12’s before 7
January 15th, 2022 at 5:59:39 AM
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What is the formula for calculating the chances of rolling a certain number of 12’s before a 7? Say 4 12’s before rolling a 7. Assuming 2 die in craps.
January 15th, 2022 at 6:42:06 AM
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The probability of exactly n 12's before a 7 is (1/7)^n*(6/7). Here is a table for n=0 to 1.
| 12's | Probability |
| 0 | 0.857142857 |
| 1 | 0.122448980 |
| 2 | 0.017492711 |
| 3 | 0.002498959 |
| 4 | 0.000356994 |
| 5 | 0.000050999 |
| 6 | 0.000007286 |
| 7 | 0.000001041 |
| 8 | 0.000000149 |
| 9 | 0.000000021 |
| 10 | 0.000000003 |
Last edited by: Wizard on Jan 16, 2022
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 15th, 2022 at 7:34:35 AM
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Quote: WizardThe probability of n 12's before a 7 is (6/7)^n*(6/7).
Er, I think you mean (1/7)^n * 6/7.
Also note that this is the probability of rolling exactly n 12s before a 7.
The probability of "rolling n 12s before a 7" is (1/7)^n, which is also 1 / 7^n.
January 15th, 2022 at 9:24:15 AM
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Yes, I didn’t specify but I do not mean n 12’s in a row. Just any chance of rolling n 12’s before the seven out.
January 15th, 2022 at 9:28:29 AM
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So would the formula (1/7)^n or (1/7)^n * 6/7?
January 15th, 2022 at 9:43:49 AM
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Quote: ThamptondmdSo would the formula (1/7)^n or (1/7)^n * 6/7?
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Again, it depends on whether you mean "exactly" n 12s, and then a 7 before rolling another 12, or not.
If it has to be exact: (1/7)^n * 6/7
If it's just "which will happen first: rolling n 12s, or one 7?", it is (1/7)^n.
January 15th, 2022 at 10:18:42 AM
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While we are at it, forgive me if this was posted before, but can we have a column for 18 yo’s before the seven?
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
January 15th, 2022 at 10:40:59 AM
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Quote: DeMangoWhile we are at it, forgive me if this was posted before, but can we have a column for 18 yo’s before the seven?
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(sigh)
I'm supposed to add an obligatory joke, under the thumbtack rules.
A bass player in a local band was having trouble deciding what to dress as for a costume party.
His girlfriend grabbed two drum sticks, tucked them in his pocket, and said "There, now you can go as a musician."
May the cards fall in your favor.
January 15th, 2022 at 12:16:54 PM
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Nice to know odds at the crapless table. 18 yo's don't need to be in a row, just before the 7.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
January 15th, 2022 at 12:36:02 PM
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Quote: DeMangoNice to know odds at the crapless table. 18 yo's don't need to be in a row, just before the 7.
Seriously? (1/4)^18, which is 1 / 68,719,476,736.
January 15th, 2022 at 1:30:08 PM
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Quote: DeMangoWhile we are at it, forgive me if this was posted before, but can we have a column for 18 yo’s before the seven?
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Can you or Mr. W calculate the odds of performing 495 different come out rolls each one different and distinct with each producing a different and distinct outcome as in the formula that defines the PL HA ar 1.41%?
tuttigym
p.s. That number should far exceed the 11 yo's in my estimation.
January 15th, 2022 at 8:27:50 PM
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Can you rephrase? I don’t understand the questionQuote: tuttigymQuote: DeMangoWhile we are at it, forgive me if this was posted before, but can we have a column for 18 yo’s before the seven?
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Can you or Mr. W calculate the odds of performing 495 different come out rolls each one different and distinct with each producing a different and distinct outcome as in the formula that defines the PL HA ar 1.41%?
tuttigym
p.s. That number should far exceed the 11 yo's in my estimation.
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It’s all about making that GTA
January 16th, 2022 at 1:38:17 AM
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For the Pass (or Come) bet there are seventeen possible ways for the bet to be resolved (ignoring the intermediary rolls that aren't needed).
2, 3, 7, 11, 12 (n/36 n=1,2,6,2,1)
4/4 4/7 (3/36*n/9 n=3,6)
5/5 5/7 (4/36*n/10 n=4,6)
6/6 6/7 (5/36*n/11 n=5,6)
similarly for 8 thru 10
Thus there aren't 495 different outcomes but their probabilities are associated with 495 (or 1980 etc) as a common factor.
2, 3, 7, 11, 12 (n/36 n=1,2,6,2,1)
4/4 4/7 (3/36*n/9 n=3,6)
5/5 5/7 (4/36*n/10 n=4,6)
6/6 6/7 (5/36*n/11 n=5,6)
similarly for 8 thru 10
Thus there aren't 495 different outcomes but their probabilities are associated with 495 (or 1980 etc) as a common factor.
January 16th, 2022 at 8:23:12 AM
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Quote: tuttigymQuote: DeMangoWhile we are at it, forgive me if this was posted before, but can we have a column for 18 yo’s before the seven?
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Can you or Mr. W calculate the odds of performing 495 different come out rolls each one different and distinct with each producing a different and distinct outcome as in the formula that defines the PL HA ar 1.41%?
tuttigym
p.s. That number should far exceed the 11 yo's in my estimation.
