# Puzzle on Ranks

Here is a puzzle from textbook : 40 Puzzles and Problems in Probability and Mathematical Statistics

Peter draws n = 100 independent realizations of a continuous rv and ranks them in increasing order from 1 to 100. Subsequently, Paula draws a single value from the same population and inserts this value into the rank order created earlier by Peter. For example, if her value is such that 50 of Peter’s draws are smaller and 50 are larger, then the rank associated with her draw would be 51 — that is, overall, her value would be the 51st in increasing order. Or, if her value is smaller than all 100 of Peter’s, then the rank 1 would be associated with it.

Derive for general n the probability that Paula’s value will occupy rank k, where 1 ≤ k ≤ n + 1.

My solution is as follows : If Paula's value is x, then

Pr(Rank = k| value = x) = $^nC_{(k-1)}[CDF(x)]^{k-1}[1-CDF(x)]^{n-k+1}$

Therefore,

Pr(Rank=k) = $\int$Pr(Rank = k| value = x) $PDF(x) dx$

Pr(Rank=k) = $\int ^nC_{(k-1)}[CDF(x)]^{k-1}[1-CDF(x)]^{n-k+1}PDF(x) dx$

Pr(Rank=k) = $\int_0^1 ~ ^nC_{(k-1)}[C]^{k-1}[1-C]^{n-k+1} dC$

Pr(Rank=k) = $^nC_{(k-1)} B(k,n-k+2)$

Pr(Rank=k) = $^nC_{(k-1)} \frac{(k-1)! (n-k+1)!}{ (n+1)!}$

Pr(Rank=k) = $\frac{n!}{(k-1)! (n-k+1)!} \frac{(k-1)! (n-k+1)!}{ (n+1)!}$

Pr(Rank=k) = $\frac{1}{ n+1}$

Is the logic correct?

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The answer is correct. The logic may or may not be correct but is certainly unnecessarily complicated. The answer is $1/(n+1)$ by symmetry. –  joriki Oct 28 '12 at 20:02
@joriki : I know the answer is 1/(n+1), but is my approach correct? I generally try to do problems with mathematical rigor and not use symmetry analogy(which is awesome. I should have thought that way!) –  damned Oct 29 '12 at 5:16
In my humble opinion, that's a skewed view of rigour and symmetry. Arguments from symmetry can be quite rigorous (of course some aren't), and I suspect the likelihood of making an error in appealing to symmetry is lower than the one of making a mistake in a calculation like the one in your answer. –  joriki Oct 29 '12 at 5:30

You may equivalently let Peter draw $n+1 = 101$ independent values and then let Paula take one of Peter's at random.
So, in the ranking of the $n+1=101$ values, Paula's can be any of them, with equal probability $\frac{1}{n+1}=\frac{1}{101}$.
The answer is right, but here is one way to see it. Since all draws are iid, there is no reason to believe that the last one should be of a given rank more than another. Imagine Paula selecting first, then the second of the $n+1$ ball, etc. So every rank is equally likely to occur.