I was reading a text and came across this problem. The wording is a little terse so I'm not clear on why the result is true. Given N balls and N urns, all of which are labeled. The balls are distributed into the urns as follows:
- A Ball is chosen from the pool of N balls, and placed into the first urn.
- A second ball is chosen from the remaining set of N-1 balls, and placed into the second bin.
- The rest are filled in a similar manner.
We know that the probability that ball 1 goes into urn 1 is $\frac{1}{n}$.
- What is the probability that ball 2 goes into urn 2?
- What is the marginal CDF of ball 2 in urn 2? (The answer is $\frac{1}{n}$, but I'm not sure how it's computed.)
For the first question, the answer is dependent on what happened at step 1. My thought is there are 2 cases.
- Case 1: Ball 2 is placed in urn 1. In this case, the probability that ball 2 is placed in urn 2 is 0, by definition.
- Case 2: Ball 2 is not placed in urn 1. In this case, there are N-1 balls remaining, so the probability must be $\frac{1}{N-1}$.
This leaves the probability at $\frac{1}{N-1}$. This approach has a problem however because we can ask this question of Urn N and the answer should be afford a similar analysis. Clearly the marginal CDF for all the urns must be the same, in order to reconcile the probability of Ball N in urn N. How do we reconcile this apparent disparity?