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I probably could...but let me ask: why? It makes sense if you are proposing something along the lines of, "Unless you can roll every combination without repetition, the use of terms like 'expected value' and 'house advantage' are meaningless, and Martingale-style systems can, and do, work," but I prefer not to listen to such nonsense.
"Expected Value," as defined by Merriam-Webster:
"the sum of the values of a random variable with each value multiplied by its probability of occurrence"
By Investopedia:
"In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values."
But even then, "expected value" and "actual results" are quite often two different things.
An easy example: when rolling two fair 6-sided dice, the "expected value" of their sum is 7. The probability of it actually happening is 1/6.
January 16th, 2022 at 9:10:51 AM
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Quote: charliepatrickThus there aren't 495 different outcomes but their probabilities are associated with 495 (or 1980 etc) as a common factor.
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Obviously, I am genuinely confused, so please help me understand the sentence above especially "their probabilities are associated with 495..... as a common factor." Example: The point established was with a hard 8. There are 5 ways to convert that point, correct? (4/4; 3/5; 5/3 etc) There are 6 ways to lose that point, i.e. 7-out, correct? (1/6; 6/1;2/5 etc) That 8 point alone has 11 possible different outcomes or resolutions, correct.? So the seventeen possible ways "for the Pass (or Come) bet" is incorrect. The 495 is the total of all possible resolutions one time. If that is incorrect, please resolve my confusion of your statements above. Thank you.
tuttigym
p.s. Addendum to the example above. The point is 2/6 or 8. The above resolutions also apply, i.e. 11 possible outcomes.
January 16th, 2022 at 9:40:08 AM
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Quote: tuttigymQuote: charliepatrickThus there aren't 495 different outcomes but their probabilities are associated with 495 (or 1980 etc) as a common factor.
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Obviously, I am genuinely confused, so please help me understand the sentence above especially "their probabilities are associated with 495..... as a common factor." Example: The point established was with a hard 8. There are 5 ways to convert that point, correct? (4/4; 3/5; 5/3 etc) There are 6 ways to lose that point, i.e. 7-out, correct? (1/6; 6/1;2/5 etc) That 8 point alone has 11 possible different outcomes or resolutions, correct.? So the seventeen possible ways "for the Pass (or Come) bet" is incorrect. The 495 is the total of all possible resolutions one time. If that is incorrect, please resolve my confusion of your statements above. Thank you.
tuttigym
p.s. Addendum to the example above. The point is 2/6 or 8. The above resolutions also apply, i.e. 11 possible outcomes.
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The overall probability of a pass line bet winning is 244 / 495. This does not mean that there are 495 distinct combinations of comeout rolls and results.
That is kind of like saying that the probability of tossing exactly two heads and two tails with four coin tosses is 3/8, so there are 8 possible outcomes.
Here's where 495 comes from:
The probability of winning with a comeout of 4 is 1/12 x 1/3 = 1/36
The probability of winning with a comeout of 5 is 1/9 x 2/5 = 2/45
The probability of winning with a comeout of 6 is 5/36 x 5/11 = 25/396
The probability of winning with a comeout of 7 is 1/6
The probability of winning with a comeout of 8 is 5/36 x 5/11 = 25/396
The probability of winning with a comeout of 9 is 1/9 x 2/5 = 2/45
The probability of winning with a comeout of 10 is 1/12 x 1/3 = 1/36
The probability of winning with a comeout of 11 is 1/18
The least common multiple of the denominators is 1980
The sum of these four numbers = 976 / 1980 = 244 / 495
You can't use 495 as a common denominator as, for example, the probability of winning with a comeout of 11 is 27.5 / 495, but you can't have 27.5 comeouts.
January 16th, 2022 at 10:42:29 AM
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Firstly the assumption most people will make is that it doesn't matter which combination of dice make any given total, so there are, for instance, 3 ways to make a 4, 4 ways to make a 5, and 5 ways to make a 6.
This means at the resolution stage
(a) 4 or 10: there are 3 ways to make the point and 6 ways to make the seven-out, so the denominator is 9. As the numerators are 3 or 6, it can be simplified to 1/3 and 2/3 = 55 and 110 / 165
(b) 5 or 9: similarly 4 to make 5, 6 to make 7, denominator is 10; simplified to 2/5 and 3/5 = 66 and 99 / 165.
(c) 6 or 8: similarly 5 to make 6, 6 to make 7, denominator is 11; simplified to 5/11 and 6/11 = 75 and 90 / 165.
The chance of natural win with 7 is 6/36, 11 is 2/36.
The chance of craps lose with 2 is 1/36, 3 is 2/36, 12 is 1/36.
The chance of point being 4 is 3/36.
The chance of point being 5 is 4/36.
The chance of point being 6 is 5/36.
You can see that to keep all the numbers whole, and if you care about which point you need, the denominator is 5940 (which is 36*165).
If you worried about specific ways to roll numbers, then you would use more rows in the table, but it would still add up to 36*165.
This means at the resolution stage
(a) 4 or 10: there are 3 ways to make the point and 6 ways to make the seven-out, so the denominator is 9. As the numerators are 3 or 6, it can be simplified to 1/3 and 2/3 = 55 and 110 / 165
(b) 5 or 9: similarly 4 to make 5, 6 to make 7, denominator is 10; simplified to 2/5 and 3/5 = 66 and 99 / 165.
