Here is an alternate way of looking at it, which does not involve the use of any specific formulas, only the fact that (as is both obvious and well-known for a geometric distribution) this process has no memory. Chances of success are independent of past history.

Instead of following single letter through the post office, think of generating a random sequence of N successes and failures, such as FFFSFFSSFFFFFFSFFSFFF..., where there are Nq failures and Np successes. (Of course, Nq and Np are only expected values, but we are going to look at very large N, so they become firm values in the limit.) Now consider the successes as positions to cut the sequence and failures as the "substance" being cut, so that the string looks like FFF|FF||FFFFF|FF|FFF.... Since a string of Nq failures is cut by Np successes into Np+1 pieces, the average length of each piece must be Nq/(Np+1), and in the limit of large N, this goes to q/p.

Since the process has no memory, one can start counting failures from ANY point (as long as the point is chosen without reference to the part ot the sequence that is further to the right), and find the same expected waiting time. For clarity, choose any particular success as the starting point. Then it is certain that we have not landed in the middle of a string of failures, so we are at the start of one. Thus the expected number of failures before the next success is simply the average mentioned above, q/p.

The time the letter spends in the post office will then by q/p + 1, since one more day is needed for the success to occur. Your way is briefer, but just for fun, I decided to try to set up a model for the probability and then find a way to look at the model that makes the answer come out without a formal analysis.

--Stuart Anderson

In all likelihood your explanation is what Monty was looking for.

>FFF|FF||FFFFF|FF|FFF.... Since a string of Nq failures is
>cut by Np successes into Np+1 pieces, the average length of
>each piece must be Nq/(Np+1), and in the limit of large N,
>this goes to q/p.

Alex

Isn't the 1/n statement better than the "classical" p/q via the following argument?

1. They are equivalent when n=q/p
2. The first statement is more elegant, having only one parameter instead of two.
3. The p/q statement implies that the probability must be a rational number. But in fact it can be any real number, as is more or less implied by 1/n. For example, n may equal pi (so q=pi, p=1?)

P.S. The p/q statement does not imply that the probability must be a rational number.

>1. They are equivalent when n=q/p
>2. The first statement is more elegant, having only one
>parameter instead of two.

I agree here!

>3. The p/q statement implies that the probability must be a
>rational number. But in fact it can be any real number, as
>is more or less implied by 1/n. For example, n may equal pi
>(so q=pi, p=1?)

Here, I see a little problem. Alex has mentioned already that p and q need not be integers, so p/q need not be rational. I would also like to point out that you can't have q=pi, p=1 since p and q are probabilities, so 0 <= p <= 1 and 0 <= q <= 1. Since p+q = 1, you would get q = pi/(pi+1) and p = 1/(pi+1).

I assume you're planning to use this on GnarlyMath. It looks like the 1/n version of the explanation above in post #3 would be just about right for your publication, better than the p/q version. Have fun with it!

--Stuart Anderson