The probability of a 50% free throw basketball shooter getting variations of 3 buckets in a row

probability

Suppose someone is a 50% basketball freethrow shooter. I allow them to properly "warm up" (practice), then I give that person 20 shots and want to know the following :

Variation A. What is the probability this person will make exactly 3 buckets in a row only once in that 20 shot round?

B = Bucket, M = Miss

example: MMBM MBMM BMMM BBBM BMBM (spaces for easier readability only).

Variation B. What is the probability this person will make at least 3 buckets in a row only once in that 20 shot round, that is, only one string of those?

I just realized that this one is basically asking what is the probability of getting between 3 and 5 in a row, but only once total. 6 in a row would not count because that would be interpreted as 2 strings of 3 in a row.

example: MMBM MBMM BMMM BBBB BMBM (5 in a row but only 1 occurrence of that)

Variation C. What is the probability this person will make 3 buckets in a row more than once in the same 20 shot round?

example: MMBM MBMM BMMM BBBB BBMM

Note in this one, what appears to be a string of 6 buckets is actually interpreted as 2 strings of non-overlapping 3 bucket events. The strings of 3 buckets can have misses in between them. This one is a special variation in that we allow consecutive occurrences of 3 in a row to count as multiple occurrences of 3 in a row, rather than 6, 9, 12… in a row. That way we don't have to "force" at least one miss in between each 3 buckets.

Assumptions are the shooter makes an honest attempt to make all shots and whenever he makes or misses, it doesn't affect his free throw % which stays at 50%.

Which of these 3 variations is the hardest to solve? How do I go about setting these up?

Best Answer

Looking at Variation A only, the possible outcomes are:

BBBM then (max $2$-streak of B's among $16$ shots)

(max $2$-streak of B's among $16$ shots) then MBBB

(max $2$-streak of B's among $k$ shots) then MBBBM then (max $2$-streak of B's among $15-k$ shots) over $k = 0$ through $15$ inclusive

Which means if $F(n)$ represents the number of sequences of length $n$ with a max B-streak of $2$ or less, then the answer is $\frac{2F(16) + \sum_{k=0}^{15} F(k)F(15-k)}{2^{20}}$

The number of max $2$-streaks can follow a recurrence:

$F(0) = 1$ (empty seq)

$F(1) = 2$ (outcomes M, B)

$F(2) = 4$ (outcomes MM, MB, BM, BB)

$F(n) = F(n-1) + F(n-2) + F(n-3)$

This allows quick computation of the result once you generate the $F$-values:

Python script:

F = [1, 2, 4]

for n in range(3, 16 + 1):
    F.append(F[n - 1] + F[n - 2] + F[n - 3])

print((2 * F[16] + sum(F[k] * F[15 - k] for k in range(16))) / 2**20)  # 0.2182178497314453
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