A game consists of two players taking turns, each time removing a limited number (1 to k) candies from either of two plates, initially containing m and n candies (m>=n).

The player who removes the last candy (or candies) wins.

The player to make the 1st move is defined by a toss of a fair coin and has the unique option, after counting the candies, either to start playing or to waive his turn and let his rival to begin.

Both players have correctly analyzed the game and play according to the best available strategy.

What is the probability that the player designated to go first decides to waive his turn?

Rem: You may assume that both m and n are 2-digit numbers and k is a one digit number over 4.

(In reply to Addressing the problem (spoiler) by Steve Herman)

There are 20,475 valid combinations of m, n, and k (given the stated limits), so I threw some code at it.

Posted by snark on 2013-06-05 02:53:22