DI Minimizing 7 Changing HE
A DI practitioner claims to be able to throw the dice and reduce a probability of a seven rolling from 1 in 6 to 1 in 8.
How does that change the HE on a 1 unit pass line bet with no odds if there is no point established? How does the HE change with a decrease in chance of come out roll wins, but increase in chance of come out roll loses, and increased chance of points being established.
How does that change the HE on a 1 unit pass line bet with no odds if there is a point established? How does the HE change due to a decrease in chance of seven out, an increase in chance of making the point, and an increase in chance of a push?
Feel free to assume either the probability of the other number combinations increase equally, or increase in proportion of their existing probabilities. Ideally, an analyses of both assumptions would be ideal.
Gene
Quote: AlanMendelsonThe house edge doesn't change.
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If the probabilities change, then the House Edge changes. The entire purpose of the DI theory is that the House Edge would change.
Gene
Quote: Mission146Quote: AlanMendelsonThe house edge doesn't change.
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If the probabilities change, then the House Edge changes. The entire purpose of the DI theory is that the House Edge would change.
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The house edge is mathematical and fixed. The edge never changes.
I assume that DIs try to avoid the seven EXCEPT on come out rolls.Quote: GenoDRPhHow does the HE change with a decrease in chance of come out roll wins, but increase in chance of come out roll loses, and increased chance of points being established. link to original post
On a come out, they don’t try to influence, or if they do, they may try to INCREASE the number of sevens.
At least that’s what I would assume.
The “mechanics” of how this would hypothetically work would be either “keeping the die on axis” or “keeping the die spinning in a correlated manner.” I expect those two “methods” would lead to different math equations because of how you spread the probabilities to other faces.
Would probably be simplest to start with an “on axis” analysis.
Say after the come out, the left die is set with a 3 on top and a 1 and 6 “off axis.” The right die is set with a 3 on top and a 2 and 5 “off axis.” For a die, the “on axis” faces would have X% of showing and “off axis” faces have Y probability of showing, where X > Y.
You can then use above to solve for X and Y where the probability of a 7 is 1/8.
You will then get a probability distribution for the other numbers based on X and Y.
For the come out, the DI would say set 1 and 6 “off axis” on both dice. Apply X and Y to get number distributions and 7 should happen some % more than 1/6.
Anyway, Mission, I hope this makes sense and helps frame the problem in an intelligible way.
Correlated spinning seems more complicated to me. Happy to think through the possible construct if people are interested in analysis.
Quote: AlanMendelsonQuote: Mission146Quote: AlanMendelsonThe house edge doesn't change.
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If the probabilities change, then the House Edge changes. The entire purpose of the DI theory is that the House Edge would change.
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The house edge is mathematical and fixed. The edge never changes.
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What is the probability, holding 10dJdQdKd, of you drawing to a Royal Flush if I remove all but the Ace of Diamonds from the deck? If the probability has changed, then what is your expected outcome of that draw?
The house edge is fixed only if the probability of each face of a die showing on top is 1/6. If that probability changes the edge changes.Quote: AlanMendelsonQuote: Mission146Quote: AlanMendelsonThe house edge doesn't change.
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If the probabilities change, then the House Edge changes. The entire purpose of the DI theory is that the House Edge would change.
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The house edge is mathematical and fixed. The edge never changes.
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Simple example, introduce loaded dice that always through a 10. The house edge of playing craps on pass line with those dice is negative infinity. It’s impossible not to hit the pass line for the throws that die is rolled for the coke out and the following roll.
Quote: AlanMendelsonQuote: Mission146Quote: AlanMendelsonThe house edge doesn't change.
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If the probabilities change, then the House Edge changes. The entire purpose of the DI theory is that the House Edge would change.
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The house edge is mathematical and fixed. The edge never changes.
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I cannot imagine a post showing less insight into math than this one!!!!! PLEASE TELL ME YOU WERE TROLLING the forum and you do not actually believe that if the frequency of certain numbers change that the house edge wouldn’t change? PLEASE!!!!!
Quote: SOOPOOQuote: AlanMendelsonQuote: Mission146Quote: AlanMendelsonThe house edge doesn't change.
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If the probabilities change, then the House Edge changes. The entire purpose of the DI theory is that the House Edge would change.
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The house edge is mathematical and fixed. The edge never changes.
