Simple random variable

Question

From a deck of $50$ cards numbered from $1$ to $50$ you randomly draw out $3$ cards. Let Y:="the third number drawn" a random variable. I must determine the distribution function of Y.

I think that $P(Y=n) = 1/50$ for each $n$ for symmetry, but the solution says another thing:

$$P(Y=n) = \frac{(n-1)(50-n)}{50 \choose 3} \quad \forall n = 2,...,49$$

I do not understand... could someone help me?

Answer 1

Your solution is correct, from what you describe. However, what is given in the solution is the distribution of the middle of the 3 drawn values, when ordered by value. The naming of "$Y$" hints to that. So maybe you misunderstood the question, or there were mutliple parts and you got the solution for the wrong one?

Answer 2

Assuming you described the problem accurately, you are right, the solution is wrong. Note that the solution claims that the probability of $Y=1$ or $Y=50$ is $0$. Why in the world would that be the case?

Answer 3

Gentlemen, you are right! I misunderstood the question. The problem introduces other two random variables I did not mention, that is X:="the highest number drawn" and Z:="the smallest number drawn". So "the third number drawn" is the one in the middle, not the one taken at the third drawing as I thought.

Thank you very much.