is the same as the probability that the team won two games out of four, and that they won another two games out of four. Since we're assuming the games are independent, we have:

P(won 2 out of last 4 and 4 out of last 8)

= P(won 2 out of last 4 and 2 out of 4 before that)

= P(won 2 out of 4) P(won 2 out of 4)

= P(won 2 out of 4)²

and we recognize that P(won 2 out of 4) = P(X = 2), where X ~ Bin(4, 0.5). Using the probability mass function from above, we find

P(X = 2) = ₄C₂ (0.5)²(0.5)²

= 4!/(2!2!) (0.5)⁴

= 3/8

So our answer is

P(won 2 out of 4)² = (3/8)²

= 27/64

≈ 0.42

Additional remarks:

(1) This problem can be solved without knowledge of random variables but its a bit more cumbersome and difficult to do so. Recognizing a quantity can be represented by a random variable is a powerful strategy you can use to solve some probability problems.

(2) If you know some set theory and the set theoretic definition of probability, we're considering a sample space where each outcome is a string of 8 wins or losses, for example one such outcome is (W, W, L, L, L, L, W, W). Then we're looking for P(A) where

A = {the team won 2 out of its last 4 games and 4 out of its last 8}.

Then we recognize A = B ∩ C.

B = {the team won 2 out of first 4 games}

C = {the team won 2 out of next 4 games}

And finally since these events are independent ,

P(A) = P(B ∩ C)

= P(B)P(C)

= P(X = 2)²

where X ~ Bin(4, 0.5).