## Thursday, April 16, 2015

### Expected Points and Variance of Points From Bonus Opportunities -- Post Three.

This is the third of three statistics problems on bonus free throws.  For all three problems go to the link below.

http://fractiontips.blogspot.com/2015/04/the-statistics-of-bonus-free-throws.html

Question:   What is the expected number and variance of free throw makes for a person with 5 bonus free throw opportunities and for a person with 10 regular (not in bonus) free throw opportunities?    Provide calculations for a 60 percent free throw shooter and a 90 percent free throw shooter?

Answer:   A previous post found the probability distribution function for a person who had one bonus free throw situation.

I used the PDF to calculate expected makes and variance of makes at 0.6 and 0.8.

The numbers in the chart for p=0.6 and p=0.8 are repeated below for your convenience.

 Calculation For Bonus Situation # Makes Prob (Make=k) p=0.6 p=0.8 0 (1-p) 0.4 0.2 1 (1-p) x p 0.24 0.16 2 p x p 0.36 0.64 E(Makes) 0.96 1.44 E(Makes^2) 1.68 2.72 Var(Makes) 0.7584 0.6464

The outcomes from the bonus free throw situations are independent and identically distributed.

This means the expected outcome of the sum of n situations is simply the sum of the expected outcome of all situations.

This also means that the variance of the sum of n situations is simply the sum of the variance of each situation.

The sum of n identical items is the same as n multiplied by one of the items.

Using these insights I create answers presented in chart below.

 Expected Makes and Variance of Makes for 5 or 10 Bonus Situations Five Bonus Situations Prob makes shot 0.6 0.8 Expected Make Per Bonus Situation 0.96 1.44 Variance One Bonus Situation 0.75 0.6464 Expected Number Makes 4.8 7.2 Variance Makes 3.75 3.232 Ten Bonus Situations Prob makes shot 0.6 0.8 Expected Make Per Bonus Situation 0.96 1.44 Variance One Bonus Situation 0.75 0.6464 Expected Number Makes 9.6 14.4 Variance Makes 7.5 6.464

Concluding Thoughts

The expected number of points from a bonus free throw situation is 1.5 times higher for a 80 percent free throw shooter than a 60 percent free throw shooter.

The variance of number of makes is around 14 percent lower for the 80 percent shooter than the 60 percent shooter.

Note from previous post the probability distributions function for the bonus free throw situation has a small right tail when the number of bonus opportunities rises.   The probability of making 10 shots on 5 bonus opportunities is p10.  The high variability for p=0.6 means that a lot of misses are likely to occur.

A useful exercise for a motivated student might be to calculate the expected value and variance of points from a bonus situation as a function in terms of p the probability of making one shot and then solving for expected points and variance of points for n bonus opportunities.

People who liked this post may also like my book on statistics and baseball.