(c) 6 or 8: similarly 5 to make 6, 6 to make 7, denominator is 11; simplified to 5/11 and 6/11 = 75 and 90 / 165.
The chance of natural win with 7 is 6/36, 11 is 2/36.
The chance of craps lose with 2 is 1/36, 3 is 2/36, 12 is 1/36.
The chance of point being 4 is 3/36.
The chance of point being 5 is 4/36.
The chance of point being 6 is 5/36.
| First Roll | Outcome | First Roll | Second Roll | Win | Lose |
|---|---|---|---|---|---|
| Natural 7 | Win | 6 | 165 | 990 | |
| Natural 11 | Win | 2 | 165 | 330 | |
| Craps 2 | Lose | 1 | 165 | 165 | |
| Craps 3 | Lose | 2 | 165 | 330 | |
| Craps 12 | Lose | 1 | 165 | 165 | |
| Point 4 | Win | 3 | 55 | 165 | |
| Point 4 | Lose | 3 | 110 | 330 | |
| Point 5 | Win | 4 | 66 | 264 | |
| Point 5 | Lose | 4 | 99 | 396 | |
| Point 6 | Win | 5 | 75 | 375 | |
| Point 6 | Lose | 5 | 90 | 450 | |
| Point 8 | Win | 5 | 75 | 375 | |
| Point 8 | Lose | 5 | 90 | 450 | |
| Point 9 | Win | 4 | 66 | 264 | |
| Point 9 | Lose | 4 | 99 | 396 | |
| Point 10 | Win | 3 | 55 | 165 | |
| Point 10 | Lose | 3 | 110 | 330 | |
| Totals | 5940 | 2928 | 3012 | ||
| Pass line (Win) | 1.414% | 2928 | |||
| Don't Pass (Wins) | 2847 | ||||
| Don't Pass (Ties) | 1.364% | 165 |
If you worried about specific ways to roll numbers, then you would use more rows in the table, but it would still add up to 36*165.
January 16th, 2022 at 3:25:30 PM
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Quote: ThatDonGuyI probably could...but let me ask: why? It makes sense if you are proposing something along the lines of, "Unless you can roll every combination without repetition, the use of terms like 'expected value' and 'house advantage' are meaningless, and Martingale-style systems can, and do, work," but I prefer not to listen to such nonsense.
First, thank you for the reply and the definitions plus sources. They are very helpful for me, and I do have questions which will follow. Next, "why"?: I am fully aware that any system that is repeated as a basis for one's play will ultimately fail at some point and create losses which can be extreme without discipline and touting such is not only nonsense but sometimes dangerous for the gullible and uninformed.
The basis for the PL HA (1.41%), as you know is what I call the rule of 495. While I understand how the math is derived, I am confident that it cannot be performed at the tables because, I believe, the odds are so long as to make this constant referral nonsense, meaningless, and totally misleading. Further, in my experience, I have never witnessed or heard of any player actually stepping up to the table, buying in, and just playing the PL as the exclusive wager for one's basis of play. If the actual odds of performing the rule of 495 were known, perhaps players might be better informed as to the reality that the HA for the game is considerably higher as it is really played. In short, the term "house advantage" is quite real but highly diluted with the droning of such unreal low percentages 1.41% just does not happen at the table.
Quote: ThatDonGuy"Expected Value," as defined by Merriam-Webster:
"the sum of the values of a random variable with each value multiplied by its probability of occurrence"
To be honest, this definition I do not comprehend. My personal math IQ is pretty insufficient.
Quote: ThatDonGuyBy Investopedia:
"In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values."
OK better, I think maybe. So, if I have a five to one advantage of winning a bet, regardless of size or regardless of win amount, can that be construed as a +EV? So the dice example is a -EV because it occurs only 18+% of the time. Am I understanding that correctly?
tuttigym
January 16th, 2022 at 3:49:29 PM
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Quote: tuttigymQuote: ThatDonGuyI probably could...but let me ask: why? It makes sense if you are proposing something along the lines of, "Unless you can roll every combination without repetition, the use of terms like 'expected value' and 'house advantage' are meaningless, and Martingale-style systems can, and do, work," but I prefer not to listen to such nonsense.
First, thank you for the reply and the definitions plus sources. They are very helpful for me, and I do have questions which will follow. Next, "why"?: I am fully aware that any system that is repeated as a basis for one's play will ultimately fail at some point and create losses which can be extreme without discipline and touting such is not only nonsense but sometimes dangerous for the gullible and uninformed.
The basis for the PL HA (1.41%), as you know is what I call the rule of 495. While I understand how the math is derived, I am confident that it cannot be performed at the tables because, I believe, the odds are so long as to make this constant referral nonsense, meaningless, and totally misleading. Further, in my experience, I have never witnessed or heard of any player actually stepping up to the table, buying in, and just playing the PL as the exclusive wager for one's basis of play. If the actual odds of performing the rule of 495 were known, perhaps players might be better informed as to the reality that the HA for the game is considerably higher as it is really played. In short, the term "house advantage" is quite real but highly diluted with the droning of such unreal low percentages 1.41% just does not happen at the table.
(Bold added, by me, for emphasis)
Okay, so what you are maintaining is that, because players do not exclusively play the Pass Line...that changes the House Edge of the Pass Line.
