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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 saw answers for this question on the forum, but I am confused why my answer is not correct. I approached this problem by determining the density function of the maximum, some $Z = Max(X_1,...,X_{10})$.

$P(Z=z) = \frac{z^{10}-(z-1)^{10}}{20^{10}}$

Then, I found the expectation by summing the product of the probabilities with values of z ranging from 1 to 20. The answer I obtained is 18.64, whilst the real answer is 210/11. I know that my approach is basically assuming sampling with replacement although in the question it is implied as without. However,I thought this shouldn't matter due to linearity of expectation?

Please let me know!

Anon
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2 Answers2

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I know that my approach is basically assuming sampling with replacement although in the question it is implied as without. However,I thought this shouldn't matter due to linearity of expectation?

I am not sure how you connected this problem with linearity of expectation. But the expected maximum would certainly be different depending on whether sampling is done with or without replacement. So you cannot assume sampling with replacement and expect to get the correct answer.

If sampling is done with replacement, maximum for a trial can be any of the numbers from $1$ to $20$. For example, if you pick the ball numbered $1$ each time, the maximum would be $1$. So, $P(Z) > 0, 1 \le Z \le 20$.

On the other hand, if sampling is done without replacement, the maximum for a trial must be $10$ or higher. So, $P(Z) = 0, 1 \le Z \le 9$.

Haris
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Expectation problems are sometimes surprisingly easy to solve because they essentially deal with averages

Consider the $20$ points making cuts to divide a line of length $1$ into $21$ equal parts to reflect where the points will fall on a $0−1$ scale, which will be $ k/21\; for\; k = 1,2,3,....20$

And ten randomly sampled points will on an average divide the line into $11$ equal parts at $1/11,2/11,3/11, ...10/11$

Thus for the highest sampled point, $k/21 = 10/11 \Rightarrow \boxed{k = 210/11}$

true blue anil
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