Sneaky Joe has just invited you as a VIP to his new casino. You know this is probably an attempt steal your money, for he always find ways to swindle people. However, you go anyway.

When you get there, he says, "Come over here and join me in a game of craps." You become slightly suspicious, but agree to come anyway. When you go over, he says, "OK, here's how we play craps in this casino, 'cause it's different here than other casinos. You have 3 dice, 2 of them are 12-sided dice and another is a 40-sided die. I will roll the 2 12-sided dice. Then you roll the 40-sided one. If your number is between (y^2-x) and (x^2-y) inclusively, being that x=the number I got from the first roll and y=the number I got on the second, you will win $10. Otherwise, you will lose $10."

"Ok," you think, "I'm pretty sure that the odds are against me, especially if it's a game that Joe made himself. But I need $30, and I only have $10." So, what's the probability of you winning $30 (as in $30 in the black, without any debt, which included the original $10 paid) from this game?

(NOTE: It can be done WITHOUT trial and error, and it is my request, though you do not have do it, that you solve this without trial and error.)

(In reply to re(2): Solution by Charlie)

The puzzle states "If your number is between (y^2-x) and (x^2-y) inclusively". I interpreted that to mean that (y^2-x) is the lower bound, and (x^2-y) is the upper bound. That is the normal understanding of the word "between". When a shop owner tells me "We're open between nine and five", I don't go to his shop at midnight, because I think he meant "between nine the following morning and five the previous afternoon". That is not what the word "between" means. 

Also, I don't see what is gained by replacing .25 by .251736111111. This is just a flooble puzzle. I think we are allowed to round in a flooble puzzle.

Edited on April 18, 2004, 8:42 am

Posted by Penny on 2004-04-18 01:09:32