Craps  Probability
A casino I played at had the 3,4,5 odds system where you were allowed 3x on the 4 and 10, 4x on the 5 and 9 and 5 x on the 6 and 8. I feel that with this "system" of placing odds, you reduce the fluctuations (with respect to standard 5x odds on all numbers) in your bankroll, and change the distribution of net gain/loss per session, i.e. you would produce a sharper peak located slightly more to the loss side than with 5x odds. Is this so, and could you put some numbers to it?
Return Table with 345X Odds
Event  Pays  Probability  Return 

Pass line win  1  0.222222  0.222222 
Pass line loss  1  0.111111  0.111111 
Point of 4 or 10 & win  7  0.055556  0.388889 
Point of 4 or 10 & lose  4  0.111111  0.444444 
Point of 5 or 9 & win  7  0.088889  0.622222 
Point of 5 or 9 & lose  5  0.133333  0.666667 
Point of 6 or 8 & win  7  0.126263  0.883838 
Point of 6 or 8 & lose  6  0.151515  0.909091 
Total  1.000000  0.014141 
The standard deviation per pass line bet is 4.915632.
1) Your preference is to count the come out roll of 12 in the calculation of the house edge on the don’t pass. If one was to choose NOT to count it, would the house edge on the pass line combined with full double odds be exactly equal to that of the house edge on the don’t pass line combined with full double odds?
2) Does the overall house edge against player x go up if player x places come bets (which will be backed up with full double odds) after betting the pass line with full double odds? i.e. player x with just a pass line with full double odds = house edge .572%, player x with same bet but places two come bets with full double odds = house edge (.572%)x(3)?
1. If we define the house edge as the expected loss per unresolved bet (not counting ties) then the house edge on the don’t pass would be 1.40%, just barely less than the 1.41% on the pass line bet. If the player can bet more money on the don’t pass side, which is the case in real but not Internet casinos, then the combined house edge favors the don’t side more the greater the multiple of odds allowed.
2. Assuming the player keeps his odds on during a come out roll then the overall house edge does not change if the player adds come bets, backed up with the odds. However if the player keeps the odds off, which is the default rule, then the overall house edge will actually go up slightly by adding come bets.
4 times without throwing a 7?
3 times without throwing a 7?
2 times without throwing a 7?
1 times without throwing a 7?
How does the math work for this? Thanks
 Player wins on come out roll: 22.22%
 Player loses on come out roll: 11.11%
 Player wins on a point: 27.07%
 Player loses on a point: 39.60%
So the player will win on a point about 1 in 3.7 rolls.
 5 place $5
 6 place $6
 8 place $6
 field $5
 total= $22
They claim the house edge is 1.136%. How is that possible if every individual bet made has a higher house edge?
Mensa Anything but Seven Combo
Number  Probability  Field  Place 5  Place 6  Place 8  Win  Return 

2  0.027778  10  0.000000  0.000000  0.000000  10  0.277778 
3  0.055556  5  0.000000  0.000000  0.000000  5  0.277778 
4  0.083333  5  0.000000  0.000000  0.000000  5  0.416667 
5  0.111111  5  7  0.000000  0.000000  2  0.222222 
6  0.138889  5  0.000000  7  0.000000  2  0.277778 
7  0.166667  5  5  6  6  22  3.666667 
8  0.138889  5  0.000000  0.000000  7  2  0.277778 
9  0.111111  5  0  0.000000  0  5  0.555556 
10  0.083333  5  0.000000  0.000000  0.000000  5  0.416667 
11  0.055556  5  0  0.000000  0.000000  5  0.277778 
12  0.027778  15  0.000000  0.000000  0.000000  15  0.416667 
Total  1  0.25 
The reason the overall house edge appears to be less than the house edge of each individual bet is because the house edge on place bets is generally measured as expected player loss per bet resolved.
However, in this case the player is only keeping the place bets up for one roll. This significantly reduces the house edge on the place bets from 4.00% to 1.11% on the 5 and 9, and from 1.52% to 0.46% on the 6 and 8.
