The game of craps is played by rolling a pair of dice. If the total comes out to 7 or 11, the shooter wins immediately. If it comes out to 2, 3, or 12, the shooter loses immediately. If any other total shows on the first roll, the player continues to roll until either his original total comes up again, in which case he wins, or a 7 comes up, in which case he loses.

What is the probability the shooter will win?

(In reply to re(5): True Solution (Flaw in common answer.) by Charlie)

Again, I said repeatedly that you are not terminating a game early.  You must complete the game.

Second, I did make a mistake in my calculations, as from the second roll on, I had inserted the value of 1/36 for both winning and losing.  It is not 1/36, but dependent on the first roll.  For example, if the first roll is a 4, then it is 3/36, a 5 is 4/36, a 6 is 5/36, a 8 is 5/36, a 9 is 4/36, and a 10 is 3/36.  Also, the losing value, 7 is 6/36.  This makes the outcome worse than I thought.

Running a state diagram for the game and computing the odds for winning, losing, and continuing at any given roll actually has the first roll as the only roll with odds better for the shooter.  Should the player be able to get out of the game at a particular roll, then the overall odds for the game would be the sum of the odds from roll 1 through n in the table below:

n W(n) L(n) C(n)
1 0.22 0.11 0.67
2 0.08 0.11 0.48
3 0.06 0.08 0.34
4 0.04 0.06 0.25
5 0.03 0.04 0.18
6 0.02 0.03 0.13
7 0.01 0.02 0.09
8 0.01 0.02 0.07
9 0.01 0.01 0.05
10 0.01 0.01 0.03
11 0.00 0.01 0.03
12 0.00 0.00 0.02
13 0.00 0.00 0.01
14 0.00 0.00 0.01
15 0.00 0.00 0.01
16 0.00 0.00 0.01
17 0.00 0.00 0.00
18 0.00 0.00 0.00

To explain this table a little, these are the odds for winning, losing, or continuing at that roll.  These are calculated by multiplying the odds of continuing from the previous roll by the odds for winning, losing, or continuing for the current roll.  To compute these, you have to separate out the states for which total was rolled on the first roll.  You can also get the same numbers by running a program which creates the total possibilities at each roll and then divides the winning/losing/continuing by the total number of possibilities at that point.  Remember to keep track of which possibilities correspond to the continue on the first roll as these determine which possibilities for the subsequent rolls are winners or continuing.  Also, don't use the 24/36 probability for continuing when multiplying for the second roll.  You have to separate out the probability for continuing for each number that causes another roll.  For example, use the 3/36 for all possibilities of 5, etc.

Now this is not to say that in a given game there is not a chance to win.  If you are looking at each individual roll, there is a chance to win, a chance to lose, and a chance to continue.  What this is saying is two things though.  First, by changing the probabilities for winning between the first and subsequent rolls, you make them dependent, not independent in calculating the overall probability of winning. The problem asked for the overall probability in winning, not the probability of winning for a given roll.  Second, despite all game lengths being equally probable if you consider N=infinity for the number of games, and if you consider a subset that is sufficiently large, then you should observe a normal distribution.  That being said, if you take a slightly smaller sample, then while the next game could be any length, if the sample indicates a significant enough deviation from the normal, then you can reasonably assume that you have a better chance of seeing a game of length such that it will help the overall distribution of game lengths to return to normal.

Finally, I will eat some crow. It does appear that upon adding up the totals in the columns for winning vs. losing, you do wind up with appx. 49.3% win and 50.7% lose  overall by the 12th game.

But you must also eat some.Notice that I have proven that should you be able to know that a game should end before the 12th roll, that you do have better odds overall.  Therefore, if you were to use the counting system and chose when to play a round of rolls (to their completion), then you can improve your odds.

Edited on September 6, 2011, 6:06 am

Posted by Joshua on 2011-09-06 04:49:05