TITLE: Random Dental Floss Odds
QUESTION [22 upvotes]: Awhile back I bought 2 identical rolls of dental floss, each with 50 uses, and picked them randomly.  Tonight, the one I picked hit the 50 use mark.  What is the expected number of uses in the remaining roll?  
With a 100000 case brute force, I get an expected answer of 8 uses.  Neat, so I have maybe a week.

What is the non-brute force answer?

REPLY [12 votes]: Model your floss use as a random walk in the plane starting at $(0,0)$ and taking a unit step to the right when you pick from roll A, and a unit step up when you pick from roll B.
The first time you use up one of the rolls, there are $k$ uses remaining in the other roll, for some $k$ between $1$ and $50$. There are two mutually exclusive possibilities: you must have arrived at the point $(50, 50-k)$ by taking a step to the right of the point $(49, 50-k)$, or you arrived at the point $(50-k, 50)$ by taking a step up from the point $(50-k, 49)$.
For the first option there are $49 + (50-k)\choose 49$ ways to reach the penultimate point, followed by a step to the right; each path has probability $(1/2)^{99-k}\cdot(1/2)$, giving an overall probability of ${99-k\choose 49}(1/2)^{100-k}$. Since same result holds for the second option, the probability that the remaining roll has $k$ uses is twice this last number:
$$p_k:={99-k\choose 49}\left(\frac12\right)^{99-k}.\tag1$$
To compute $E(K)$, the expected number of uses in the remaining roll, you can evaluate the summation $\sum k p_k$, as in @joriki's approach. Alternatively you can use the following device to obtain a closed form expression: Rearrange (1) into the recursion relation
$$
2[50-k]p_k = [100-(k+1)]p_{k+1},\tag2
$$
valid for all $k=1,\ldots,50$. Sum both sides of (2) over all $k$:
$$
2[50 - E(K)] = 100(1-p_1) - (E(K)-p_1).\tag3
$$
Finally rearrange (3) to obtain
$$E(K)=99p_1= 99{98\choose49}\left(\frac12\right)^{98}\approx7.96.$$