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I cannot imagine a post showing less insight into math than this one!!!!! PLEASE TELL ME YOU WERE TROLLING the forum and you do not actually believe that if the frequency of certain numbers change that the house edge wouldn’t change? PLEASE!!!!!
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I guess I'm using the wrong definitions. But the way I determine the house edge is by the numbers of combinations for each pairing. That to me is the house edge.
Quote: AlanMendelson
I guess I'm using the wrong definitions. But the way I determine the house edge is by the numbers of combinations for each pairing. That to me is the house edge.
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I don't think you're using the wrong definition. The number of combinations for each roll of a pair of dice represents the probability of rolling that result. The fundamental goal of dice influencing is to change the probabilities (by increasing or decreasing the likelihood of particular results) to make the probabilities NOT that. If that can be done, then the goal would be to demonstrate that it can be done well-enough to shift the probabilities such that one would gain an advantage.
In other words, there's literally zero point to attempting to influence the dice (unless you just set them because it looks neat) if you're not trying to influence the probabilities to be something other than what they are otherwise.
Quote: DJTeddyBearI assume that DIs try to avoid the seven EXCEPT on come out rolls.Quote: GenoDRPhHow does the HE change with a decrease in chance of come out roll wins, but increase in chance of come out roll loses, and increased chance of points being established. link to original post
On a come out, they don’t try to influence, or if they do, they may try to INCREASE the number of sevens.
At least that’s what I would assume.
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Under the conditions in my hypothetical - possibly rolling a 7 less than 1 in 8 times - change the ratio of come out roll winners on a 7 or 11 to come out roll losers on a 2,3 or 12 or establishing a point? Is it more advantageous to roll randomly on the come out or use DI on the come out, factoring in the possibility of hitting a point for a win in a subsequent roll?
Gene
Quote: unJonI don’t think the OP question is the exactly correct to analyze. To be more practical, the question would have the DI attempt to throw more 7s on the come out and less 7s after.
The “mechanics” of how this would hypothetically work would be either “keeping the die on axis” or “keeping the die spinning in a correlated manner.” I expect those two “methods” would lead to different math equations because of how you spread the probabilities to other faces.
Would probably be simplest to start with an “on axis” analysis.
Say after the come out, the left die is set with a 3 on top and a 1 and 6 “off axis.” The right die is set with a 3 on top and a 2 and 5 “off axis.” For a die, the “on axis” faces would have X% of showing and “off axis” faces have Y probability of showing, where X > Y.
You can then use above to solve for X and Y where the probability of a 7 is 1/8.
You will then get a probability distribution for the other numbers based on X and Y.
For the come out, the DI would say set 1 and 6 “off axis” on both dice. Apply X and Y to get number distributions and 7 should happen some % more than 1/6.
Anyway, Mission, I hope this makes sense and helps frame the problem in an intelligible way.
Correlated spinning seems more complicated to me. Happy to think through the possible construct if people are interested in analysis.
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The OP stands by his hypothetical of a hypothetical dice influencer attempting to roll a seven no greater than 1 roll in 8 for all rolls, under the conditions he mentioned in the original, not unintelligible post.
Gene
Quote: GenoDRPhQuote: unJonI don’t think the OP question is the exactly correct to analyze. To be more practical, the question would have the DI attempt to throw more 7s on the come out and less 7s after.
The “mechanics” of how this would hypothetically work would be either “keeping the die on axis” or “keeping the die spinning in a correlated manner.” I expect those two “methods” would lead to different math equations because of how you spread the probabilities to other faces.
Would probably be simplest to start with an “on axis” analysis.
Say after the come out, the left die is set with a 3 on top and a 1 and 6 “off axis.” The right die is set with a 3 on top and a 2 and 5 “off axis.” For a die, the “on axis” faces would have X% of showing and “off axis” faces have Y probability of showing, where X > Y.
You can then use above to solve for X and Y where the probability of a 7 is 1/8.
You will then get a probability distribution for the other numbers based on X and Y.
For the come out, the DI would say set 1 and 6 “off axis” on both dice. Apply X and Y to get number distributions and 7 should happen some % more than 1/6.
Anyway, Mission, I hope this makes sense and helps frame the problem in an intelligible way.
Correlated spinning seems more complicated to me. Happy to think through the possible construct if people are interested in analysis.
