## Thursday, April 16, 2015

### The difference between regular and bonus free throws -- post two.

Regular and Bonus Free Throws --- Post Two

Question:   Consider two basketball situations.   Situation one is a player with four guaranteed free throws.   Situation two is a player with two bonus free throw situations.  (Recall that under the bonus the player needs to make the first shot to get the second shot.)

What are the possible outcomes in terms of shots made and the likelihood of made shots in the two situations?   Evaluate these two situations under the assumption that each free throw is independent and identically distributed.   Evaluate the situations  for two free throw shooters, one with a sixty percent likelihood of making each shot and the other with a 80 percent success rate.

Analysis:

Four independent foul shots

When the shooter is not in the bonus outcomes are binomially distributed.

See Wikipedia for an explanation of the binomial distribution.

The binomial distribution is easily calculated in Excel.   In this problem we set the number of successes from 0 to 4, the success probability at 0.6 and 0.8, the number of trials at 4, and cumulative is set to false.

 Likelihoods for four Independent Free Throws 0.6 0.8 0 0.0256 0.0016 1 0.1536 0.0256 2 0.3456 0.1536 3 0.3456 0.4096 4 0.1296 0.4096 Check 1 1 Expected Makes 2.4 3.2 Variance Makes 0.96 0.64

Two independent shots in the bonus:

Outcomes range from 0 makes to 4 makes when a player has two bonus free throws.   However, two shots in the bonus is much more pressure than four guaranteed shots.   The player who misses the first shot does not get a second one.  Two misses on two first shots results in 0 makes.   By comparison, when the player gets four independent shots he or she has to miss all four to have 0 makes.

I derived the pdf for one bonus shot in a previous post.

Below I review this result.

There are three outcomes

Outcome one is miss on the first shot probability (1-p).

Outcome two is make first shot miss on second shot probability (1-p)*p.

Outcome three is make on first and second shot probability p2.

Now how many combinations are possible when there are two bonus shots?

Hint:  When two coins are tossed there are four outcomes and when two die are thrown there are 36 outcomes.

There are 9 combinations of outcomes when a player has two bonus foul opportunities.

The outcome from the first bonus situation is independent of the outcome from the second bonus situation.

This means that the probability of the outcome from the first shot multiplied by the probability of an outcome from the second shot is equal to probability of  both the first and second outcomes occurring.

For example the probability of zero makes on the first bonus situation and zero makes on the second bonus situation is (1-p) x (1-p).

Below are the simplified formulas for the nine combinations of outcomes evaluated for a shooter than makes 60% of his or her free throws.

 Outcomes for two bonus free throws for a 60 percent free throw shooter Total shots made First shot Outcome Prob first shot outcome eval p at 0.6 Second shot outcome eval p at 0.6 P(first)* p(second) 0 0 (1-p) 0.4 0 0.4 0.1600 1 0 (1-p) 0.4 1 0.24 0.0960 2 0 (1-p) 0.4 2 0.36 0.1440 1 1 (1-p)*p 0.24 0 0.4 0.0960 2 1 (1-p)*p 0.24 1 0.24 0.0576 3 1 (1-p)*p 0.24 2 0.36 0.0864 2 2 p*p 0.36 0 0.4 0.1440 3 2 p*p 0.36 1 0.24 0.0864 4 2 p*p 0.36 2 0.36 0.1296 1.0000

Below are the nine combinations for p=0.80.

 Outcome for two bonus free throws for a 80 percent free throw shooter Total shots made First shot Outcome Prob first shot outcome eval p at 0.8 Second shot outcome eval p at 0.8 P(first)* p(second) 0 0 (1-p) 0.2 0 0.2 0.0400 1 0 (1-p) 0.2 1 0.16 0.0320 2 0 (1-p) 0.2 2 0.64 0.1280 1 1 (1-p)*p 0.16 0 0.2 0.0320 2 1 (1-p)*p 0.16 1 0.16 0.0256 2 1 (1-p)*p 0.16 1 0.64 0.1024 2 2 p*p 0.64 0 0.2 0.1280 3 2 p*p 0.64 1 0.16 0.1024 4 2 p*p 0.64 2 0.64 0.4096 1.0000

Note from the two charts above there is one way to make no free throws and one way to make four free throws.  Two combinations can make one free throw and another two combinations can make three free throws.  There are three combinations that result in two made shots.

Groups can be combined.

The chart below provides a list of outcomes total shots made (0 to 4) and likelihood of each total for the two bonus free throws for the 60 percent free throw shooter and the 80 percent free throw shooter.

 Likelihoods for Two Bonus Attempts 60% Free Throw Shooter 80% Free Throw Shooter 0 0.1600 0.0400 1 0.1920 0.0640 2 0.3456 0.2816 3 0.1728 0.2048 4 0.1296 0.4096 1.00 1.00 E(Shots Made) 1.92 2.88

The likelihood of making all four shots is identical for the four foul and two bonus situations.

In many respects, the value of the 80% free throw shooter over a 60% free throw shooter is much more obvious in the two-bonus situation than the four-shot situation.  The 80% free throw shooter has a 4% shot of missing both initial shots on two bonus situations.  The range in expected shots made is wider for the two-bonus situation than the four-independent-shot situation.