For you purists who think I am inconsistent in measuring the house edge on place bets as per bet resolved (or ignoring ties) then I invite you to visit my craps appendix 2 where all craps bets are measured per roll (including ties).
x = 3.375758 + 0.60404*x
0.39596*x = 3.375758
x = 8.52551
Low Bet
Total  Combinations  Probability  Pays  Return 
Hard 6,8  2  0.055556  2  0.111111 
Soft 6,8  8  0.222222  1  0.222222 
7  6  0.166667  1  0.166667 
All other  20  0.555556  1  0.555556 
Total  36  1  0.055556 
House Edge in Craps According to Seven Probability
Seven Probability  Pass House Edge  Don’t Pass House Edge 
15.000%  0.666%  3.499% 
15.333%  0.202%  3.024% 
15.667%  0.237%  2.574% 
16.000%  0.652%  2.148% 
16.333%  1.044%  1.744% 
16.667%  1.414%  1.364% 
17.000%  1.762%  1.005% 
17.333%  2.089%  0.667% 
17.667%  2.395%  0.349% 
18.000%  2.682%  0.051% 
18.333%  2.949%  0.227% 
For the first bet, this formula would call for a pass line bet of $90.91, but I rounded it up to $100. Then a point was rolled, I think a 6 or 8. On the second roll the shooter sevened out. So the entire grand was lost in two rolls. It apparently didn’t make for very entertaining television and that story never made the air.
Two questions I can anticipate being asked would be (1) why did I have them bet the pass as opposed to the don’t pass, and (2) why didn’t I bet $91 on the line and $910 on the odds, adding the extra dollar out of my own pocket. To answer the first question, I think that for purposes of going for a quick big win the pass line is better. While the overall house edge is less on the don’t pass, I felt it would have taken more rolls to achieve the $5,000 goal, thus exposing more money to the house edge. To answer the second question, there is not much difference between 9x odds and 10x odds and I thought it would look better on television to be betting only black chips, at least to start.
Place and Buy Bets in Crapless Crapspass and buying oddsin Crapless Craps
Bet  Pays  Prob. Win  House Edge 
Place 2, 12  11 to 2  0.142857  0.071429 
Place 3,11  11 to 4  0.25  0.0625 
Buy 2, 12 (commision only on wins)  119 to 20  0.142857  0.007143 
Buy 3,11 (commision only on wins)  59 to 20  0.25  0.0125 
Buy 2, 12 (commision always)  119 to 21  0.142857  0.047619 
Buy 3,11 (commision always)  59 to 21  0.25  0.047619 
We can see from my analysis of the Fire Bet that the probability of a shooter making all six points is 0.000162435. So, the value of the promotion per shooter is $4,000 × 0.000162435 = 0.649739.
The next question to be asked is what is the expected loss per shooter. The house edge on the pass line bet is 7/495 = 1.414141%. The tricky part is how many pass line bets will a shooter make, on average.
There are four possible states the shooter can be in. Let's define each one as the expected number of future pass line bets for that shooter.
 A = Come out roll
 B = Point of 4 or 10 made
 C = Point of 5 or 9 made
 D = Point of 6 or 8 made
Here are the equations showing the probability of each state leading to the next state.
A = 1 + (12/36)*A + (6/36)*B + (8/36)*C + (10/36)*D
B = (1/3)*A
C = (2/5)*A
D = (5/11)*A
A little algebra results in A = 2.525510, the number of pass line bets made per shooter.
So, the expected loss per $5 shooter is $5*2.525510*0.0141414 = 0.178571.
The expected amount bet by the shooter is $5*2.525510=$12.627551.
Finally, the expected return is the expected win divided by the expected bet: (0.6497390.178571)/12.627551 = 3.73127%. So the house edge is 3.73%.
In Crapless Craps the 3 and 11 pay 11 to 4. Using the same formula, t=3, and a=2.75, so the house edge is 0.25/4 = 6.25%.
If you rephrase the question to be what is the probability of rolling five 6’s before a 12, then the answer is (6/7)^{5} = 46.27%. With four rolls it is (6/7)^{4} = 53.98%. So there is no number of 7’s before a 12 that is exactly 50/50. If you’re looking for a good sucker bet, suggest you can either roll four 7’s before a 12, or a 12 before five 7’s.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
 Sevening out.
