Thread: Steaks in sports: calculating the expected number of streaks

Having some trouble with the material, so I just wanted to double check this question.

A basketball player makes 95% of his free throws. If he attempts 350 free throws in a season, what is the expected number of streaks of 50 or more consecutive free throws made?

I used a formula (p^k)(1+(n-k)q), where I figured p= .95, q= 1-p = .05, n = 350, k = 50. The answer I got was E(k) ≈ 1.23. Did I go about this correctly?

Hey MN1987.

You can't use the binomial exactly if you are looking at consecutive streaks since the binomial only looks at getting n successes and it doesn't care about the order of the way that you get it.

Basically you need to consider the number of possibilities of getting 50 or more consecutive free throws. Now with 350 you will have at maximum 6 sets of a streak of 50 at the most and one streak of 350 at the minimum (in terms of streaks).

Can you explain where you got that formula from and how it was derived?

I think MN1987's solution is correct. The formula he quotes is easy to derive using the indicator variable method and additivity of expectation, given the assumption that all the trials are independent. There has been considerable controversy about the validity of this assumption in sports.

I'll post the derivations here for commentary:

Expected number of streaks of at least k free throws from a total of n free throws:

g(x) = probability a streak lasting at least 'k' free throws ends in free throw 'x'

g(1) = g(2) = ... g(k) = 0
g(k+1)=g(k+2)=...=g(n)=p...pq=(p^k)q
g(n+1)=probability a streak of at least 'k' free throws includes free throw 'n' = p...p=p^k

Expected number of streaks of k free throws from a total of n free throws = (p^k)(1+[n-k]q)

Thanks for that MN1987.