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Dalton wants to measure the effectiveness of a treatment for a particular type of plant and, being a good scientist, decides to have a control group that does not have the treatment.

He is able to measure the health of a plant and, without treatment, the health score is a number between 0 and 1, with a flat distribution.

He has 15 plants which get the treatment and he scores them as $a_1<a_2<\ldots<a_{15}$. Similarly, he scores the plants in the control group and they get scores of $c_1<c_2<\ldots<c_{15}$.

(A word about this: each plant gets a score, is sorted by this score, and then is labelled according to the sorted score.)

Now, Mr Dalton notes that in the 15 cases, the plant that got the treatment was ahead of the corresponding plant that did not 13 times. In other words, $a_k>c_k$ 13 times (and less twice).

"Ah ha!" cries Mr Dalton. "My treatment is effective!"

"Mmm," says Mr Darwin. "Not so fast. What is the probability that this result could have arisen by chance? That, after all, is the point of a control!"

What is the probability that $a_k>c_k$ 13 or more times if both groups were just randomly sampled?

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  • $\begingroup$ If it is 14 instead of 13, I guess one could make use of a Catalan number. I still don't know how to derive to 13 by not computing a table of 2x15x15 cells hmm.. (edit: oof, the Catalan may work only if it is 15 ><) $\endgroup$ – athin Apr 27 at 1:01
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The probability is:

3/16, or 18.75%

Reasoning:

For any number of plants in each group $n$, if $x$ is the number of times $a_k > c_k$, each possible value of $x$ in the range [0,n] occurs with equal possibility.

Because:

The exact scores don't matter, what matters is the ordering of all $2n$ plants, which we can look at as a balanced bracket problem or a Dyck path. For any $x$, a Dyck path of length $2n$ with $x$ "flaws", equivalent to a group of plants where $a_k > c_k$ $x$ times, is the $n$th Catalan number. This means there are the same number of orderings for each value of $x$, and each ordering occurs with equal probability.

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  • $\begingroup$ It's not clear to me from what you said why this swapping operation means all the levels are the same. There's 15 ways to 14 for each of the orderings with 15 and there's 14 ways to 15 from each of the orderings with 14 etc. Can you clarify? $\endgroup$ – Dr Xorile Apr 27 at 16:59
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    $\begingroup$ That's a good point. I'm still pretty sure of my conclusion, but I'll try to come up with a more rigorous argument. $\endgroup$ – user34258 Apr 27 at 17:47

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