 Repeating a point already made (4 to 9).
 Rolling a 10 on the come out roll, and then making it.
We need to quantify the second and third probabilities only. The shooter will eventually make a point, and then eventually make it or seven out. The probability that the point established and then made is 4 to 9 is:
(3/24)×(3/9) + (4/24)×(4/10) + (5/24)×(5/11) + (5/24)×(5/11) + (4/24)×(4/10) = 0.364394.
The probability of establishing a 10 point and then making it is (3/24)*(1/3) = 0.041667.
Let p be the probability of making a 10 point before sevening out. If the player makes any other point, he is right back to where he started from. So...
p = 0.364394 × p + 0.041667
p × (10.364394) = 0.041667
p = 0.041667/(10.364394)
p = 0.065554
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
If the probability of an event is p, then the expected number of times it will happen before failure is p/(1p). So, the expected number of points per shooter is 0.406061/(10.406061) = 0.683673.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
This question was asked at TwoPlusTwo.com, and was answered correctly by BruceZ. The following solution is the same method as that of BruceZ, who deserves proper credit. It is a difficult answer, so pay attention.
First, consider the expected number of rolls to obtain a total of two. The probability of a two is 1/36, so it would take 36 rolls on average to get the first 2.
Next, consider the expected number of rolls to get both a two and three. We already know it will take 36 rolls, on average, to get the two. If the three is obtained while waiting for the two, then no additional rolls will be needed for the 3. However, if not, the dice will have to be rolled more to get the three.
The probability of a three is 1/18, so it would take on average 18 additional rolls to get the three, if the two came first. Given that there is 1 way to roll the two, and 2 ways to roll the three, the chances of the two being rolled first are 1/(1+2) = 1/3.
So, there is a 1/3 chance we'll need the extra 18 rolls to get the three. Thus, the expected number of rolls to get both a two and three are 36+(1/3)×18 = 42.
Next, consider how many more rolls you will need for a four as well. By the time you roll the two and three, if you didn't get a four yet, then you will have to roll the dice 12 more times, on average, to get one. This is because the probability of a four is 1/12.
What is the probability of getting the four before achieving the two and three? First, let's review a common rule of probability for when A and B are not mutually exclusive:
pr(A or B) = pr(A) + pr(B)  pr(A and B)
You subtract pr(A and B) because that contingency is double counted in pr(A) + pr(B). So,
pr(4 before 2 or 3) = pr(4 before 2) + pr(4 before 3)  pr(4 before 2 and 3) = (3/4)+(3/5)(3/6) = 0.85.
The probability of not getting the four along the way to the two and three is 1.0  0.85 = 0.15. So, there is a 15% chance of needing the extra 12 rolls. Thus, the expected number of rolls to get a two, three, and four is 42 + 0.15*12 = 43.8.
Next, consider how many more rolls you will need for a five as well. By the time you roll the two to four, if you didn't get a five yet, then you will have to roll the dice 9 more times, on average, to get one, because the probability of a five is 4/36 = 1/9.
What is the probability of getting the five before achieving the two, three, or four? The general rule is:
pr (A or B or C) = pr(A) + pr(B) + pr(C)  pr(A and B)  pr(A and C)  pr(B and C) + pr(A and B and C)
So, pr(5 before 2 or 3 or 4) = pr(5 before 2)+pr(5 before 3)+pr(5 before 4)pr(5 before 2 and 3)pr(5 before 2 and 4)pr(5 before 3 and 4)+pr(5 before 2, 3, and 4) = (4/5)+(4/6)+(4/7)(4/7)(4/8)(4/9)+(4/10) = 83/90. The probability of not getting the four along the way to the two to four is 1  83/90 = 7/90. So, there is a 7.78% chance of needing the extra 7.2 rolls. Thus, the expected number of rolls to get a two, three, four, and five is 43.8 + (7/90)*9 = 44.5.
Continue with the same logic, for totals of six to twelve. The number of calculations required for finding the probability of getting the next number before it is needed as the last number roughly doubles each time. By the time you get to the twelve, you will have to do 1,023 calculations.
