Letter Chatters!

Let's call A the first person. At each step, the probability for a person different from A of not sending a letter to A is $\frac{498}{499}\times\frac{497}{498}=\frac{497}{499}$ , since this person sends their letters to two distinct uniformly random people.

Now the key step is to realize that the probability of A not receiving a letter is just $(\frac{497}{499})^N$ , being $N$ the number of times a person (different from A) sends their two letters. Note that two people send their two letters at the second stage, four people at the third stage, eight people at the fourth stage, and so on until $2^{11}$ people send their two letters at the twelfth stage. It's important to notice that at each step it is possible that one person has received more than one letter so they have to send more than two, but this is irrelevant for the argument above. Hence $N=2+2^2+2^3+\ldots+2^{11}=2^{12}-2,$ from where we deduce that the desired probability is $\alpha=\left(\frac{497}{499}\right)^{2^{12}-2}\approx 7.235\cdot 10^{-8}.$ The number we've been asked for is $[10^{10}\alpha]=\boxed{723}$ .

we can find it by probality A can't send it a is 498/499 X 407/498=497/499,Now the key step is to realize that the probability of A not receiving a letter is just ,(497/499)N being N the number of times a person (different from A) sends their two letters.2 people send it to 2 letters at second one. THEN 2+2X2+2X2X2.....2X2X2X2X2X2X2X2X2X2X2X2 -2 then (497/499) 2X2X2X2X2X2X2X2X2X2X2X2 -2 so 10x10x10x10x10x10x10x10x10x10 infinity =723