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The OP stands by his hypothetical of a hypothetical dice influencer attempting to roll a seven no greater than 1 roll in 8 for all rolls, under the conditions he mentioned in the original, not unintelligible post.
Gene
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LOL. If you found my post unintelligible, wait until you see Mission’s response and math!
Quote: unJon
LOL. If you found my post unintelligible, wait until you see Mission’s response and math!
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I should double down on this and post it in extremely poor French.
Quote: Mission146Quote: unJon
LOL. If you found my post unintelligible, wait until you see Mission’s response and math!
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I should double down on this and post it in extremely poor French.
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Would you prefer my response in my equally poor Spanish or my equally poor Italian?
Gene
Quote: unJonQuote: GenoDRPhQuote: unJonI don’t think the OP question is the exactly correct to analyze. To be more practical, the question would have the DI attempt to throw more 7s on the come out and less 7s after.
The “mechanics” of how this would hypothetically work would be either “keeping the die on axis” or “keeping the die spinning in a correlated manner.” I expect those two “methods” would lead to different math equations because of how you spread the probabilities to other faces.
Would probably be simplest to start with an “on axis” analysis.
Say after the come out, the left die is set with a 3 on top and a 1 and 6 “off axis.” The right die is set with a 3 on top and a 2 and 5 “off axis.” For a die, the “on axis” faces would have X% of showing and “off axis” faces have Y probability of showing, where X > Y.
You can then use above to solve for X and Y where the probability of a 7 is 1/8.
You will then get a probability distribution for the other numbers based on X and Y.
For the come out, the DI would say set 1 and 6 “off axis” on both dice. Apply X and Y to get number distributions and 7 should happen some % more than 1/6.
Anyway, Mission, I hope this makes sense and helps frame the problem in an intelligible way.
Correlated spinning seems more complicated to me. Happy to think through the possible construct if people are interested in analysis.
link to original post
The OP stands by his hypothetical of a hypothetical dice influencer attempting to roll a seven no greater than 1 roll in 8 for all rolls, under the conditions he mentioned in the original, not unintelligible post.
Gene
link to original post
LOL. If you found my post unintelligible, wait until you see Mission’s response and math!
link to original post
https://www.youtube.com/watch?v=-z0o6hqEnP0
Math is math!
If I'm playing craps alone and I have a 5% players edge, whats the houses edge?Quote: AlanMendelsonThe house edge doesn't change.
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Math? We ain't got no math, we don't need no math! I don't have to show you any stinking math!Quote: GenoDRPhQuote: unJonQuote: GenoDRPhQuote: unJonI don’t think the OP question is the exactly correct to analyze. To be more practical, the question would have the DI attempt to throw more 7s on the come out and less 7s after.
The “mechanics” of how this would hypothetically work would be either “keeping the die on axis” or “keeping the die spinning in a correlated manner.” I expect those two “methods” would lead to different math equations because of how you spread the probabilities to other faces.
Would probably be simplest to start with an “on axis” analysis.
Say after the come out, the left die is set with a 3 on top and a 1 and 6 “off axis.” The right die is set with a 3 on top and a 2 and 5 “off axis.” For a die, the “on axis” faces would have X% of showing and “off axis” faces have Y probability of showing, where X > Y.
You can then use above to solve for X and Y where the probability of a 7 is 1/8.
You will then get a probability distribution for the other numbers based on X and Y.
For the come out, the DI would say set 1 and 6 “off axis” on both dice. Apply X and Y to get number distributions and 7 should happen some % more than 1/6.
Anyway, Mission, I hope this makes sense and helps frame the problem in an intelligible way.
Correlated spinning seems more complicated to me. Happy to think through the possible construct if people are interested in analysis.
link to original post
The OP stands by his hypothetical of a hypothetical dice influencer attempting to roll a seven no greater than 1 roll in 8 for all rolls, under the conditions he mentioned in the original, not unintelligible post.
Gene
link to original post
LOL. If you found my post unintelligible, wait until you see Mission’s response and math!
link to original post
https://www.youtube.com/watch?v=-z0o6hqEnP0
Math is math!
link to original post
Quote: GenoDRPhSo as not to hijack an existing thread, here's a hypothetical:
A DI practitioner claims to be able to throw the dice and reduce a probability of a seven rolling from 1 in 6 to 1 in 8.
How does that change the HE on a 1 unit pass line bet with no odds if there is no point established? How does the HE change with a decrease in chance of come out roll wins, but increase in chance of come out roll loses, and increased chance of points being established.
