Question 558903
Let N1 and N2 be the numbers chosen by players 1 and 2.


Note that we have three probabilities: P(N2 > N1), P(N1 > N2) and P(N1 = N2). Exactly one of these happens, so the sum of these three probabilities is 1.


By symmetry, we can say that P(N2 > N1) = P(N1 > N2) because the players' numbers are independent of each other, and {N1, N2} = {9,4} and {N1, N2} = {4,9} have the same probability of occurring, for example.


To find P(N1 = N2), simply fix N1 (e.g. suppose N1 = 6). Then P(N1 = N2) = 1/10 (one number out of ten is equal to P1).


Hence we have P(N2 > N1) + P(N1 > N2) + 1/10 = 1 --> P(N2 > N1) + P(N1 > N2) = 9/10. Since these two probabilities are equal, P(N2 > N1) equals half of 9/10, or 9/20.