Simply put, if the House Edge of the, 'Any 7,' bet is 16.67%, then playing Pass Line AND Any 7 (as one example) increases the House Edge against you on your total action, but no amount put on Any 7 bets changes the House Edge of the Pass Line bet taken alone.
(Middle part removed by me as irrelevant)
Quote: ThatDonGuyBy Investopedia:
"In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values."
Quote: TuttigymOK better, I think maybe. So, if I have a five to one advantage of winning a bet, regardless of size or regardless of win amount, can that be construed as a +EV? So the dice example is a -EV because it occurs only 18+% of the time. Am I understanding that correctly?
tuttigym
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This is just more of the usual from you. Probability of winning being greater than 50% does not yield a positive expectation. It's the same question as betting $18 on black at Roulette and then individually betting 12 red numbers at $1 apiece. You have $30 in total action, the probability that you will profit on the single spin is 30/38 (double zero assumed), but that does not make the bet +EV.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 16th, 2022 at 3:53:06 PM
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Quote: ThatDonGuyQuote: WizardThe probability of n 12's before a 7 is (6/7)^n*(6/7).
Er, I think you mean (1/7)^n * 6/7.
Also note that this is the probability of rolling exactly n 12s before a 7.
The probability of "rolling n 12s before a 7" is (1/7)^n, which is also 1 / 7^n.
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Yes, you're right, to both.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 16th, 2022 at 4:20:38 PM
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Quote: ThatDonGuyThe overall probability of a pass line bet winning is 244 / 495. This does not mean that there are 495 distinct combinations of comeout rolls and results.
That is kind of like saying that the probability of tossing exactly two heads and two tails with four coin tosses is 3/8, so there are 8 possible outcomes.
Here's where 495 comes from:
The probability of winning with a comeout of 4 is 1/12 x 1/3 = 1/36
The probability of winning with a comeout of 5 is 1/9 x 2/5 = 2/45
The probability of winning with a comeout of 6 is 5/36 x 5/11 = 25/396
The probability of winning with a comeout of 7 is 1/6
The probability of winning with a comeout of 8 is 5/36 x 5/11 = 25/396
The probability of winning with a comeout of 9 is 1/9 x 2/5 = 2/45
The probability of winning with a comeout of 10 is 1/12 x 1/3 = 1/36
The probability of winning with a comeout of 11 is 1/18
The least common multiple of the denominators is 1980
The sum of these four numbers = 976 / 1980 = 244 / 495
You can't use 495 as a common denominator as, for example, the probability of winning with a comeout of 11 is 27.5 / 495, but you can't have 27.5 comeouts.
You have been very patient and accommodating. I greatly appreciate your efforts to help me. Thank you.
My simple mind looks at 36 ways to roll the dice as a starting place so that all win/loss possibilities for me mathematically are derived at that place. Therefore, since there are only two ways to roll an eleven (11), my mind interprets the probability or odds of winning at 2/36 or 5.5%. The winning of the bet is number specific and therefore as a one roll event, win or lose, dismisses any other possibilities or rolled numbers.
The Place numbers 4, 5, 6, 8, 9, & 10 again are win/loss specific although not one roll bets. Therefore, the only relevant numbers to winning or losing is that number or the 7 (after come out). All other tossed numbers during that point play are moot and do not affect the outcome of that point. My mind and the casino pay table basically states that the 4 point has 3 ways to win and 6 ways to lose (3/6) or two to one for and against with the payout at 2X for the win and 1/2 (50%) for the loss (Don't odds). That is why, for me, there is a great disconnect of what many have tried to convey to me. But it all starts with 36 ways to roll the dice.
tuttigym
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January 16th, 2022 at 4:35:55 PM
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Quote: Mission146Okay, so what you are maintaining is that, because players do not exclusively play the Pass Line...that changes the House Edge of the Pass Line.
There you go again making up conclusions that have NOT been stated and are DISTORTIONS of my words. You need to brush up on your critical thinking and reading skills
Quote: Mission146Simply put, if the House Edge of the, 'Any 7,' bet is 16.67%, then playing Pass Line AND Any 7 (as one example) increases the House Edge against you on your total action, but no amount put on Any 7 bets changes the House Edge of the Pass Line bet taken alone.
Those are your words and conclusions, not mine.
tuttigym
January 16th, 2022 at 4:51:24 PM
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Quote: tuttigymQuote: Mission146Okay, so what you are maintaining is that, because players do not exclusively play the Pass Line...that changes the House Edge of the Pass Line.
There you go again making up conclusions that have NOT been stated and are DISTORTIONS of my words. You need to brush up on your critical thinking and reading skillsQuote: Mission146Simply put, if the House Edge of the, 'Any 7,' bet is 16.67%, then playing Pass Line AND Any 7 (as one example) increases the House Edge against you on your total action, but no amount put on Any 7 bets changes the House Edge of the Pass Line bet taken alone.
Those are your words and conclusions, not mine.
tuttigym
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I know they are my conclusions and not yours; start with the conclusions are right, so that they are more likely to be mine becomes a given.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
January 16th, 2022 at 8:30:47 PM
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Quote: tuttigymQuote: ThatDonGuyBy Investopedia:
"In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values."