How does that change the HE on a 1 unit pass line bet with no odds if there is a point established? How does the HE change due to a decrease in chance of seven out, an increase in chance of making the point, and an increase in chance of a push?
Feel free to assume either the probability of the other number combinations increase equally, or increase in proportion of their existing probabilities. Ideally, an analyses of both assumptions would be ideal.
Gene
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DISCLAIMER: THIS POST HAS BEEN SUBSTANTIALLY EDITED TO REMOVE MISTAKES DUE TO GOING TOO FAST AND ALSO THE FACT THAT I AM AN IDIOT.
First of all, I apologize for the delay, but my big project took a little longer than anticipated.
Secondly, I'm going to answer this, but I do want to somewhat modify the parameters. The key to being a good snake oil salesman is that the product you are offering must superficially sound viable, with that, I think 1 in 8 sevens is entirely too high. It wouldn't even take a very large sample of tosses to prove it is mathematically unlikely that the claimant does that long-term, so I seriously doubt anyone would even make that claim.
Superficially, it looks like just a 4.1% change of outcome, but look at it over 100 rolls with 12.5 (actual) compared to 16.6 (normally expected) and that example would be saying that it effectively reduces the frequency of sevens by about 25%.
What I am going to do is reduce this to a more reasonable 1 in 6.5, which would change the expected percentage from 16.6667% to about 15.3846%.
The second modification I am going to make is that we are just going to assume that the CO roll is thrown such as to not change anything vis-a-vis the long-term probabilities. After all, a PL bet enjoys its only advantage (in terms of probability of winning as opposed to losing) on the CO anyway. Now, one might argue, "But, I try to roll more sevens on the CO, but I don't care," since this is all hypothetical anyway.
The one thing that I am keeping is that we will treat the increase to other combinations as equal. I get that dice influencers say they are trying to hit certain other numbers with greater frequency, but in the non-hypothetical world, we have yet to see any actual mathematical evidence to back the assertion that they can even reduce sevens, much less increase the probability of other numbers.
With that, the probabilities on the CO are:
Snake Eyes: 2.7778%
Three: 5.5556%
Four: 8.3333%
Five: 11.1111%
Six: 13.8889%
Seven: 16.6667%
Eight: 13.8889%
Nine: 11.1111%
Ten: 8.3333%
Yo: 5.5556%
Midnight: 2.7778%
Okay, so what we have are the following:
Immediate Loss: 11.1111%
Immediate Win: 22.2222%
Point Established (4, 10): 16.6667%
Points Established (5,9): 22.2222%
Point Established (6, 8) 27.7778%
Any Point Established: 66.6667%
Therefore, our expectation on the Pass Line bet (taken alone) is simply:
.222222 - .111111 = .111111
Thus, on a bet of one unit, we have an expected gain of 11.1111% looking at the Come Out roll in isolation.
The point being established, we only care about sevens and that point number. With that, we have the following expected losses based on the probability of arriving at that Point Number (via the CO) to begin with and then comparing to sevens:
Point of Four or Ten: 3/36 * (3/9 - 6/9) = -0.02777777777 * 2 = -0.05555555555
Point of Five or Nine: 4/36 * (4/10 - 6/10) = -0.02222222222 * 2 = -.00444444444
Point of Six or Eight: 5/36 * (5/11 - 6/11) = -0.01262626262 * 2 = -0.02525252525
Now, we just total these losses and then subtract the result from our Come Out advantage:
.111111 - (.0555555555555 + .0444444444444 + .02525252525) = -0.01414152524
Thus, reflecting the House Edge of the Pass Line of about 1.414%.
That's the Pass Line bet in a nutshell.
Okay, so what we want to do is reduce the odds of a seven to 1 in 6.5 rather than 1 in 6, which will result in the following probability shift:
(1/6) - (1/6.5) = 0.01282051282
Okay, so we want sevens to decrease by that and everything else to increase proportionately. It's important to note that, given the intended impact of dice setting, these things are not meant to increase proportionately to one another, but there's really no way to measure the success rate of something that has not been demonstrated to be successful in the first place. For that reason, we can't know the exact probabilities of what should shift away from sevens to other numbers.
Also, we have taken away from sevens, so we do not want to add to sevens. For that reason, what we will do is take this change in percentage and multiply it by the fact that the sevens must also be something else and then add those together.
