More Info

Hopefully this program will be useful and help you figure out the best way for you to play.  If you have comments or suggestions, please e-mail me at paul@web-expressioncraps (replace the craps with .com when you email, otherwise the spambots may be able to grab my address off of this page).

The chances of winning a decision of any of the following bets and the values used in this program are:

Bet Win Prob. House Edge (%)
Pass Line 0.4929292929293 1.41
Don't Pass  0.492987012987 1.4
Place 6/8 0.4545454545455 1.52
Place 5/9 0.4 4
Place 4/10  0.3333333333333 6.67
Field* 0.4444444444444 5.55
Hard 6/8 0.09090909090909  9.09
Hard 4/10 0.1111111111111 11.1
Any Craps 0.1111111111111 11.1
Any Seven 0.1666666666667 16.67

*  Refers to Field bets with a double two and twelve, not a triple two or twelve (which can be found in some casinos and have a lower house edge).

House edge is a completely different calculation and depends on the payout relative to the chance of a win/loss.  Note that the probability of various bets cannot be directly compared in helping you decide the smartest way to play – while some events are quite unlikely (such as the consecutive appearances of 2/12 or Any Craps), the payout on these bets is much higher than most others.  Most players feel that the house edge should be a major consideration in helping you decide which bets are the best to make.  The chance of losing a decision of any of the above bets is simply 1 minus the probability of a win (see above).  The first question on the form (number of consecutive wins/losses) is calculated by raising the chance of a win/loss to the power of the number of decisions.  The second and third questions are a little more complicated to calculate – the result is given by this formula:

where win is the chance of winning the bet, loss is the chance of losing the bet, i is the number of wins/losses, and n is the total number of decisions.  The above formula gives the result for the probability of winning i decisions out of n; to find the chance of losing i out of n, you need to switch the values for "win" and "loss".

When you want to find the probability of winning i decisions or more out of n, the above formula becomes a subset of a larger calculation; since there is more than one way the outcome can occur, you need to sum over the set of probabilities from i to n:

Again, this works for a certain number or more of wins, to calculate a certain number or more of losses, simply switch the "win" and "loss" values.  And if you find any bugs, please contact me; I had to jump through a few hoops to put the results in easily readable form (to round off, convert from scientific notation to decimals for medium numbers, limit the number of decimals, eliminate trailing zeros, and add commas when displaying large numbers).  If you calculate a probablility that comes to about 1 in 350 trillion, believe it or not it's probably right! :-)  However, you're welcome and encouraged to check the math, please let me if you find any errors.  Be aware that for certain requests, the result is simply too huge to be calculated – even for numbers of decisions in the low hundreds, a calculation may be performed whose result exceeds 10^300 (10 to the 300th power).  This generates an error, but the program will tell you when the numbers you're using are too big.  Sorry about that.  Thanks to Stacy, Vasi, Steen (for the WinCraps House Advantage table), and Pravda for their help with formulas and odds.

A few other interesting figures I came across – on average, there are:

3.4 rolls per decision on the Pass/Don't Pass

120 rolls per hour

35.3 decisions per hour on Pass/Don't Pass

Yo eleven!