All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Play or Pass (Posted on 2013-05-28)
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.

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Addressing the problem (spoiler) Comment 4 of 4 |

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

Given the mod(k+1) strategy, I calculated the percentage of these combinations where the first player passes, depending on k. The percentage rises from 10.99% with k=9 (450/4095) to 17.58% with k=5 (720/4095).

Overall, the first player will pass for 13.88% of the 20,475 combinations (2841/20475).

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

 Search: Search body:
Forums (0)