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Shaq is shooting free throws. He makes his first shot, then misses his second.

Confidence is a huge factor in how well Shaq plays. This means that for each subsequent shot, the probability he makes it is equal to the fraction of shots he has made so far.

For example, there is a $\frac12$ chance he makes his third shot. If he makes that, there is a $\frac23$ chance he will make his fourth.

After 101 shots (including the first two), what is the probability that Shaq sank less than 21 baskets?

Side note: Symmetry implies that Shaq will make half of his free throws on average, which matches his career average of 52.7% pretty closely.

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    $\begingroup$ He was lucky to make the first shot. Otherwise he would never hit again. $\endgroup$ – Sleafar Sep 28 '15 at 19:13
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    $\begingroup$ @Sleafar I'd say he was incredibly unlucky to miss the second, since he should have had a 100% chance at that point ;) $\endgroup$ – Set Big O Sep 28 '15 at 19:48
  • $\begingroup$ @Geobits Hehe, that's true. $\endgroup$ – Sleafar Sep 28 '15 at 19:52
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    $\begingroup$ In my opinion, this question is on-topic because the result is unexpectedly simple. $\endgroup$ – f'' Sep 28 '15 at 20:51
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After $n$ shots, it is equally likely that Shaq has made any number of shots from $1$ to $n-1$.

Proof by induction:

If $n=2$, then Shaq has made $1$ shot, which is the entire range from $1$ to $2-1$.

Otherwise:

  • For Shaq to make $1$ shot out of $n$, he must make $1$ shot out of $n-1$ and then miss the next shot. This has a probability of $\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.
  • For Shaq to make $n-1$ shots out of $n$, he must make $n-2$ shots out of $n-1$ and then make the next shot. This has a probability of $\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.
  • For Shaq to make $k$ shots out of $n$, with $1<k<n-1$, he either makes $k-1$ shots out of $n-1$ and makes the next shot, or he makes $k$ out of $n-1$ and misses the next. The total probability of these is $\frac{1}{n-2}*\frac{k-1}{n-1}+\frac{1}{n-2}(1-\frac{k}{n-1})=\frac{1}{n-2}(\frac{k-1}{n-1}+1-\frac{k}{n-1})=\frac{1}{n-2}*\frac{n-2}{n-1}=\frac{1}{n-1}$.

So the probability that Shaq makes $m$ shots out of $n$, for $1\le m\le n-1$, is always $\frac{1}{n-1}$. The probability that Shaq makes between $1$ and $20$ shots inclusive out of $101$ is $\frac{20}{100}=\boxed{\frac{1}{5}}$.

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This is equivalent to Polya's urn model (https://en.wikipedia.org/wiki/P%C3%B3lya_urn_model). From the canonical starting state of one ball of each colour the resulting distribution at any stage is uniform, i.e. Shaq has an equal chance of making any number of free throws between 1 and 100.
So the probability that Shaq makes 20 or fewer free throws is $\frac{20}{100}=0.2$.

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Here is a quick Monte Carlo simulation I wrote to verify the result of $\frac{1}{5}$:

Python 3.4:

from random import random
def made_less_than_21():
    baskets = 1
    total = 2
    for shot in range(99):
        if random() < baskets/total:
            baskets += 1
        total += 1
    return baskets < 21
less_than_21 = 0
trials = 100000
for trial in range(trials):
    if made_less_than_21():
        less_than_21 += 1
print ("Answer is about", less_than_21/trials)

$\implies answer \approx 0.200622$

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