OK better, I think maybe. So, if I have a five to one advantage of winning a bet, regardless of size or regardless of win amount, can that be construed as a +EV? So the dice example is a -EV because it occurs only 18+% of the time. Am I understanding that correctly?
tuttigym
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No, you are not.
Expected value = the sum of (the probability of a particular result x the value of that result) over all possible results.
An easy example: betting 1 on red in double-zero roulette.
The probability of winning 1 is 18/38; the probability of losing 1 is 20/38.
EV = (18/38 x 1) + (20/38 x (-1)) = -2/38 = -1/19, or about -5.263%.
It is a little harder with the pass line in craps (BTW, I play only pass line and odds), because of the multiple conditions. However, let me borrow most of CharliePatrick's table from a previous post:
| First Roll | Outcome | First Roll | Second Roll | Win | Lose |
|---|---|---|---|---|---|
| Natural 7 | Win | 6 | 165 | 990 | |
| Natural 11 | Win | 2 | 165 | 330 | |
| Craps 2 | Lose | 1 | 165 | 165 | |
| Craps 3 | Lose | 2 | 165 | 330 | |
| Craps 12 | Lose | 1 | 165 | 165 | |
| Point 4 | Win | 3 | 55 | 165 | |
| Point 4 | Lose | 3 | 110 | 330 | |
| Point 5 | Win | 4 | 66 | 264 | |
| Point 5 | Lose | 4 | 99 | 396 | |
| Point 6 | Win | 5 | 75 | 375 | |
| Point 6 | Lose | 5 | 90 | 450 | |
| Point 8 | Win | 5 | 75 | 375 | |
| Point 8 | Lose | 5 | 90 | 450 | |
| Point 9 | Win | 4 | 66 | 264 | |
| Point 9 | Lose | 4 | 99 | 396 | |
| Point 10 | Win | 3 | 55 | 165 | |
| Point 10 | Lose | 3 | 110 | 330 | |
| Totals | 5940 | 2928 | 3012 |
The probability of a particular row's result = (the number in the First Row column divided by 36) x (the number in the Second Row column divided by 165). Note that the numbers in the Win and Loss columns = the First Row value x the Second Row value, so the probability of a row's result = the number in the win (or loss, as appropriate) column / 5940, which is 36 x 165.
EV = (1 x (the sum of (each numbers in the Win column / 5940))) + (-1 x (the sum of (each number in the Loss column / 5940)) =
2928 / 5040 - 3012 / 5940 = -84 / 5940 = -7 / 495, or about 1.414%.
Note that this does not mean that every set of 5940 consecutive comeouts will have exactly 2928 wins and 3012 losses; "expected" does not mean "guaranteed." If it did, then here is a Guaranteed Surefire Cannot Miss System:
1. Track 5939 consecutive comeouts.
2. Count the number of wins; if there have been 2928, the next comeout has to be a loss, and if there have been only 2927, the next comeout has to be a win. Bet accordingly. "What if it's neither of those two?" Then you have a set of 5940 consecutive comeouts that don't have 2928 wins, but this system works on the assumption that This Cannot Happen.
3. But wait - there's more! In order for the 5940 consecutive comeouts starting with #2 to have 2928 wins, comeout #5941 has to have the same result as comeout #1. For that matter, #5942 has to have the same result as #2, #5943 has to have the same result as #3, and so on.
January 16th, 2022 at 9:30:53 PM
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What are the odds of twelve 7s before seven 12s ?
It’s all about making that GTA
January 17th, 2022 at 6:06:59 AM
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Quote: Ace2What are the odds of twelve 7s before seven 12s ?
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Over 99%? Just a guess….
January 17th, 2022 at 6:34:18 AM
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Quote: Mission146Okay, so what you are maintaining is that, because players do not exclusively play the Pass Line...that changes the House Edge of the Pass Line.
You posted the above. I DID NOT "MAINTAIN" or infer what you wrote. You did NOT state what YOUR conclusion was. The post was dishonest and mis-leading and sought to discredit covertly.
Quote: Mission146I know they are my conclusions and not yours; start with the conclusions are right, so that they are more likely to be mine becomes a given.
What does that mean???
tuttigym
January 17th, 2022 at 7:02:51 AM
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Quote: tuttigymQuote: Mission146I know they are my conclusions and not yours; start with the conclusions are right, so that they are more likely to be mine becomes a given.
What does that mean???
tuttigym
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I'm going to conclude it means Mission is reasonably confident in his 15th grade math skills.
May the cards fall in your favor.
January 17th, 2022 at 7:13:43 AM
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Quote: Ace2What are the odds of twelve 7s before seven 12s ?
