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?
Best Answer
The result is linked to the Law of large numbers, which basically states that the more trials you do of something, the closer you will get to the expected probability. So after 10,000 trials, I would expect to be proportionally closer to 6000 than being close to 60 after 100 trials.
The point here is that in all these cases the proportion is $\frac23$ - i.e. greater than the expected probability of 60%. Since for larger numbers we will be closer to the expected probability of 60%, that means that we are less likely to be above $\frac23$.
If you were to change the parameters slightly - and ask what is the probability of success in more than 55% of cases, for example, then you'd see the complete opposite happening.