0.01282051282 + (0.01282051282 * 1/6) = 0.01495726495
With that out of the way, we simply multiply our other results' normal probabilities by the above, then add what the probability would normally be into that, and hopefully, the sum of all probabilities after we have done so will be something very close to 1.
Snake Eyes: 2.7778%---> (1/36 * 0.01495726495) + 1/36 = 0.02819325735
Three: 5.5556%---> (2/36 * 0.01495726495) + 2/36 = 0.05638651471
Four: 8.3333%---> (3/36 * 0.01495726495) + 3/36 = 0.08457977207
Five: 11.1111%---> (4/36 * 0.01495726495) + 4/36 = 0.11277302943
Six: 13.8889%---> (5/36 * 0.01495726495) + 5/36 = 0.14096628679
Seven: 16.6667--->0.15384615384
Eight: 13.8889%---> 0.14096628679
Nine: 11.1111%---> 0.11277302943
Ten: 8.3333%---> 0.08457977207
Yo: 5.5556%---> 0.05638651471
Midnight: 2.7778%---> 0.02819325735
SUM: 0.02819325735 + 0.02819325735 + 0.05638651471 + 0.05638651471 + 0.08457977207 + 0.08457977207 + 0.11277302943 + 0.11277302943 + 0.14096628679 + 0.14096628679 + 0.15384615384 = 0.99964387454
I'd assume this is slightly off of one due to rounding. It's certainly a hell of a lot closer than it was last time.
Okay, so now what we will do is look at our situations the same way, so we only care about those numbers and seven.
Four or Ten + Seven Combined Probability:
0.08457977207 + 0.15384615384 = 0.23842592591
Four and Ten Solved:
(0.08457977207/0.23842592591) - (0.15384615384/0.23842592591) = -0.29051530996 * 6/36 = -0.04841921832
Five and Nine Combined Probability:
0.15384615384 + 0.11277302943 = 0.26661918327
Five and Nine Solved:
(0.11277302943/.26661918327) - (0.15384615384/.26661918327) = -0.15405164739 * 8/36 = -0.03423369942
Six and Eight Combined Probability:
0.15384615384 + 0.14096628679 = 0.29481244063
Six and Eight Solved:
(.14096628679/.29481244063) - (.15384615384/.29481244063) = -0.04368834307 * 10/36 = -0.01213565085
As we can see, the disadvantage of these situations has been reduced due to the reduced frequency of sevens thus necessitating an increased frequency of these other numbers. Will it change our results overall to an expected positive, let's find out:
.111111 - 0.04841921832 - 0.03423369942 - 0.0121356085 = 0.01632247376
The result is a player advantage of 1.632247376%, which would be even more insurmountable than the, 'Normal,' disadvantage on the Pass Line bet is over the long-term.
Also, keep in mind we made the following other assumptions:
1.) We roll, 'Normally,' on the Come Out.
2.) While we reduce the frequency of sevens to 1 in 6.5, we are not disproportionately increasing the frequency of other numbers. This is important to note because people who purport to be dice influencers will often state that they are trying for sixes and eights. Since nobody has ever demonstrated that they can reduce the frequency of sevens to 1 in 6.5, much less demonstrated what doing so will increase their expected sixes and eights to, I really don't care.
3.) This is Pass Line only. If you want to know what this does to Odds Bets, it would be trivial to incorporate the Odds into what I have already done. Alternatively, I can always do so later.*
*Obviously, given the above assumptions, Odds bets would be made at an advantage if you could reduce the frequency of sevens (without reducing the frequency of the desired number) at all.
DISCLAIMER: THIS POST HAS BEEN SUBSTANTIALLY EDITED TO REMOVE MISTAKES DUE TO GOING TOO FAST AND ALSO THE FACT THAT I AM AN IDIOT.
worth emphasizingQuote: Mission146[snips]
In conclusion, this would reduce the House Edge of the Pass Line bet to 0.79829%,
Of course we want to know! Thanks for the work, Mission. I don't think the math is over my head, exactly, but it's always a chore to try to follow what someone else has done. Most here are also pondering that. But thanks!Quote:... roughly, given the following parameters:
1.) We roll, 'Normally,' on the Come Out.
2.) While we reduce the frequency of sevens to 1 in 6.5, we are not disproportionately increasing the frequency of other numbers. This is important to note because people who purport to be dice influencers will often state that they are trying for sixes and eights.
