# Probability rank-n

## Main Question or Discussion Point

Hello
Let y1,..yn be drawn from a normal distribution width known parameters.
Let x be drawn from another normal distribution with known parameters.
If the set {y1,...,yn, x} is ordered, what is the probability that x appears in the "k" position?

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mathman
It depends very much on the parameters of these distributions. You need to be a little more specific. The simplest example is all having the same distribution, then the probability is the same for all positions.

It depends very much on the parameters of these distributions. You need to be a little more specific. The simplest example is all having the same distribution, then the probability is the same for all positions.
I know that the probability depends on the parameters of both distributions. If both have the same distribution, it will give an uniform. If x has a higher mean, it will give an exponential distribution.

I am looking for a formula that, taking into account the gaussian parameters of both distributions, computes the desired probability.

P(X appears in kth position) where k is an element from {0,1,2,....,n}
= P(X higher than k number of Yi)
= (n C k) [P(X> Yi)]k[P(X< Yi)]n-k

P(X appears in kth position) where k is an element from {0,1,2,....,n}
= P(X higher than k number of Yi)
= (n C k) [P(X> Yi)]k[P(X< Yi)]n-k
Thanks for your answer, but it may not be that simple. If that formula was true, then such probability for distributions with the sama parameters will give a normal, instead of a uniform, which I actually confirmed by simulation

Have you found the answer then?

P(X appears in kth position) where k is an element from {0,1,2,....,n}
= P(X higher than k number of Yi)
= (n C k) [P(X> Yi)]k[P(X< Yi)]n-k
Not quite correct - the events {X>Yi} are only conditionally independent, however conditioning on X will get a similar expression

P(X higher than k number of Yi)
$$=\int_{-\infty}^{\infty}f(x)(^nC_k)F(x)^k(1-F(x))^{n-k}dx$$
where F is the cdf of the Yi and f is the pdf of X, which reduces to 1/(n+1) when X has the same distribution as Y.