37th Swedish 1997 Problem 4 A and B play a game as follows. Each throws a dice. Suppose A gets x and B gets y. If x and y have the same parity, then A wins. If not, then they make a list of all two digit numbers ab ≤ xy with 1 ≤ a, b ≤ 6. Then they take turns (starting with A) replacing two numbers on the list by their (non-negative) difference. When just one number remains, it is compared to x. If it has the same parity A wins, otherwise B wins. Find the probability that A wins.
3/4 Solution Consider first the list procedure. a + b and a - b have the same parity, so the sum of the numbers keeps the same parity during the replacement process. In other words, the parity of the last remaining number is fixed by the list of starting numbers, it is not influenced by the choices that A and B make. Thus the chance of A winning is 1/2. A die has three odd numbers 1, 3, 5 and three even numbers 2, 4, 6, so the chance that x and y are the same parity is 1/2. If they are of opposite parity, then we have just shown that A has 1/2 chance of winning. Hence overall A has 1/2 + 1/4 = 3/4 chance of winning. 37th Swedish 1997 © John Scholes jscholes@kalva.demon.co.uk 19 January 2004 Last corrected/updated 19 Jan 04