Craps Odds
January 12th, 2016 at 11:34:28 AM
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Great site. I wish I knew of it years ago.
I know that there is no betting system that will give you an advantage over the casinos. I am just looking for some math help.
What are the odds of crapping out before a place bet hits? I know the odds of hitting a 7 five times in a row is 0.129% (0.1667 to the 5th power). However, I do not know how to calculate the odds of a certain number hitting before five 7's are rolled. I have also learned from this site that the average 3.3 rolls before a 7 hits. I am wondering how to do the math to calculate the odds of place bets hitting before a 7.
Thanks for the help.
I know that there is no betting system that will give you an advantage over the casinos. I am just looking for some math help.
What are the odds of crapping out before a place bet hits? I know the odds of hitting a 7 five times in a row is 0.129% (0.1667 to the 5th power). However, I do not know how to calculate the odds of a certain number hitting before five 7's are rolled. I have also learned from this site that the average 3.3 rolls before a 7 hits. I am wondering how to do the math to calculate the odds of place bets hitting before a 7.
Thanks for the help.
January 12th, 2016 at 12:07:25 PM
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I don't think you mean "crapping out" - I think you mean "sevening out". "Crapping out" means rolling 2, 3, or 12 on the comeout roll.
The "odds" of sevening out before a place bet hits depend on the number.
For a 4 or 10, you will seven out 2/3 of the time.
For a 5 or 9, you will seven out 3/5 of the time.
For a 6 or 8, you will seven out 6/11 of the time.
How did I figure this out? Easy.
Suppose you place the 5. The only numbers that matter to this bet are 5 (you win) and 7 (you lose).
On any particular roll, of the 36 ways to roll the two dice, 6 are a 7, and 4 are a 5, so you lose 6 / (6 + 4) = 3/5 of the time.
The "odds" of sevening out before a place bet hits depend on the number.
For a 4 or 10, you will seven out 2/3 of the time.
For a 5 or 9, you will seven out 3/5 of the time.
For a 6 or 8, you will seven out 6/11 of the time.
How did I figure this out? Easy.
Suppose you place the 5. The only numbers that matter to this bet are 5 (you win) and 7 (you lose).
On any particular roll, of the 36 ways to roll the two dice, 6 are a 7, and 4 are a 5, so you lose 6 / (6 + 4) = 3/5 of the time.
January 12th, 2016 at 1:38:17 PM
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Quote: vphammerI know the odds of hitting a 7 five times in a row is 0.129% (0.1667 to the 5th power).
it's 0.0129% I believe
I believe you are making things too hard. You are focusing on how long you can roll before you roll a 7, but all that matters is that there are 6 ways to roll a 7 and, say, in the case of a 6 or 8 to resolve, 5 ways to roll. Odds are thus 6:5 against, or 6/11 chances the place bet fizzles, 5/11 chances you win the place bet. Thatdonguy was trying to answer your question the way you approached it, but is all that what you are after?
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
January 13th, 2016 at 2:37:15 PM
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I was looking at a Martindale system where I could place bet the six or 8 and have a bankroll with $400. That way I would be able to double up six times before I lost my bankroll. I know this is somewhat foolish as you would be betting $192 to win a total of $11. I was just wondering what the true odds were of rolling six 7s before a six or eight was rolled. I know the odds of rolling a 7 five times in a row is approximately 0.13%. I assume the true odds of rolling five 7s before a six or eight was rolled would be higher. I just wanted to know how high - would it be around 5%, 10% ??? Thanks
January 13th, 2016 at 4:14:01 PM
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Quote: vphammerI was looking at a Martindale system where I could place bet the six or 8 and have a bankroll with $400. That way I would be able to double up six times before I lost my bankroll. I know this is somewhat foolish as you would be betting $192 to win a total of $11. I was just wondering what the true odds were of rolling six 7s before a six or eight was rolled. I know the odds of rolling a 7 five times in a row is approximately 0.13%. I assume the true odds of rolling five 7s before a six or eight was rolled would be higher. I just wanted to know how high - would it be around 5%, 10% ??? Thanks
Wink Martindale would never bet like this....
The odds of rolling five 7's before a 6 is 6/11 to the 5th power. Or around 4.8%
January 14th, 2016 at 9:27:37 AM
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I am still unsure of the math.
The odds of anything but a 7 being rolled is 30/36 = 83.3%. The odds of rolling a 6 are 5/36 = 13.9%. It would seem that if you don't seven-out, that rolling a 6 with the 30 remaining combinations is 5/30 = 16.67%, which is the same odds of rolling a seven. According to other posts, the average roll before a seven out is 3.3 times.
Does this indicate I have 3.3 chances that a 16.7% chance that a 6 will be rolled for each person that throws the dice before sevening out? And if this happen over five different people throwing the dice before sevening out - what would the math look like?
The odds of anything but a 7 being rolled is 30/36 = 83.3%. The odds of rolling a 6 are 5/36 = 13.9%. It would seem that if you don't seven-out, that rolling a 6 with the 30 remaining combinations is 5/30 = 16.67%, which is the same odds of rolling a seven. According to other posts, the average roll before a seven out is 3.3 times.
Does this indicate I have 3.3 chances that a 16.7% chance that a 6 will be rolled for each person that throws the dice before sevening out? And if this happen over five different people throwing the dice before sevening out - what would the math look like?
January 14th, 2016 at 10:59:03 AM
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Quote: vphammerI am still unsure of the math.
The odds of anything but a 7 being rolled is 30/36 = 83.3%. The odds of rolling a 6 are 5/36 = 13.9%. It would seem that if you don't seven-out, that rolling a 6 with the 30 remaining combinations is 5/30 = 16.67%, which is the same odds of rolling a seven.
When you put it that way, yes - but you don't have the luxury of knowing in advance that you're not going to seven out, so "the remaining 30 combinations" don't apply. Each dice roll has the same set of 36 possible results.
There are always 6 ways to roll a 7; there are always 5 ways to roll a 6. Since the only rolls that affect a place bet on 6 are 6 and 7, we can ignore the other ones. Among the 11 rolls that affect the bet, 5 of them are winners, and 6 of them are losers, so the probability of the bet winning is 5/11.
The average number of rolls is calculated like this (click on the button):
11/36 of the time, the first roll is a 6 or 7, so there would be 1 roll.
25/36 of the time, the first roll is not 6 or 7, so a second roll is needed, which comes up 6 or 7 11/36 of the time; this means that 25/36 x 11/36 of the time, there would be 2 rolls.
25/36 x 25/36 of the time, the first two rolls are neither 6 nor 7, so a third roll is needed, which comes up 6 or 7 11/36 of the time; this means that 25/36 x 25/36 x 11/36 of the time, there would be 3 rolls, and so on.
The expected number of rolls is (11/36 x 1) + (25/36 x 11/36 x 2) + (25/36 x 25/36 x 11/36 x 3) + (25/36 x 25/36 x 25/36 x 11/36 x 4) + ...
= 11/36 x (1 + (25/36 x 2) + (25/36 x 25/36 x 3) + (25/36 x 25/36 x 25/36 x 4) + ...)
= 11/36 x ( (1 + 25/36 + (25/36)2 + + (25/36)3 + ...)2
= 11/36 x (1 / (1 - 25/36))2
= 11/36 x (1 / (11/36))2
= 1 / (11/36) = 36/11 = about 3.272727 rolls
Note that this does not include the comeout roll (which was 6 in this case).