Here is the general rule for pr(A or B or C or ... or Z)
pr(A or B or C or ... or Z) =
pr(A) + pr(B) + ... + pr(Z)
 pr (A and B)  pr(A and C)  ...  pr(Y and Z) Subtract the probability of every combination of two events
+ pr (A and B and C) + pr(A and B and D) + ... + pr(X and Y and Z) Add the probability of every combination of three events
 pr (A and B and C and D)  pr(A and B and C and E)  ...  pr(W and X and Y and Z) Subtract the probability of every combination of four eventsThen keep repeating, remembering to add probability for odd number events and to subtract probabilities for an even number of events. This obviously gets tedious for large numbers of possible events, practically necessitating a spreadsheet or computer program.
The following table shows the the expected number for each step along the way. For example, 36 to get a two, 42 to get a two and three. The lower right cell shows the expected number of rolls to get all 11 totals is 61.217385.
Expected Number of Rolls Problem
Highest Number Needed  Probability  Expected Rolls if Needed  Probability not Needed  Probability Needed  Expected Total Rolls 

2  0.027778  36.0  0.000000  1.000000  36.000000 
3  0.055556  18.0  0.666667  0.333333  42.000000 
4  0.083333  12.0  0.850000  0.150000  43.800000 
5  0.111111  9.0  0.922222  0.077778  44.500000 
6  0.138889  7.2  0.956044  0.043956  44.816484 
7  0.166667  6.0  0.973646  0.026354  44.974607 
8  0.138889  7.2  0.962994  0.037006  45.241049 
9  0.111111  9.0  0.944827  0.055173  45.737607 
10  0.083333  12.0  0.911570  0.088430  46.798765 
11  0.055556  18.0  0.843824  0.156176  49.609939 
12  0.027778  36.0  0.677571  0.322429  61.217385 
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
I was playing craps at one of your advertisers and got 38% too many sevens. I suspect they are cheating. Here is my full roll history: 7,5,7,2,4,6,8,7,9,4,9,6,6,6,5,12,7,11,8,4,7,7,9,5,12,5,11,5,8,1,7,7,6,6,6,5,5,9,8,10,9,7,7,11,8,9,3,7,6,10,6,7,8,7,8,6,6,5,5,9,6,7. I think you should quit endorsing this cheating casino!
In 61 rolls the expected number of sevens is 61×(1/6) = 10.17. You had 14. The probability of exactly 14 sevens is 7.96% and the probability of 14 or more is 12.77%. So, nothing unusual there. I also did a chisquared test on every roll. I know that it isn't very kosher to do a chisquared test on such a small sample, so take the results with a grain of salt. Here are the results:
ChiSquared Test on 61 Dice Rolls.
Dice total  Actual Observations 
Expected Observations 
ChiSquared Statistic 


2  1  1.69  0.284608  
3  1  3.39  1.683971  
4  3  5.08  0.853825  
5  9  6.78  0.728597  
6  12  8.47  1.468944  
7  14  10.17  1.445355  
8  7  8.47  0.255829  
9  7  6.78  0.007286  
10  2  5.08  1.870219  
11  3  3.39  0.044627  
12  2  1.69  0.055100  
Total  61  61.00  8.698361 
The bottom right cell shows a chisquared statistic of 8.70. The probability of a statistic that high or higher with ten degrees of freedom is 56.09%. These results were close to the peak of the bell curve, so the casino easily passes the chisquared randomness test.
The answer is 219.149467.
There are two ways I can think of to solve this. The first is with a Markov Chain. The following table shows the expected rolls needed from any given state of the 128 possible.
