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 (1p).
Outcome two is make first shot miss on second shot probability
(1p)*p.
Outcome three is make on first and second shot probability p^{2}.
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 (1p) x (1p).
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

(1p)

0.4

0

0.4

0.1600

1

0

(1p)

0.4

1

0.24

0.0960

2

0

(1p)

0.4

2

0.36

0.1440

1

1

(1p)*p

0.24

0

0.4

0.0960

2

1

(1p)*p

0.24

1

0.24

0.0576

3

1

(1p)*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

(1p)

0.2

0

0.2

0.0400

1

0

(1p)

0.2

1

0.16

0.0320

2

0

(1p)

0.2

2

0.64

0.1280

1

1

(1p)*p

0.16

0

0.2

0.0320

2

1

(1p)*p

0.16

1

0.16

0.0256

2

1

(1p)*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

Comments:
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 twobonus situation
than the fourshot 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 twobonus situation than the fourindependentshot
situation.
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