The Busy Busy Robot

Since a box can hold up to 100 phones, which is larger than 26, we may assume each box has infinite capacity, so each phone can be placed in any of the boxes with equal probability, and the phones are mutually independent (where a phone is placed doesn't affect other phones).

Fix the box where the 26th phone is placed. The probability that any one phone is not placed in this box is $34/35$ , so the probability that the previous 25 phones are not placed in this box is simply $(34/35)^{25} \approx 0.48447549 \ldots$ , so the answer is $\boxed{4845}$ .

Let there be $N$ boxes, and let $p_{n,j}$ , for $1 \le j \le n$ be the probability that, after $n$ phones have been placed, that there are $j$ boxes containing at least one phone. We shall restrict attention to the case $n < N \le 100$ , to avoid complications of running out of empty boxes, or filling any of the boxes.

It is clear that $p_{1,1} = 1$ , and that $p_{2,1} = \tfrac{1}{N}$ and $p_{2,2} = \tfrac{N-1}{N}$ . Also note that $p_{n,1} \; = \; p_{n-1,1} \times \tfrac{1}{N} \hspace{2cm} p_{n,n} \; = \; p_{n-1,n-1} \times \tfrac{N+1-n}{N} \hspace{2cm} n \ge 2$ so that $p_{n,1} \; = \; \frac{1}{N^{n-1}} \; = \; \frac{N}{N^n} \;,\; p_{n,n} \; = \; \frac{(N-1)(N-2)\cdots(N+1-n)}{N^{n-1}} \; = \; \frac{N!}{(N-n)! N^n} \hspace{2cm} n \ge 1$ More generally, if $2 \le j \le n-1$ , then $p_{n,j} \; = \; p_{n-1,j} \times \frac{j}{N} + p_{n-1,j-1} \times \frac{N+1-j}{N}$ By induction on $n$ , we can show that $p_{n,j} \; = \; \frac{j!}{N^n} \left\{\begin{array}{c} n \\ j \end{array} \right\} \left(\begin{array}{c}N\\j\end{array}\right) \hspace{2cm} 1 \le j \le n < N \le 100$ where $\left\{\begin{array}{c}n \\ j \end{array}\right\}$ is the Stirling number of the second kind.

Thus the probability that the $(n+1)$ st phone is placed into an empty box is $P(n) \; =\; \sum_{j=1}^n p_{n,j} \times \frac{N-j}{N} \; = \; \frac{1}{N^{n+1}}\sum_{j=1}^n (N-j) j! \left\{ \begin{array}{c} n \\ j \end{array} \right\} \left(\begin{array}{c}N \\j \end{array} \right) \hspace{2cm} n < N \le 100$ With $N=35$ , we compute $X \; = \; P(25) \; = \; \frac{193630125104980427932766033374162714624}{ 399669593472470313551127910614013671875}$ and since $10000X = 4844.75...$ , the answer is $\boxed{4845}$ .