Bernardo randomly picks 3 distinct numbers from the set
(1; 2; 3; 4; 5; 6; 7; 8; 9)
and arranges them in descending order to form a 3-digit number.
Silvia randomly picks 3
distinct numbers from the set
(1; 2; 3; 4; 5; 6; 7; 8) and also arranges them in descending order
to form a 3-digit number.
What is the probability that Bernardo's number is larger than
Silvia's number?
Source: AMO
1/3 of the time Bernardo will pick a 9, and his number will be larger.
Otherwise, Bernardo and Silvia each have C(8,3) = 8*7*6/3*2*1 = 56 different possibilities. Their numbers are equal with probability 1/56, and if they are not equal then Bernardo's is greater half of the remaining 55/56 of the time.
Altogether, the requested probability is 1/3 + (2/3)*(1/2)*(55/56) = (1/3)*(1+55/56) = 111/168 which is slightly less than 2 times out of 3 (112/168). It would be exactly 2 out of 3 if they picked new numbers in the event of a tie.
Analytical Solution (spoiler)
