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There are $20$ balls in an urn labeled from $1$ to $20$. You randomly pick $10$ balls out of this urn. What is the expected maximum value of the $10$ balls you picked out?

I was able to solve the problem using quite tedious combinatorics as shown below. Is there any other method to solve it?

My Solution: $$\frac{20\cdot\binom{19}{9} + 19\cdot\binom{18}{9} +\dots+10\cdot\binom{9}{9}}{\binom{20}{10}} = \frac{210}{11} $$

Devansh Agarwal
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3 Answers3

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You can use binomial coefficient identities to simplify the fraction: $$ \frac{\sum_{k=10}^{20} k \binom{k-1}{9}}{\binom{20}{10}} = \frac{\sum_{k=10}^{20} 10 \binom{k}{10}}{\binom{20}{10}} = \frac{10 \binom{21}{11}}{\binom{20}{10}} = \frac{10 \cdot\frac{21}{11}\binom{20}{10}}{\binom{20}{10}} = \frac{210}{11} $$

RobPratt
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Distribute the $20$ numbers in ascending order uniformly on a $0-1$ scale, so they're at $\frac{k}{21},$ for $k=1,2,3,...20$

On the other hand, in similar vein, the sampled numbers are at $\frac1{11}, \frac2{11}, \frac3{11} ... \frac{10}{11}$ on an average.

[The $20$ numbers partition a line of length $1$ into $21$ equal segments, and the $10$ samples on an average are expected to divide it into $11$ equal segments, each of which have the same distribution, and we want to find out at what point on the line the highest sampled point falls]

Thus for sampling $10$ numbers, $\frac{10}{11} = \frac{k}{21}, k = \frac{210}{11}$

If only $3$ numbers had been sampled, for example, we would directly get $k$ = $\frac{3\cdot21}{4} = \frac{63}{4}$

Note that the actual highest numbers could have been anything, we are dealing with expected values

true blue anil
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    Thank you for the approach. Just to confirm that I'm getting the concept right, can I generalise the approach as follows? If n balls are numbered 1 to n, and m (m < n) balls are picked randomly then for the expected max value E => $\frac{E}{n+1} = \frac{m}{m+1}$ => $E = \frac{(n+1)⋅m}{m+1}$ – Devansh Agarwal Oct 02 '23 at 21:39
  • That is correct. – true blue anil Oct 03 '23 at 04:54
  • Hi I was reading this answer and completely agree, but I'm confused why my original argument failed. The numbers between 1 and 20 are partitioned by a random choice of 10 samples. It follows then that if we order those from smallest to biggest, the size of the consecutive intervals should be equal segments. The total length (sum) of all the segments should be $19$, hence each segments should be of size $\frac{19}{11}$. Actually from your answer, this should be $\frac{10}{11}$. Where does the mistake arise from? – Jerry Mahajan Dec 09 '24 at 07:07
  • @Jerry Mahajan: The argument I have used can be considered to be a simplified approach to expectation problems, often used as a shortcut. See a more formal approach using indicator variables by Andre Nicolas along with the simplified $0-1$ scale approach used there (like me) by Michael Hardy, and the bend into a circle approach by Joriki. – true blue anil Dec 09 '24 at 08:01
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There is a clever method using linearity of expectation. I got this argument from two of joriki's answers: one about cards in a deck and one about the continuous version of points on a line.

First, imagine that there are $21$ balls in a circle. We make an ordered selection $11$ of these balls. We then delete the first selected ball, and rearrange the remaining $20$ balls into a line, such that clockwise nieghbor of the deleted ball is at the right end of the line. We now have a linear arrangement of $20$ balls, where $10$ of them have been selected, which is equivalent to the setup for your problem. However, the extra information about the circle will be useful.

For the circle, the $21-11=10$ unselected balls will be broken into $11$ circular segments by the selected balls (some of these segments might be empty). By symmetry, each segment has the same expected length, so the length of each segment is $10/11$ on average. Note that the number of balls which are above the maximum ball is exactly equal to the length of the segment which is clockwise of the deleted ball. We conclude that the largest selected ball has on average $10/11$ balls to the right of it, so the expected position of the largest ball is $$ 20-\frac{10}{11}=\frac{210}{11}. $$

Mike Earnest
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  • Thanks for the unique insight. Can you please explain "Note that the number of balls which are above the maximum ball is exactly equal to the length of the segment which is clockwise of the deleted ball." ? I understood everything up until that point. – Devansh Agarwal Oct 02 '23 at 21:23
  • Just to clarify, since someone has given a downvote to my answer today, I didn't give a downvote to you. Regards. – true blue anil Oct 26 '24 at 18:11