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 12sided dice and another is a 40sided die. I will roll the 2 12sided dice. Then you roll the 40sided one. If your number is between (y^2x) and (x^2y) 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.)
Actually the odds are with you in an individual game. Maybe Joe is not as bright as he is greedy.
In one playing of the game there are 144 possible outcomes on his two rolls of the 12sided dice. (Actually he needs only one, since he rolls at successive times.) I assume all the dice are marked from 1 to the number of sides each has. On some combinations of his rolls it would be impossible for you to win, such as 8 and 12, which give 52 and 136 as the bounds, none of which is on your die.
If he rolls 7 and 12, however, for example, the limits are 37 and 137, four of which (37, 38, 39 and 40) each have a 1/40 chance of winning, giving you a 4/40 chance of winning or 1/10, under the condition that he in fact rolled a 7 and a 12.
If he rolls a 1 and a 12, for example, you definitely will win as the range of winners is 11 to 143. All 40 of your numbers fall in that range.
Each conditional probability (conditioned upon what Joe's rolls are) is a multiple of 1/40, or .025, and the 144 equally likely values are:
1 2 3 4 5 6 7 8 9 10 11 12
1 0.000 0.075 0.200 0.375 0.600 0.875 1.000 1.000 1.000 1.000 1.000 1.000
2 0.075 0.025 0.175 0.350 0.575 0.850 1.000 1.000 1.000 1.000 1.000 1.000
3 0.200 0.175 0.025 0.225 0.475 0.775 0.975 1.000 1.000 1.000 1.000 1.000
4 0.375 0.350 0.225 0.025 0.275 0.575 0.800 0.825 0.850 0.875 0.900 0.925
5 0.600 0.575 0.475 0.275 0.025 0.325 0.575 0.600 0.625 0.650 0.675 0.700
6 0.875 0.850 0.775 0.575 0.325 0.025 0.300 0.325 0.350 0.375 0.400 0.425
7 1.000 1.000 0.975 0.800 0.575 0.300 0.000 0.000 0.025 0.050 0.075 0.100
8 1.000 1.000 1.000 0.825 0.600 0.325 0.000 0.000 0.000 0.000 0.000 0.000
9 1.000 1.000 1.000 0.850 0.625 0.350 0.025 0.000 0.000 0.000 0.000 0.000
10 1.000 1.000 1.000 0.875 0.650 0.375 0.050 0.000 0.000 0.000 0.000 0.000
11 1.000 1.000 1.000 0.900 0.675 0.400 0.075 0.000 0.000 0.000 0.000 0.000
12 1.000 1.000 1.000 0.925 0.700 0.425 0.100 0.000 0.000 0.000 0.000 0.000
These average .5026041666666667, making that your probability of winning any one game. (This is 2895/(144*40) = 193/384.)
However, you're trying to be up by two games, net, without ever falling even one game, net, behind, as you'd be broke if you were behind by even one game. This is called a "gambler's ruin" problem, as you are either ruined (broke) or gain the goal amount.
Call the probability of winning an individual game, .502504..., P. Call the probability of gaining your goal from the state of being even with Joe P0. Call the probability of gaining your goal from being one game ahead P1.
Then, P0 = P * P1, since P is your probability of getting to being one ahead from being even and once you're one behind, you're out, but P1 = (1P)*P0 + P, since getting to the next level, 2 ahead, is gaining your goal.
Thus P0 = P*((1P)*P0 + P)
Solving this for P0, we get P0 = P^2/(1P+P^2) = .3368115522682268, or more exactly, 37249/110593. So you have just over 1 chance in 3 of winning $20 from Joe versus losing your last $10 to Joe.

Posted by Charlie
on 20040417 11:59:44 