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P(12 7s, 0 12s) = (6/7)^12 = 6^12 / 7^12 = 6^12 / 7^18 x 7^6
P(12 7s, 1 12, ending with a 7) = C(12,1) (6/7)^12 (1/7)^1 = 12 6^12 / 7^13 = 6^12 / 7^18 x 12 x 7^5
P(12 7s, 2 12s, ending with a 7) = C(13,2) (6/7)^12 (1/7)^2 = 78 6^12 / 7^14 = 6^12 / 7^18 x 78 x 7^4
P(12 7s, 3 12s, ending with a 7) = C(14,3) (6/7)^12 (1/7)^3 = 364 6^12 / 7^15 = 6^12 / 7^18 x 364 x 7^3
P(12 7s, 4 12s, ending with a 7) = C(15,4) (6/7)^12 (1/7)^4 = 1365 6^12 / 7^16 = 6^12 / 7^18 x 1365 x 7^2
P(12 7s, 5 12s, ending with a 7) = C(16,5) (6/7)^12 (1/7)^5 = 4368 6^12 / 7^17 = 6^12 / 7^18 x 4368 x 7
P(12 7s, 6 12s, ending with a 7) = C(17,6) (6/7)^12 (1/7)^6 = 12,376 6^12 / 7^18 = 6^12 / 7^18 x 12,376
P(12 7s before 7 12s) = 6^12 / 7^18 x (117,649 + 201,684 + 187,278 + 124,852 + 66,885 + 30,576 + 12,376)
= 741,300 x 6^12 / 7^18
= about 110 / 111
January 17th, 2022 at 7:35:25 AM
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Mr. TDG: Your hard work is evident and appreciated. Your math IQ is many levels above mine. If I were in the same room with you having this conversation with the charts, I could ask questions, point by point, that would help me to fully understand what you are trying to convey.
I am sorry. I just cannot grasp the totality of the concept. It is not your fault; it is my deficiency.
Having said that, let me ask:
1. The roulette example does not show a positive edge to the player (+50%) from the start, i.e., 18 chances to win vs 38 chances to lose. My example in craps is 30 chances to win vs only 6 chances to lose or a positive possible outcome favoring the player at 5 to 1. While there are no guarantees as we both have stated in our examples, My mind clearly believes my example favors the player greatly over your example of winning a bet. So, the straight up question is if given these two choices for a one wager win, which would you choose?
2. The table (thank you CharleyPatrick), for me, is even more confusing. I do not know where the "second roll" numbers come from and what purpose it is they serve. The Come Out natural winners and losers are one roll bet outcomes while the point bets can be finalized with no definitive number of rolls. Hense my confusion there.
How does one come up with 5940 rolls exactly?
3. The "Win"/"Lose" columns really baffle me because I do not know how they are determined based on the number of rolls exhibited.
As I said, a conversation at a table with a beverage of choice and a relaxed, informal atmosphere would, more than likely, allow me to more fully understand the concepts.
Thank you for your patience and endurance.
tuttigym
I am sorry. I just cannot grasp the totality of the concept. It is not your fault; it is my deficiency.
Having said that, let me ask:
1. The roulette example does not show a positive edge to the player (+50%) from the start, i.e., 18 chances to win vs 38 chances to lose. My example in craps is 30 chances to win vs only 6 chances to lose or a positive possible outcome favoring the player at 5 to 1. While there are no guarantees as we both have stated in our examples, My mind clearly believes my example favors the player greatly over your example of winning a bet. So, the straight up question is if given these two choices for a one wager win, which would you choose?
2. The table (thank you CharleyPatrick), for me, is even more confusing. I do not know where the "second roll" numbers come from and what purpose it is they serve. The Come Out natural winners and losers are one roll bet outcomes while the point bets can be finalized with no definitive number of rolls. Hense my confusion there.
How does one come up with 5940 rolls exactly?
3. The "Win"/"Lose" columns really baffle me because I do not know how they are determined based on the number of rolls exhibited.
As I said, a conversation at a table with a beverage of choice and a relaxed, informal atmosphere would, more than likely, allow me to more fully understand the concepts.
Thank you for your patience and endurance.
tuttigym
January 17th, 2022 at 8:53:32 AM
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Quote: ThatDonGuyQuote: Ace2What are the odds of twelve 7s before seven 12s ?
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P(12 7s, 0 12s) = (6/7)^12 = 6^12 / 7^12 = 6^12 / 7^18 x 7^6
P(12 7s, 1 12, ending with a 7) = C(12,1) (6/7)^12 (1/7)^1 = 12 6^12 / 7^13 = 6^12 / 7^18 x 12 x 7^5
P(12 7s, 2 12s, ending with a 7) = C(13,2) (6/7)^12 (1/7)^2 = 78 6^12 / 7^14 = 6^12 / 7^18 x 78 x 7^4
P(12 7s, 3 12s, ending with a 7) = C(14,3) (6/7)^12 (1/7)^3 = 364 6^12 / 7^15 = 6^12 / 7^18 x 364 x 7^3
P(12 7s, 4 12s, ending with a 7) = C(15,4) (6/7)^12 (1/7)^4 = 1365 6^12 / 7^16 = 6^12 / 7^18 x 1365 x 7^2
P(12 7s, 5 12s, ending with a 7) = C(16,5) (6/7)^12 (1/7)^5 = 4368 6^12 / 7^17 = 6^12 / 7^18 x 4368 x 7
P(12 7s, 6 12s, ending with a 7) = C(17,6) (6/7)^12 (1/7)^6 = 12,376 6^12 / 7^18 = 6^12 / 7^18 x 12,376
P(12 7s before 7 12s) = 6^12 / 7^18 x (117,649 + 201,684 + 187,278 + 124,852 + 66,885 + 30,576 + 12,376)
= 741,300 x 6^12 / 7^18
= about 110 / 111
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I am very proud of my ‘more than 99%’ guess! So it is ever so slightly more than 99%!