3.) This is Pass Line only. If you want to know what this does to Odds Bets, and if an overall advantage could be produced by incorporating them, it would be trivial to incorporate the Odds into what I have already done. Alternatively, I can always do so later.*
I was thinking this: I believe it is a principle that if the HE on any bet cannot be changed to +EV, then adding a zero edge bet means it can't be flipped to +EV then either [edited]. However, then I read the below
so instead of betting along with the 'good shooter' a player should endeavor instead to find a different player who isn't using up all his odds betting potential and bet behind his number. There are a lot of those people!Quote:*Obviously, given the above assumptions, Odds bets would be made at an advantage if you could reduce the frequency of sevens (without reducing the frequency of the desired number) at all.
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A first time player is just as likely to have a monster roll as the most experienced shooter
Under the same or similar hypothetical conditions, how infrequent must a 7 roll to reduce the HE on a 1 unit pass line bet with no odds behind to reduce the HE on that bet from 1..41% to zero?
Gene
Breakeven on no odds passline I get 1/6.23 of the time 7.
Mission I don’t know where you messed up (or if I messed up, but I just double checked my Excel).
* As I said in my first (possibly unintelligible) post in this thread, I don’t think raising the probability of rolling other numbers is the right way to analyze the issue. For instance, if the DI in theory was setting so to get a 1 and 6 not to show one of the die, and was semi-successful in that, then the chance of a 2 or 12 would of course fall also, in addition to the chance of 7. And the chance of other numbers would increase more than ratably.
Quote: GenoDRPhI thank you for your hard work and promptitude in answering the question asked, as modified. I chose a single, 1 unit pass line bet with no odds behind because that's a common bet, and has a low house edge. I figured that would be a relatively simple analysis.
Under the same or similar hypothetical conditions, how infrequent must a 7 roll to reduce the HE on a 1 unit pass line bet with no odds behind to reduce the HE on that bet from 1..41% to zero?
Gene
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The change we made cut the HE almost in half, so I’d say that they would have to roll slightly fewer than 1 in 7 sevens, all else being equal and proportionate. You’re welcome! I’ll do Odds tomorrow based on 1 in 6.5 and actually do it right; though the oversight is negligible.
Quote: unJonI get a player edge of 1.6459% on a flat passline if you roll sevens 1/6 of the time on come out and 1/6.5 of the time after. Assuming the extra probability is spread among the other outcomes ratably.*
Breakeven on no odds passline I get 1/6.23 of the time 7.
Mission I don’t know where you messed up (or if I messed up, but I just double checked my Excel).
* As I said in my first (possibly unintelligible) post in this thread, I don’t think raising the probability of rolling other numbers is the right way to analyze the issue. For instance, if the DI in theory was setting so to get a 1 and 6 not to show one of the die, and was semi-successful in that, then the chance of a 2 or 12 would of course fall also, in addition to the chance of 7. And the chance of other numbers would increase more than ratably.
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I’m not sure, but I just noticed it would be a player ADVANTAGE of 0.79829% per my math because I started with +HE and subtracted from that. I had somewhere to go and was going too fast anyway. Editing that post now.
If I made a mistake, do you think you’d be able to find it? I feel like you’d have better fortune with mine than me trying to check your Excel; I probably wouldn’t even understand what was being done.
Quote: Mission146Quote: unJonI get a player edge of 1.6459% on a flat passline if you roll sevens 1/6 of the time on come out and 1/6.5 of the time after. Assuming the extra probability is spread among the other outcomes ratably.*
Breakeven on no odds passline I get 1/6.23 of the time 7.
Mission I don’t know where you messed up (or if I messed up, but I just double checked my Excel).
* As I said in my first (possibly unintelligible) post in this thread, I don’t think raising the probability of rolling other numbers is the right way to analyze the issue. For instance, if the DI in theory was setting so to get a 1 and 6 not to show one of the die, and was semi-successful in that, then the chance of a 2 or 12 would of course fall also, in addition to the chance of 7. And the chance of other numbers would increase more than ratably.
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I’m not sure, but I just noticed it would be a player ADVANTAGE of 0.79829% per my math because I started with +HE and subtracted from that. I had somewhere to go and was going too fast anyway. Editing that post now.
