Answer to Question #150083 in Statistics and Probability for Amir

Question

At a party, there are 100 cats. Each pair of cats flips a coin, and they shake paws if and only if the coin comes up heads. It is known that exactly 4900 pairs of cats shook paws. After the party, each cat is independently assigned a “happiness index” uniformly at random in the interval [0,1]. We say a cat is practical if it has a happiness index that is strictly greater than the index of every cat with which it shook paws. The expected value of the number of practical cats is m/n, where m and n are positive integers with gcd(m,n) = 1. Compute 100m+n.

Answer 1

Number of pairs = 99+98+97+...+2+1+0 = 99/2*100 = 4950

Number of shakes = 4900

Let arrange cats by happiness increasing.

P(first cat is practical) = 1

P(2nd cat is practical) = 50/4950 (1 definite shake didn't happen)

P(3rd cat is practical) = 50*49/(4950*4949) (2 definite shakes didn't happen)

...

P(51th cat is practical) = 50*49*..*1/(4950*4949*...*4941) (50 definite shakes didn"t happen)

P(nth cat is practical) = 50!/(50-(n-1))!/(4950!/(4950-(n-1))!)

The expected value of the number of practical cats =

$\sum_{n=0}^{50}50!/(50-n)!/(4950!/(4950-n)!)=\sum_{n=0}^{50}50!*(4950-n)!/((50-n)!*4950!)$ =

=4951/4901

m=4951

n=4901

100m+n = 495100+4901=500001