Fire Bet — Markov Chain
Point 4 Made 
Point 5 Made 
Point 6 Made 
Point 8 Made 
Point 9 Made 
Point 10 Made 
Expected Rolls 

No  No  No  No  No  No  219.149467 
No  No  No  No  No  Yes  183.610129 
No  No  No  No  Yes  No  208.636285 
No  No  No  No  Yes  Yes  168.484195 
No  No  No  Yes  No  No  215.452057 
No  No  No  Yes  No  Yes  177.801038 
No  No  No  Yes  Yes  No  203.975216 
No  No  No  Yes  Yes  Yes  160.639243 
No  No  Yes  No  No  No  215.452057 
No  No  Yes  No  No  Yes  177.801038 
No  No  Yes  No  Yes  No  203.975216 
No  No  Yes  No  Yes  Yes  160.639243 
No  No  Yes  Yes  No  No  211.272344 
No  No  Yes  Yes  No  Yes  170.911638 
No  No  Yes  Yes  Yes  No  198.520513 
No  No  Yes  Yes  Yes  Yes  150.740559 
No  Yes  No  No  No  No  208.636285 
No  Yes  No  No  No  Yes  168.484195 
No  Yes  No  No  Yes  No  196.113524 
No  Yes  No  No  Yes  Yes  149.383360 
No  Yes  No  Yes  No  No  203.975216 
No  Yes  No  Yes  No  Yes  160.639243 
No  Yes  No  Yes  Yes  No  189.938796 
No  Yes  No  Yes  Yes  Yes  137.865939 
No  Yes  Yes  No  No  No  203.975216 
No  Yes  Yes  No  No  Yes  160.639243 
No  Yes  Yes  No  Yes  No  189.938796 
No  Yes  Yes  No  Yes  Yes  137.865939 
No  Yes  Yes  Yes  No  No  198.520513 
No  Yes  Yes  Yes  No  Yes  150.740559 
No  Yes  Yes  Yes  Yes  No  182.290909 
No  Yes  Yes  Yes  Yes  Yes  121.527273 
Yes  No  No  No  No  No  183.610129 
Yes  No  No  No  No  Yes  136.890807 
Yes  No  No  No  Yes  No  168.484195 
Yes  No  No  No  Yes  Yes  113.177130 
Yes  No  No  Yes  No  No  177.801038 
Yes  No  No  Yes  No  Yes  126.849235 
Yes  No  No  Yes  Yes  No  160.639243 
Yes  No  No  Yes  Yes  Yes  98.046264 
Yes  No  Yes  No  No  No  177.801038 
Yes  No  Yes  No  No  Yes  126.849235 
Yes  No  Yes  No  Yes  No  160.639243 
Yes  No  Yes  No  Yes  Yes  98.046264 
Yes  No  Yes  Yes  No  No  170.911638 
Yes  No  Yes  Yes  No  Yes  113.931818 
Yes  No  Yes  Yes  Yes  No  150.740559 
Yes  No  Yes  Yes  Yes  Yes  75.954545 
Yes  Yes  No  No  No  No  168.484195 
Yes  Yes  No  No  No  Yes  113.177130 
Yes  Yes  No  No  Yes  No  149.383360 
Yes  Yes  No  No  Yes  Yes  80.208000 
Yes  Yes  No  Yes  No  No  160.639243 
Yes  Yes  No  Yes  No  Yes  98.046264 
Yes  Yes  No  Yes  Yes  No  137.865939 
Yes  Yes  No  Yes  Yes  Yes  53.472000 
Yes  Yes  Yes  No  No  No  160.639243 
Yes  Yes  Yes  No  No  Yes  98.046264 
Yes  Yes  Yes  No  Yes  No  137.865939 
Yes  Yes  Yes  No  Yes  Yes  53.472000 
Yes  Yes  Yes  Yes  No  No  150.740559 
Yes  Yes  Yes  Yes  No  Yes  75.954545 
Yes  Yes  Yes  Yes  Yes  No  121.527273 
Yes  Yes  Yes  Yes  Yes  Yes  0.000000 
Briefly, the expected rolls from any given state is the expected rolls until point is either made or lost (5.063636) plus the expected number of rolls if the player advances to a further state, divided by the probability of not advancing in state.
The other method uses integral calculus. First calculate the expected rolls for each possible outcome to happen. Then take the dot product of the probability of each event and average rolls to get the average rolls to resolve a pass line bet, which the lower right corner shows is 3.375758 = 557/165.