January 17th, 2022 at 9:41:38 AM
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Quote: tuttigymMr. TDG: Your hard work is evident and appreciated. Your math IQ is many levels above mine. If I were in the same room with you having this conversation with the charts, I could ask questions, point by point, that would help me to fully understand what you are trying to convey.
I am sorry. I just cannot grasp the totality of the concept. It is not your fault; it is my deficiency.
Having said that, let me ask:
1. The roulette example does not show a positive edge to the player (+50%) from the start, i.e., 18 chances to win vs 38 chances to lose. My example in craps is 30 chances to win vs only 6 chances to lose or a positive possible outcome favoring the player at 5 to 1. While there are no guarantees as we both have stated in our examples, My mind clearly believes my example favors the player greatly over your example of winning a bet. So, the straight up question is if given these two choices for a one wager win, which would you choose?
You are confusing expected value with probability of winning.
Let me give you another example: suppose the casino had this game - you bet $1000, and you roll a 6-sided die; if you roll a 1-5, you win $1, but if you roll a 6, you lose your $1000. You have a 5/6 chance of winning, but obviously you don't play it as the risk far outweighs the reward.
This is what Expected Value is - a comparison of risk versus reward.
If you are interested in a bet with a high chance of winning, then go to the roulette wheel and bet 1 on each number from 1 to 35. You have a 35/38 chance of winning on a double-zero wheel; you gain 35 on the winning bet and lose 34 on the 34 losing bets combined, so you win 1 on the total bet. True, if any of the other three numbers come up, you lose 35, but that doesn't seem to be important to you; only the fact that you will win 92% of the time does.
As to which I would choose: you will have to refresh my memory as to which bet you are referring to that has 30 chances to win. Given the choice between the Pass Line in craps (EV = -1.414%) and pretty much any bet in double-zero roulette (EV = -5.263%), I will choose craps every time.
If you are interested in determining the probability of starting with some bankroll B and being able to get X ahead without busting the entire bankroll first, that is a completely different discussion called the Gambler's Ruin Problem.
Quote: tuttigym
2. The table (thank you CharleyPatrick), for me, is even more confusing. I do not know where the "second roll" numbers come from and what purpose it is they serve. The Come Out natural winners and losers are one roll bet outcomes while the point bets can be finalized with no definitive number of rolls. Hense my confusion there.
How does one come up with 5940 rolls exactly?
The "second number" is the probability, divided by 165, that this bet wins or loses.
Quote: tuttigym
3. The "Win"/"Lose" columns really baffle me because I do not know how they are determined based on the number of rolls exhibited.
Remember when I said that expected value is the sum of every product of (the probability of an event happening and the result of that event)?
Let's use rolling a point of 4 as an example.
The probability of rolling a 4 on a comeout is 1/12.
Once you have a point of 4, the only numbers that matter are 4 and 7. There are 3 ways to roll a 4 and 6 ways to roll a 7, so the probability of rolling a 4 before a 7, which wins the bet, is 3 / (3 + 6) = 1/3, and the probability of rolling a 7 before a 4, which loses the bet, is 6 / (3 + 6) = 2/3.
The portion of the EV that comes from winning with a point of 4 = the probability of rolling, and then making, a point of 4, multiplied by the value of the win. The probability = 1/12 x 1/3 = 1/36, and the value is 1, so the product is 1/36.
Similarly, the portion of the EV that comes from losing with a point of 4 = the probability of rolling, and then missing, a point of 4, multiplied by the value of the loss. The probability = 1/12 x 2/3 = 1/18, and the value is -1, so the product is -1/18.
Now, do this for every comeout roll, keeping in mind that some probabilities - for example, winning with a comeout roll of 2 - are zero.
This is what Charlie's chart does, but it multiplies the numbers by constant values so that they all appear as integers.
The "first number" is the probability of rolling that number in the comeout, multiplied by 36.
The "second number" is the probability that, given that the particular point number for that row has been established, you will win or lose the bet, multiplied by 165. For example, for a point of 4, the "second number" for a win is 1/3 x 165 = 65, and the "second number" for a loss is 2/3 x 165 = 110. 165 was chosen because it is the smallest number that, when you multiply all of the "make or miss the point" probabilities by it, turns them all into integers.
The "win" number equals the second number if the row represents a win, and zero if it represents a loss.
The "loss" number equals the second number if the row represents a loss, and zero if it represents a win.
The reason those two columns exist is to show what the sums of the winning and losing results are.
I will try expressing it in a different way:
Each result is the product of three values: the probability of rolling that number on the comeout, the probability of winning or losing, as appropriate for that result, once the point has been established, and the value of that result (1 for a win, or -1 for a loss; if you are calculating Don't Pass, it would be -1 for a win and 1 for a loss, but remember that a comeout of 12 has a result of 0).
