From the above, the probability of rolling 7 is 6/36 = 1/6 and the probability of rolling an 11 is 2/36 = 1/18. These are the numbers that win on the first throw.

Next, lets assume the point thrown on the first roll is a 4, what is the probability of throwing it again before a 7?

Let pr(x) stand for the probability of event x happening on any given roll. The answer is:

pr(4) +
pr(anything other than 4 and 7) * pr(4) +
pr(anything other than 4 and 7)² * pr(4) +
pr(anything other than 4 and 7)³ * pr(4) +
pr(anything other than 4 and 7)⁴ * pr(4) +
+ ...

Pr(4) = 3/36 = 1/12, pr(anything other than 4 and 7) = 1-3/36-6/36 = 27/36 = 3/4.

pr(rolling a 4 before a 7)
= 1/12 + (3/4 * 1/12) + ((3/4)² * 1/12) + ((3/4)³ * 1/12) + ...
= 1/12 * sum for i = 0 to infinity of (3/4)ⁱ
= 1/12 * (1/(1-3/4))
= 1/12 * 4 = 1/3.

The probability of rolling a 10 is the same as the probability of rolling a 4 so pr(rolling a 10 before a 7) also equals 1/3.

Next assume the point thrown is a 5.

Pr(5) = 4/36 = 1/9, pr(anything other than 5 and 7) = 1-4/36-6/36 = 26/36 = 13/18.

pr(rolling a 5 before a 7)
= 1/9 + (13/18 * 1/9) + ((13/18)² * 1/9) + ((13/18)³ * 1/9) + ...
= 1/9 * sum for i = 0 to infinity of (13/18)ⁱ
= 1/9 * (1/(1-13/18))
= 1/9 * 18/5 = 2/5.

The probability of rolling a 9 is the same as the probability of rolling a 5 so pr(rolling a 9 before a 7) also equals 2/5.

Next assume the point thrown is a 6.

Pr(6) = 5/36, pr(anything other than 6 and 7) = 1-5/36-6/36 = 25/36.

pr(rolling a 6 before a 7)
= 5/36 + (25/36 * 5/36) + ((25/36)² * 5/36) + ((25/36)³ * 5/36) + ...
= 5/36 * sum for i = 0 to infinity of (25/36)ⁱ
= 5/36 * (1/(1-25/36))
= 5/36 * 36/11 = 5/11.

The probability of rolling an 8 is the same as the probability of rolling a 6, so pr(rolling an 8 before a 7) also equals 5/11.

Now we're ready to add all this together...

The probability of winning the pass line bet is:
pr(7) + pr(11) + pr(4)*pr(4 before a 7) + pr(5)*pr(5 before a 7) + pr(6)*pr(6 before a 7) + pr(8)*pr(8 before a 7) + pr(9)*pr(9 before a 7) + pr(10)*pr(10 before a 7)
= 6/36 + 1/18 + (1/12 * 1/3) + (1/9 * 2/5) + (5/36 * 5/11) + (5/36 * 5/11) + (1/9 * 2/5) + (1/12 * 1/3)
= 6/36 + 1/18 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36
= 330/1980 + 110/1980 + 55/1980 + 88/1980 + 125/1980 + 125/1980 + 88/1980 + 55/1980
= 976/1980 = 244/495.

The expected return on this wager is the product of the probability of winning and the ratio of what you keep (including your original wager) to the original bet if you do win (in this case 2). Thus, the expected return is 244/495 * 2 = 488/495 =~ 98.59%.

The house advantage is what the casino gets to keep on average, which is 1 minus the expected return, which equals 1-(488/495) = 7/495 =~ 1.41%

What would be the combined house edge in craps if you could buy x times odds on the 6 and 8, y times on the 5 and 9, and z times on the 4 and 10?

The answer is (7/495) / (1 + ((5x+4y+3z)/18))

What would be the combined house edge in craps if you could lay x times odds on the 6 and 8, y times on the 5 and 9, and z times on the 4 and 10?

The answer is (3/220) / (1 + ((x+y+z)/3)).

What would be the combined house edge in craps if I bet 1 unit on both the pass and don't pass and would buy a multiple of u times odds on the 6 and 8, v on the 5 and 9, w on the 4 and 10, and would lay a multiple of x on the 6 and 8, y on the 5 and 9, and z on the 4 and 10?

