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?

The common answer would be correct, if the game didn't depend on what came before.  However, as is stated in the puzzle, craps does depend on what comes before, at least for the first roll, and this is where the common answer of roughly 50/50 odds is wrong.  This is also why the house wins.  Even mathematicians get this one wrong on occasion.

The short answer is this, yes, over time, the odds of winning versus losing do resolve to roughly 50/50, but suprisingly, this does not occur until around 12 rolls!  This means that with some patience, you can come out on top (as long as you stop playing before the end of time, lol--more on this at the end).

Standard two six sided die probability:
2 3 4 5 6 7 8 9 10 11 12  =Sum
1 2 3 4 5 6 5 4 3   2   1   =Probability (out of 36)

n=number of rolls
W(0)=C(0)=L(0)=0% (For simplicity.)

R(n)=Overall probability of winning considering a given number n rolls.

W(n)=Probability of winning for a given roll.
Computed as W(n)=Wr(n)+W(n-1)*C(n-1).

L(n)=Probability of losing for a given roll.
Computed as L(n)=Lr(n)+W(n-1)*C(n-1).

C(n)=Probability of not losing or winning on a given roll.
Computed as C(n)=1-W(n)-L(n)+C(n-1), except where n=0.

Wr(n), Lr(n), and Cr(n) are the probabilities considered for the current roll only.

For the first roll, the shooter wins if he rolls 7 or 11, loses if he rolls 2, 3, or 12.  Otherwise, he continues to roll until the first number rolled comes up again, in which case he wins, or a 7 comes up in which case he loses.

Given this, Wr(1)=(6+2)/36=8/36 or approximately 22.22%.
Likewise, Lr(1)=(1+2+1)/36=4/36 or approximately 11.11%.
Also, Cr(1)=(36-8-4)/36=24/36 or appoximately 66.67%.

Now, if the roll was to continue to a second roll (a 66.67% probability), then you have:

W(2)=Wr(2)+W(1)*C(1)
=1/36+8/36*24/36=(36+8*24)/(36*36)

L(2)=Lr(2)+L(1)*C(1)
=1/36+4/36*24/36=(36+4*24)/(36*36)

C(2)=Cr(2)+C(1)*C(1)
=10/36+24/36*24/36=(360+24*24)/(36*36)

The probabilites Wr(n), Lr(n), and Cr(n) don't change from this point forward, so you can keep using the formulas above to calculate the following table of probabilities:

n    W        L        C
0    0.00    0.00    0.00
1    0.22    0.11    0.67
2    0.18    0.10    0.72
3    0.15    0.10    0.74
4    0.14    0.10    0.75
5    0.14    0.11    0.76
6    0.13    0.11    0.76
7    0.13    0.11    0.76
8    0.12    0.11    0.76
9    0.12    0.11    0.76
10  0.12    0.11    0.76
11  0.12    0.11    0.76
12  0.12    0.12    0.76
13  0.12    0.12    0.76
14  0.12    0.12    0.76
15  0.12    0.12    0.76

From this, it is simple to see that while the overall odds for the game settle to roughly 50/50 for winning or losing, the odds are disproportionately better for shorter runs.  Specifically, any run less than 12 rolls are actually in the favor of the shooter winning.

This allows for an advantage to someone with patience.  Just count the number of rolls for each play. Subtract 12 from this number and add to your total.  For example, a set of runs such as 5, 18, 9, 12, 15,18, would give the following values, -7, +6, -3, 0, +3, and +6.  This gives a sum total of +5.

Positive is good, negative is bad, and you play at the level you are comfortable with.

Over time, you should come out somewhat ahead, but play on to infinity and the odds return in favor of the house, as the distributions of short and long runs are supposedly equal, meaning that over time, the greater
number of long runs (those over 12) become more likely to occur. After all, there are many more numbers greater than 12, than there are between 1 and 12!

The problem with any system such as this is that people who use it tend to ignore the fact that short term analysis and long term analysis are different.  These equations for this analysis ignore the long term implications of run length distribution, and as such, can only be used with a finite set of example runs to predict the next outcome.  In other words, Know When To Get Out!

And remember it's a game.  Don't lose your heads. :)

Edited on September 2, 2011, 10:09 pm

Posted by Joshua on 2011-09-02 22:06:04