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 Solution by Penny)

The number of ways shown add up to 1450. Out of 12*12*40 that comes out to .251736111111..., rather than .25.

A larger problem is that it consists solely of situations in which Joe's first throw is higher than or equal to his second.  I'm sure if the second is higher than the first, it would count the same as if it were the other way around.  Counting those cases, we'd have 2895.  Out of 12*12*40, that's the .502604166666666... from my post.

Prob of WW then becomes .5026^2 = .2526
Prob of WLWW = [(0.5026)^3]*(0.4974) =0.063151
Prob of WLWLWW = [(0.5026)^4]*[(.4974)^2] = 0.015787
etc.

coming out to the .336811... I had posted.

Posted by Charlie on 2004-04-18 00:43:16