The answer is (7/495 + 3/220) / ( 2 + ((5u+4v+3w)/18) + ((x+y+z)/3) ).

Here is how I bet in craps (insert very long and complicated explanation of system here). What is my overall house edge?

I'm not big on calculating the house edge on a mix of bets. The house edge of each of them is already listed on my site. If you really want to know then take a weighted average of the individual house edge of each bet according to how much you bet on it in a session.

What is your opinion of betting equal amounts on the pass and don't pass lines in craps and taking full odds on the pass line? Wouldn't the pass and don't pass cancel each other out and leave only the odds, allowing me to gamble with no house edge?

No, this is not a good idea. The pass and don't pass do not cancel each other out. If you roll a 12 the pass will lose and the don't pass will push, for a net loss. To illustrate the increase in the house edge in this strategy lets assume the casino allows 5X odds. The house advantage by betting the pass line and full odds would be (7/495)/(1+5*2/3) = 0.326% . The house advantage by betting both the pass and don't pass lines and full double odds would be (7/495+27/1925)/(2+5*2/3) =0.528% . You can see the house advantage is about 62% greater by betting both the pass and don't pass lines. By adding the don't pass bet into the equation you are mixing into the formula another bet with a house advantage which brings down your overall expected return.

I understand that the house edge on the pass line and come is 1.414% but what is the edge by following a strategy that can have a combination of both active at the same time? For example, betting on the pass line and on come until three different points are covered.

The answer is still 1.414%. There is no way to fool the odds. Playing around with strategy can effect short term results but the house will get its cut over the long run no matter how you play a game of pure luck like craps.

Your combined house edge table between the don't pass and laying odds disagrees with every gambling writer I have seen address the topic, including the legendary John Scarne. Are you sure you're right?

Yes, I'm positive. Most authors make at least one of three mistakes in their calculations. The first mistake is ignoring ties in calculating the house edge of the don't pass bet. This results in a house edge of 1.403%, instead of 1.364%. Some may say this is just a difference of opinion in how it should be calculated. However, I am firmly of the opinion that ties should be counted. The second mistake is incorrect rounding. The third and most serious mistake is ignoring the fact that the player is allowed to bet more laying the odds than taking the odds. For example, if 2X odds are allowed and the player bets $10 on the don't pass and the point is 4 then he can lay $40 on the odds for an odds win of $20. Most other gambling writers base the odds bet on just the bet amount, or $20 in the previous example. John Scarne, for example, makes all three mistakes. First, he ignores ties. Ignoring ties, the house edge on the don't pass should be 1.403%. Second, he rounds incorrectly to 1.402%. Third, he doesn't lay the full odds but only an amount equal to the product of the don't pass bet and the odds multiple. Doing it this way based on a don't pass house edge of 1.402% should result in a combined house edge of 0.601% but he somehow gets 0.591%, probably due to more bad rounding. For this example the correct answer is (3/220)/(1+2*((6/36)*2+(8/36)*1.5+(10/36)*1.2)) = 0.455%. I don't mean to disrespect Scarne. He is an early pioneer in the mathematics of gambling and he had a large influence on me at an early age. The book I quote from, "Scarne's New Complete Guide to Gambling", was originally written in 1961, and all calculations were probably done without a calculator. His mistakes are mostly forgivable but replicating them over and over by modern writers is not.

What is your position on dice control?

For those who dont understand the question, there are many believers that a skilled shooter can influence the dice enough to overcome the house edge. This is done with certain dice settings, and gently throwing them, in an effort to keep them on axis, and bounce as little as possible.

I still have yet to see evidence that anybody has had long-term success setting dice. However, many people I respect are believers, including my hero, Stanford Wong. I will reluctantly admit that maybe there are a handful of people who truly can influence the dice. However, for every one of them, I believe there are at least a hundred who think they can, but cant. I see people at the Red Rock casino practicing dice control all the time, and Im happy to bet against them. I strongly feel the casinos make more from self-perceived dice setters than they lose to them.

Let me say formally that Im asking for evidence that anyone out there can influence the dice beyond a 99% confidence interval. If Im convinced, I would be happy to plug that persons book, web site, or lessons.

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