Winning with a comeout of 2: 1/36 x 0 x 1 = 0
Losing with a comeout of 2: 1/36 x 1 x (-1) = -1/36
Winning with a comeout of 3: 1/18 x 0 x 1 = 0
Losing with a comeout of 3: 1/18 x 1 x (-1) = -1/18
Winning with a comeout of 4: 1/12 x 1/3 x 1 = 1/36
Losing with a comeout of 4: 1/12 x 2/3 x (-1) = -1/18
Winning with a comeout of 5: 1/9 x 2/5 x 1 = 2/45
Losing with a comeout of 5: 1/9 x 3/5 x (-1) = -1/15
Winning with a comeout of 6: 5/36 x 5/11 x 1 = 25/396
Losing with a comeout of 6: 5/36 x 6/11 x (-1) = -5/66
Winning with a comeout of 7: 1/6 x 1 x 1 = 1/6
Losing with a comeout of 7: 1/6 x 0 x (-1) = 0
Winning with a comeout of 8: 5/36 x 5/11 x 1 = 25/396
Losing with a comeout of 8: 5/36 x 6/11 x (-1) = -5/66
Winning with a comeout of 9: 1/9 x 2/5 x 1 = 2/45
Losing with a comeout of 9: 1/9 x 3/5 x (-1) = -1/15
Winning with a comeout of 10: 1/12 x 1/3 x 1 = 1/36
Losing with a comeout of 10: 1/12 x 2/3 x (-1) = -1/18
Winning with a comeout of 11: 1/18 x 1 x 1 = 1/18
Losing with a comeout of 11: 1/18 x 0 x (-1) = 0
Winning with a comeout of 12: 1/36 x 0 x 1 = 0
Losing with a comeout of 12: 1/36 x 1 x (-1) = -1/36
The total EV is the sum of these 22 numbers, which is -7/495.
Note that 495 does not appear in any of the 22 numbers themselves; it just happens that when you add them up and reduce the fraction to lowest terms, the denominator is 495.
January 17th, 2022 at 2:55:40 PM
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Mr.TDG: Yes, you are correct. I was confusing EV with probability of winning. My personal ultimate approach mathematically (4th grade style) is probability of winning. I try to create betting patterns that give me an edge over the house without regard to the EV. I basically disregard EV, even though I know the actual loss potential of the event, to concentrate or focus on the table, the shooters, and the play. With that being stated, my play goes to both sides (right side and "dark side) and will include hedging. It does not work all the time, but with a moderate buy-in, I can play for a period of time that is comfortable and rewarding. So, thank you for ending my confusion.
Rhetorical question: You stated that you are basically a right side bettor (PL+odds). Your math acumen is off the charts in my estimation. Mr W. is a "Dark" side player. Which of you gets the edge? Have you ever played along side one another?
Your explanation of point play, for me, is probability which allows me the luxury of using my 4th grade arithmetic without regard and the "confusion" of EV. I believe many players shy away from the game and the real simplicity it can offer if they can stay "out of the weeds" of HA/HE which in my view are misleading with the extremely low percentages that are offered within the confines of this forum.
Charlie's chart using integers is just too much information, and, for me as well as the vast majority of potential players, would have them running and screaming into the night.
Again, thank you for your hard work, patience, due diligence, and humility. It is much appreciated.
tuttigym
Rhetorical question: You stated that you are basically a right side bettor (PL+odds). Your math acumen is off the charts in my estimation. Mr W. is a "Dark" side player. Which of you gets the edge? Have you ever played along side one another?
Your explanation of point play, for me, is probability which allows me the luxury of using my 4th grade arithmetic without regard and the "confusion" of EV. I believe many players shy away from the game and the real simplicity it can offer if they can stay "out of the weeds" of HA/HE which in my view are misleading with the extremely low percentages that are offered within the confines of this forum.
Charlie's chart using integers is just too much information, and, for me as well as the vast majority of potential players, would have them running and screaming into the night.
Again, thank you for your hard work, patience, due diligence, and humility. It is much appreciated.
tuttigym
January 17th, 2022 at 3:01:47 PM
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Quote: DieterI'm going to conclude it means Mission is reasonably confident in his 15th grade math skills.
Mr. Dieter: This forum is extremely lucky to have you, a certified "MSL" * member in its ranks.
Thank you.
tuttigym
*Mission 146 as a Second Language
January 17th, 2022 at 3:09:07 PM
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Quote: tuttigymRhetorical question: You stated that you are basically a right side bettor (PL+odds). Your math acumen is off the charts in my estimation. Mr W. is a "Dark" side player. Which of you gets the edge? Have you ever played along side one another?
He has the edge, as Don't Pass is a slightly better bet than Pass. I don't like playing DP because I am turned off by the fact that odds bets with DP pay less than even money.
As for whether we have played alongside each other: no. I usually don't come across the other forum regulars while I'm in Vegas, especially as I don't gamble that much - for example, the last time I was here, the only gambling I did was in VP, and even then, it was less than an hour.
January 17th, 2022 at 3:20:16 PM
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I have a similar feeling about laying odds. 5:6 is close to even money and 2:3 isn’t so bad, but I don’t like laying 2:1, especially when I have a large flat bet
It’s all about making that GTA
January 17th, 2022 at 4:03:13 PM
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Funny. I do too and recognize it as irrational but still feel it. I always think now that I’ve survived the come out, my DP is +EV, why should I dilute it with odds . . .
. . . of course I then go make a don’t come.
. . . of course I then go make a don’t come.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 18th, 2022 at 6:00:25 AM
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Quote: tuttigymQuote: DieterI'm going to conclude it means Mission is reasonably confident in his 15th grade math skills.
Mr. Dieter: This forum is extremely lucky to have you, a certified "MSL" * member in its ranks.
Thank you.
tuttigym
*Mission 146 as a Second Language
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You'll forgive me for not having the time to also be an English tutor, I'm sure.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
