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 Sneaky Joe's Craps (Posted on 2004-04-17)
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.)

 No Solution Yet Submitted by Victor Zapana Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 2 of 11 |
I came up with my odds of winning \$30 as 0.07 out of 1.00.

Let X be Joe's first toss, Y his second, and Z my roll. The winning combinations for me are:

X=2, Y=1, Z=1,2,3
X=2, Y=2, Z=2
X=3, Y=1, Z=1,2,3,4,5,6,7,8
X=3, Y=2, Z=1,2,3,4,5,6,7
X=3, Y=3, Z=6
X=4, Y=1, Z=1,2,3,....,15
X=4, Y=2, Z=1,2,3,....,14
X=4, Y=3, Z=5,6,7,8,9,10,11,12,13
X=4, Y=4, Z=12
X=5, Y=1, Z=1,2,3,...,24
X=5, Y=2, Z=1,2,3,....,23
X=5, Y=3, Z=4,5,6,...,22
X=5, Y=4, Z=11,12,13,...,21
X=5, Y=5, Z=20
X=6, Y=1, Z=1,2,3,...,35
X=6, Y=2, Z=1,2,3,...,34
X=6, Y=3, Z=3,4,5,...,33
X=6, Y=4, Z=10,11,12,...,32
X=6, Y=5, Z=19,20,21,...,31
X=6, Y=6, Z=30
X=7, Y=1, Z=1,2,3,...,40
X=7, Y=2, Z=1,2,3,...,40
X=7, Y=3, Z=2,3,4,...,40
X=7, Y=4, Z=9,10,11,...,40
X=7, Y=5, Z=18,19,20,...,40
X=7, Y=6, Z=29,30,31,...,40
X=8, Y=1, Z=1,2,3,...,40
X=8, Y=2, Z=1,2,3,...,40
X=8, Y=3, Z=1,2,3,...,40
X=8, Y=4, Z=8,9,10,...,40
X=8, Y=5, Z=17,18,19,...,40
X=8, Y=6, Z=28,29,30,...,40
X=9, Y=1, Z=1,2,3,...,40
X=9, Y=2, Z=1,2,3,...,40
X=9, Y=3, Z=1,2,3,...,40
X=9, Y=4, Z=7,8,9,...,40
X=9, Y=5, Z=16,17,18,...,40
X=9, Y=6, Z=27,28,29,...,40
X=9, Y=7, Z=40
X=10, Y=1, Z=1,2,3,...,40
X=10, Y=2, Z=1,2,3,...,40
X=10, Y=3, Z=1,2,3,...,40
X=10, Y=4, Z=6,7,8,...,40
X=10, Y=5, Z=15,16,17,...,40
X=10, Y=6, Z=26,27,28,...,40
X=10, Y=7, Z=39,40
X=11, Y=1, Z=1,2,3,...,40
X=11, Y=2, Z=1,2,3,...,40
X=11, Y=3, Z=1,2,3,...,40
X=11, Y=4, Z=5,6,7,...,40
X=11, Y=5, Z=14,15,16,...,40
X=11, Y=6, Z=25,26,27,...,40
X=11, Y=7, Z=38,39,40
X=12, Y=1, Z=1,2,3,...,40
X=12, Y=2, Z=1,2,3,...,40
X=12, Y=3, Z=1,2,3,...,40
X=12, Y=4, Z=4,5,6,...,40
X=12, Y=5, Z=13,14,15,...,40
X=12, Y=6, Z=24,25,26,...,40
X=12, Y=7, Z=37,38,39,40

If each of the above X,Y,Z combinations has likelihood =
(1/12)*(1/12)*(1/40), then they all sum to my odds of winning a specific game, which are 0.25. The odds of my losing a game are then 0.75. In order to win \$30, I have to win two consecutive games before I lose all my money and cannot continue. I can win \$30 in the following ways:

odds of WW = (0.25)^2 = 0.06
odds of WLWW = [(0.25)^3]*(0.75) = 0.01
odds of WLWLWW = [(0.25)^4]*[(0.75)^2] = 0.00
odds of WLWLWLWW = [(0,25)^5]*[(0.75)^3] = 0.00
etc.

Total odds of my winnng \$30 are 0.07

Edited on April 17, 2004, 9:29 pm
 Posted by Penny on 2004-04-17 21:03:03

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