TITLE: Free throw interview question
QUESTION [42 upvotes]: I recently had an interview question that posed the following... Suppose you are shooting free throws and each shot has a 60% chance of going in (there are no "learning" or "depreciation" effects, all have the some probability no matter how many shots you take).
Now there are three scenarios where you can win $1000

Make at least 2 out of 3
Make at least 4 out of 6
Make at least 20 out of 30

My initial thought is that each are equally appealing as they all require the same percentage of free throw shots.  However when using a binomial calculator (which this process seems to be) the P (X > x) seems to be the highest for scenario 1.  Is this due to the number of combinations?

REPLY [10 votes]: Because your expected number of successes is less than the number required in order to win, you need to get a bit lucky in order to succeed. That's much harder for large samples: if you had to get 2/3 of a thousand throws, you would be in very bad shape; whereas even getting 1/1 out of a single throw is a 60% chance.
So the intuitive explanation is: you lose in the long term, so you need variance to win. The smaller the sample size, the bigger the impact variance has.