On any given day, there are 10^9 people with a 1% chance of going to a
hotel, so there are 10^7 people going to 10^5 hotels, for an average of 100
people per hotel.
How many pair meetings are there on day 1? At each hotel, there are C(100,2)
or 4950 meetings. Since there are 10^5 hotels, the total number of pair
10^5*4950 or 495,000,000, which we'll round off to
How many pairs of people exist in total? That's C(10^9,2), which we'll round
This means there's a
(5*10^8)/(5*10^17) = 10^-9 chance that two randomly
selected people will meet in a hotel.
In other words, given two randomly selected people, there's a (1-10^-9)
chance they won't meet in a hotel today.
Reminder: (1-p)^q ~ 1 - pq for small values of p (binomial theorem)
What are the chances they won't meet over 1000 days (as given in the
problem). This is:
(1-10^-9)^1000 ~ 1 - 1000*10^-9 = 1-10^-6
What are the chances they'll meet exactly once over 1000 days? That's
1000*(1-10^-9)^999*(10^-9) ~ 1000*(1 - 999*10^-9)*10^-9 ~
1000*(1-10^-6)*10^-9 = 10^-6*(1-10^-6) = 10^-6 - 10^-12
Thus, the chance they'll meet 0 or exactly 1 times is the sum of those two
Thus, the chance they'll meet two or more times is
So, the chance that any randomly selected couple will meet 2 or more times
in the same hotel over a period of 1000 days is
Since there are
5*10^17 possible couples (as above), the expected number of
couples that will randomly meet in the same hotel two or more times in 1000
5*10^17*10^-12 = 5*10^5 or 500,000 couples.
This is twice the answer the book gets. I believe the book is incorrect in
"The chance that they will visit the same hotel on two different given days
is the square of this number, 10^-18"
and that they undercount by a factor of 2, but it's possible there's an
error in my own calculation.
Hopefully, however, this clarifies the orders of magnitude involved.