numbers of rolls per craps decision
July 4th, 2016 at 7:29:16 PM
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Hi Wizard....
I can't get my head around the math for calculating rolls/decision in craps.
For example:
I would expect the number of rolls to decision a place bet 6 to be 11/36 ie. 5 ways to win and 6 ways to lose out of 36 combinations in the dice or 11/36= .30555555= 3.2727 rolls, yet according to numbers I've seen elsewhere the answer is 4.65 rolls. That is from a casino management text book. The same source puts hard 6 and 8 at the 3.27 rolls which is 11/36 and consistent with 11 results for a decision per 36 combinations on each of those bets (1 way to win and 10 ways to lose out of 36 combinations= 11/36). Same source puts hard 4 and 10 @ 4 rolls, equivalent to 9/36 which again concurs with the number of decisions per each of those bets per 36 combinations (1 way to win and 8 ways to lose out of 36 combinations= 9/36).
Actually the rolls/decision in the text book across all bets agree with my logic except the place bets listed as follows:
Place 4 and 10= 5.68 rolls (not my 9/36= 4 rolls)
Place 5 and 9= 5.115 rolls (not my 10/36= 3.6 rolls)
Place 6 and 8= 4.65 rolls (not my 11/36= 3.27 rolls)
Help!
Thanks in advance.
I can't get my head around the math for calculating rolls/decision in craps.
For example:
I would expect the number of rolls to decision a place bet 6 to be 11/36 ie. 5 ways to win and 6 ways to lose out of 36 combinations in the dice or 11/36= .30555555= 3.2727 rolls, yet according to numbers I've seen elsewhere the answer is 4.65 rolls. That is from a casino management text book. The same source puts hard 6 and 8 at the 3.27 rolls which is 11/36 and consistent with 11 results for a decision per 36 combinations on each of those bets (1 way to win and 10 ways to lose out of 36 combinations= 11/36). Same source puts hard 4 and 10 @ 4 rolls, equivalent to 9/36 which again concurs with the number of decisions per each of those bets per 36 combinations (1 way to win and 8 ways to lose out of 36 combinations= 9/36).
Actually the rolls/decision in the text book across all bets agree with my logic except the place bets listed as follows:
Place 4 and 10= 5.68 rolls (not my 9/36= 4 rolls)
Place 5 and 9= 5.115 rolls (not my 10/36= 3.6 rolls)
Place 6 and 8= 4.65 rolls (not my 11/36= 3.27 rolls)
Help!
Thanks in advance.
July 5th, 2016 at 6:52:04 AM
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if something has a 50-50 chance, 1/2, does that mean 2 rolls? 1/3 means 3 rolls? You seem to have jumped to that conclusion. Too simple I think.
Others can help you better, I'll shut up.
Others can help you better, I'll shut up.
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
July 5th, 2016 at 9:55:27 AM
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I'm realizing now that the deviation from the straight up probabilities of place bet decisions (e.g. 11/36 ways for a place bet 6) obviously arises from the condition of them being off on the come-out roll, although I'm still working on how it could be a different increase depending on which place bet as illustrated below:
Place bets working on the come-out:
6 and 8: 11/36= 3.27 rolls/decison
5 and 9: 10/36= 3.6 rolls/decision
4 and 10: 9/36= 4 rolls/decision
Place bets off on the come-out (figures are per casino management text book):
6 and 8: 4.65 rolls/decision (1.38 more than working on the come-out)
5 and 9: 5.12 rolls/decision (1.52 more than working on the come-out)
4 and 10: 5.68 rolls/decision (1.68 more than working on the come-out)
Surely the number of rolls occurring on the come-out is constant as should be the average number of rolls between points being made (where place bets are then off coming out), so the variation in the number of rolls/decision according to which place bet remains unresolved at this point, the text book's numbers being erroneous notwithstanding.
Place bets working on the come-out:
6 and 8: 11/36= 3.27 rolls/decison
5 and 9: 10/36= 3.6 rolls/decision
4 and 10: 9/36= 4 rolls/decision
Place bets off on the come-out (figures are per casino management text book):
6 and 8: 4.65 rolls/decision (1.38 more than working on the come-out)
5 and 9: 5.12 rolls/decision (1.52 more than working on the come-out)
4 and 10: 5.68 rolls/decision (1.68 more than working on the come-out)
Surely the number of rolls occurring on the come-out is constant as should be the average number of rolls between points being made (where place bets are then off coming out), so the variation in the number of rolls/decision according to which place bet remains unresolved at this point, the text book's numbers being erroneous notwithstanding.
July 5th, 2016 at 11:26:22 AM
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I think your numbers would be right if place bets were still good on a comeout. If you have a point of 6, the expected number of rolls to resolve it is 3.2727. However, if you place the 6 and the point is, say, 8, you have to allow for the fact that, if 8 is rolled before 6 or 7, then there will be one or more additional rolls needed to establish a new point, during which the place bet is off.
However, for some reason, when I simulate this, I get numbers that are slightly more than 1 less than what the textbook says. For example, I get 4.59 rolls to resolve a place bet on 4 or 10.
However, for some reason, when I simulate this, I get numbers that are slightly more than 1 less than what the textbook says. For example, I get 4.59 rolls to resolve a place bet on 4 or 10.
July 5th, 2016 at 11:32:45 AM
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Some questions to examine:
What is the conditional probability of actually getting back to the comeout roll so you can turn your place bets off, given that you have a place 6 bet?
Is it the same for when you have a place 4 bet?
Are you assuming that you don't place the point number? (And you clearly aren't because if you placed the point number, you don't have an unresolved bet to turn off on the next comeout roll)
What is the conditional probability of actually getting back to the comeout roll so you can turn your place bets off, given that you have a place 6 bet?
Is it the same for when you have a place 4 bet?
Are you assuming that you don't place the point number? (And you clearly aren't because if you placed the point number, you don't have an unresolved bet to turn off on the next comeout roll)
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
July 5th, 2016 at 5:52:53 PM
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Obviously it matters not what place bets are on the layout, once the point is made the number of rolls to reestablish a point is constant:
24/36 come-out rolls establishes a point= 1.5 rolls to establish a point
Therefore, including the come-out rolls to reestablish the point so the place bets work again:
Place bet 6 and 8: 3.27 rolls per bet is working decision + 1.5 rolls per come-out= 4.77 rolls
5 and 9: 3.6 rolls per bet is working decision + 1.5 rolls per come-out= 5.10 rolls
4 and 10: 4 rolls per bet is working decision + 1.5 rolls per come-out= 5.50 rolls
Text book numbers:
Place bet 6 and 8: 4.65 rolls
5 and 9: 5.12 rolls
4 and 10: 5.68 rolls
Close, but no cigar?
We need The Wizard on this lol?
24/36 come-out rolls establishes a point= 1.5 rolls to establish a point
Therefore, including the come-out rolls to reestablish the point so the place bets work again:
Place bet 6 and 8: 3.27 rolls per bet is working decision + 1.5 rolls per come-out= 4.77 rolls
5 and 9: 3.6 rolls per bet is working decision + 1.5 rolls per come-out= 5.10 rolls
4 and 10: 4 rolls per bet is working decision + 1.5 rolls per come-out= 5.50 rolls
Text book numbers:
Place bet 6 and 8: 4.65 rolls
5 and 9: 5.12 rolls
4 and 10: 5.68 rolls
Close, but no cigar?
We need The Wizard on this lol?
