[Math] 2 out of 3 or 4 out of 6 free-throw problem – how does it mathematically make sense

expectationprobability

Question:

Suppose you are shooting free throws and each shot has a 60% chance of going in (there is no "learning" effect and "depreciation" effect, all have the some probability no matter how many shots you take).

Now there are two scenarios where you can win $1000

  • Make at least 2 out of 3

  • Make at least 4 out of 6

Which do you choose?

Thought Process:

I understand that this problem depends on how good you are at shooting free throws. If your free throw percentage (60%) is less than the expected probability (2/3 = 66.6%) for large numbers you will likely be closer to the expected probability of 60% and less likely to be closer to the desired number of 66.6%.

Assume 60% chance for a free throw:

  • $(0.6^2)(0.4){3 \choose 2} = 43.2\%$

  • $(0.6^4)(0.4^2){6 \choose 4} = 31.1\%$

Therefore you are more likely to hit 2/3 than 4/6 if your free throw percentage is 60%. I understand how this works but now,

let's say you are an 80% chance free throw shooter:

Assume 80% chance for a free throw:

  • $(0.8^2)(0.2){3 \choose 2} = 38.4\%$

  • $(0.8^4)(0.2^2){6 \choose 4} = 24.58\%$

Shouldn't I have a higher probability of 4/6 than 2/3 according to this logic? Why isn't my math working out?

Best Answer

You are not accounting for the at least part of the question. For example, your $(0.6^2)(0.4)\binom{3}{2}$ is the probability of getting exactly 2 out of 3 in. You must also include the chance of getting all 3 in.

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