Probability & Statistics

Basketball Scores

A player makes her first throw, misses her second, then scores at the rate she has scored so far. After 100 throws, how likely is she to have made exactly 50?

solvedhard2 min

A basketball player takes 100 free throws. She makes the first and misses the second. From the third throw on, the chance she makes a throw equals the fraction of her throws so far that she has made. What is the probability she ends with exactly 50 makes?

Reveal solutionHide solution

#The rule is a Polya urn

After two throws she sits at one make and one miss. Put one red ball (a make) and one blue ball (a miss) in an urn. Drawing a ball, noting its colour, and returning it with one more of the same colour reproduces the rule exactly. After kk throws the urn holds kk balls, MkM_k of them red, and the next throw is a make with probability Mk/kM_k / k, the fraction made so far. This is the standard Polya urn.

#Uniform by induction

The claim is that for every n2n \ge 2, the number of makes MnM_n is uniform on {1,2,,n1}\{1, 2, \dots, n-1\}.

The base case is n=2n = 2, where M2=1M_2 = 1 with certainty and {1,,n1}={1}\{1,\dots,n-1\} = \{1\}. For the step, assume MnM_n is uniform on {1,,n1}\{1,\dots,n-1\}, so each value has probability 1n1\tfrac{1}{n-1}. The next throw lifts the count by one with probability Mn/nM_n/n and leaves it with probability 1Mn/n1 - M_n/n, so for any jj,

P(Mn+1=j)=P(Mn=j)(1jn)+P(Mn=j1)j1n.(1)\PP(M_{n+1} = j) = \PP(M_n = j)\Big(1 - \tfrac{j}{n}\Big) + \PP(M_n = j-1)\,\tfrac{j-1}{n}. \tag{1}

For an interior j{2,,n1}j \in \{2,\dots,n-1\} both terms appear and collapse,

1n1[(1jn)+j1n]=1n1n1n=1n.(2)\frac{1}{n-1}\left[\Big(1 - \tfrac{j}{n}\Big) + \tfrac{j-1}{n}\right] = \frac{1}{n-1}\cdot\frac{n-1}{n} = \frac{1}{n}. \tag{2}

At the ends j=1j = 1 and j=nj = n only one term survives, and each still gives 1n\tfrac{1}{n}. So Mn+1M_{n+1} is uniform on {1,,n}\{1,\dots,n\}, closing the induction.

#Read it off

At n=100n = 100 the makes are uniform on {1,,99}\{1,\dots,99\}, the forced first make and second miss ruling out 00 and 100100. Every tally is equally likely, so

P(M100=50)=199.(3)\PP(M_{100} = 50) = \frac{1}{99}. \tag{3}
1/99 each15099makes after 100 throws
The Polya urn flattens the distribution completely. Every final tally from 1 to 99 is equally likely, so landing on exactly 50 makes carries probability 1/99.

The surprise is that 50 is no more likely than 1 or 99. The reinforcement keeps whatever lead it stumbles into early, smearing the final count flat across its whole range.