Fire Bet — Expected Rolls
Event  Probability  Average Rolls  Expected Rolls 

Point 4 win  0.027778  5  0.138889 
pt 5 win  0.044444  4.6  0.204444 
pt 6 win  0.063131  4.272727  0.269743 
pt 8 win  0.063131  4.272727  0.269743 
pt 9 win  0.044444  4.6  0.204444 
pt 10 win  0.027778  5  0.138889 
pt 4 loss  0.055556  5  0.277778 
pt 5 loss  0.066667  4.6  0.306667 
pt 6 loss  0.075758  4.272727273  0.323691 
pt 8 loss  0.075758  4.272727273  0.323691 
pt 9 loss  0.066667  4.6  0.306667 
pt 10 loss  0.055556  5  0.277778 
Come out roll win  0.222222  1  0.222222 
Come out roll loss  0.111111  1  0.111111 
Total  1.000000  3.375758 
From there we can get the expected rolls between any given point winning:
 Rolls between a point of 4 winning = (3/36)*(3/9)*5*(557/165) = 6684/55 = apx 121.527273.
 Rolls between a point of 5 winning = (4/36)*(4/10)*4.6*(557/165) = 1671/21 = apx 75.954545.
 Rolls between a point of 6 winning = (5/36)*(5/11)*(47/11)*(557/165) = 6684/125 = apx 53.472.
The expected rolls for a 10, 9, and 8 point winner are the same as for 4, 5, and 6, respectively.
Let's say that instead of a point4 winner happening on a discrete basis, it follows an exponential distribution with a mean of 6684/55. The probability such a random variable lasts x units of time without happening is exp(x/(6684/55)) = exp(55x/6684).
The probability it has happened within x units of time, at least once, is 1exp(55x/6684).
If we represent all six points as continuous variables, then the probability all six have happened within x units of time is (1exp(55x/6684))^2 * (1exp(22x/1671))^2 * (1exp(125x/6684))^2.
The probability at least one of the six events not happening within x units of time is 1  (1exp(55x/6684))^2 * (1exp(22x/1671))^2 * (1exp(125x/6684))^2.
We can get the expected time for all six events to happen by integrating the above from 0 to infinity.
Using this integral calculator gives an answer of 8706865474775503638338329687/39730260732259873692189000 = apx 219.1494672902.
Why this works is harder to explain, so please take that part on faith.
I know a dice influencer who claims to have recorded the following rolls in craps. The roller claims he objective was to hit the inside numbers (4, 5, 6, 8, 9, and 10). Can you analyze his results?
Craps Data
Dice Total 
Actual Results 

2  710 
3  1,366 
4  2,132 
5  2,831 
6  3,487 
7  3,963 
8  3,590 
9  2,894 
10  2,136 
11  1,409 
12  709 
Sum  25,227 
First, let's add a column to the table to show the expected tally of each total, assuming a fully random roll.
Craps Data with Expectations
Dice Total 
Actual Results 
Expected Results 

2  710  700.75 
3  1,366  1,401.50 
4  2,132  2,102.25 
5  2,831  2,803.00 
6  3,487  3,503.75 
7  3,963  4,204.50 
8  3,590  3,503.75 
9  2,894  2,803.00 
10  2,136  2,102.25 
11  1,409  1,401.50 
12  709  700.75 
Sum  25,227  25,227.00 
You didn't ask me how to analyze the data, so I'll do it a few different ways.
A chisquared test has a chisquared statistic of 21.43009, with 10 degrees of freedom. The probability of data this skewed, or more, is 1.83%.
Looking at just the inside numbers, which you mentioned was the goal, the total achieved was 12,802, while the expected total would be 25,227 × (2/3) = 12613.5. This excess of inside numbers is 2.52 standard deviations above expectations. The probability of such an excess number, or more, is 0.59%.
I couldn't help but notice the lack of sevens. In 25,227 rolls, the expected sevens are 25,227 × (1/6) = 4204.5. The shooter had 3,963. That is 4.08 standard deviations away from expectations. The probability of such a shortfall is 0.0000225, or one in 44,392.
However, I must say it's usually easy to look at historic data and find something fishy about it. Then again, avoiding sevens is an intrinsic goal to the dice influencer.
The scientific way to test whether the shooter can influence the dice is to state the goal BEFORE data is collected.