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.
If we assume that m and n are two digit numbers, then the exact probability depends on how they were selected.
For example, let k = 1 (yes, I know that this is less than 4). Then the strategy is to waive if the total of the two piles is even, and play if the total of the two piles are odd. It is a parity situation.
Does this mean that the probability of waiving = 50%? Well, it depends.
If m and n are chosen independently and randomly, then their total is equally likely to be odd or even.
However, try picking n first (using a uniform distribution from 10 to 99) and then picking m (using a uniform distribution from n to 99). Now, the total is slightly more likely to be even, and the probability of waiving is slightly higher than 50%. Specifically, if n is even then there is a 50% chance that m is also even. But, if n is odd, then there is a greater than 50% chance that m is odd. Putting it all together in Excel, I get a probability of slightly over 51.6%. This is a lot higher than I expected, but it is heavily influenced by cases where n = 99 or even 97 or 95.
And yes, I have not figured out the general strategy. I guess that the probability would be 1/(k+1), if m and n are allowed to be any number greater than 9. This guess has no basis at all, but maybe I got lucky.
Avoiding the problem