If I made a mistake, do you think you’d be able to find it? I feel like you’d have better fortune with mine than me trying to check your Excel; I probably wouldn’t even understand what was being done.
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Closing the gap.
The impact of odds is incredible. Huge player edge if you could do 1/6.5. I get breakeven around 1/6.037.
Haven’t double checked the odds calcs yet though.
Quote: Mission146Quote: unJonI get a player edge of 1.6459% on a flat passline if you roll sevens 1/6 of the time on come out and 1/6.5 of the time after. Assuming the extra probability is spread among the other outcomes ratably.*
Breakeven on no odds passline I get 1/6.23 of the time 7.
Mission I don’t know where you messed up (or if I messed up, but I just double checked my Excel).
* As I said in my first (possibly unintelligible) post in this thread, I don’t think raising the probability of rolling other numbers is the right way to analyze the issue. For instance, if the DI in theory was setting so to get a 1 and 6 not to show one of the die, and was semi-successful in that, then the chance of a 2 or 12 would of course fall also, in addition to the chance of 7. And the chance of other numbers would increase more than ratably.
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I’m not sure, but I just noticed it would be a player ADVANTAGE of 0.79829% per my math because I started with +HE and subtracted from that. I had somewhere to go and was going too fast anyway. Editing that post now.
If I made a mistake, do you think you’d be able to find it? I feel like you’d have better fortune with mine than me trying to check your Excel; I probably wouldn’t even understand what was being done.
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In the beginning part of your post you calculate pass line HE at -0.139394. That’s an order of magnitude too high. The HE is -0.01414.
Quote: Mission146
Thus, on a bet of one unit, we have an expected gain of 11.1111% looking at the Come Out roll in isolation.
The point being established, we only care about sevens and that point number. With that, we have the following expected losses in the following cases:
Point of Six or Eight: (5/11 - 6/11) * 2 = --0.18182
Point of Five or Nine: (4/10 - 6/10) * 2 = -.4
Point of Four or Ten: (3/9) - (6/9) * 2 = -0.66667
Finally, to get our total expected loss from each situation, we must multiply by the probability of that situation occurring on the CO roll:
Point of Six or Eight: -.18182 * .277778 = -0.050506
Point of Five or Nine: -.4 * .22222 = -0.088888
Point of Four or Ten: -.66667 * .16667 = -0.111111
Now, we just total these losses and then subtract the result from our Come Out advantage:
.111111 - (.050506 + .088888 + .111111) = -0.139394 (Error is due to rounding)
Quote truncated. Take the 4/10 calculation where you’ve doubled the HE impact. I think you doubled it twice when you combined 4 and 10 for convenience.
Contribution from the 4 is:
3/36 * (3/9 - 6/9) = -0.0277777
So contribution for 4/10 is double that: -0.05555556
You doubled it again to get to -0.1111111111
Quote: AxelWolfYou guys are over here trying to calculate what tiny edge can be achieved via DI if possible. Meanwhile, We have someone with a HUGE advantage on roulette.
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These calculations are good to see. You don't get rich with DI. It's a matter of small wins.
The misinformed think DIs are walking away with bags of cash.
Gene
Flat passline betting, after come out need to throw a little more than 6 fewer sevens in 1,000 rolls.
With max 3/4/5 odds, after come out need to throw 1 fewer seven in 1,000 rolls.
All goes to Ace2 point about how variance of the max odds better can keep them positive so easily.
Quote: AxelWolfYou guys are over here trying to calculate what tiny edge can be achieved via DI if possible. Meanwhile, We have someone with a HUGE advantage on roulette.
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When he either takes the Pepsi Challenge, or gets booted from a B&M casino, then I'll start caring....
Gene
Gene
Quote: unJonQuote: Mission146
Thus, on a bet of one unit, we have an expected gain of 11.1111% looking at the Come Out roll in isolation.
The point being established, we only care about sevens and that point number. With that, we have the following expected losses in the following cases:
Point of Six or Eight: (5/11 - 6/11) * 2 = --0.18182
Point of Five or Nine: (4/10 - 6/10) * 2 = -.4
Point of Four or Ten: (3/9) - (6/9) * 2 = -0.66667
Finally, to get our total expected loss from each situation, we must multiply by the probability of that situation occurring on the CO roll:
Point of Six or Eight: -.18182 * .277778 = -0.050506
Point of Five or Nine: -.4 * .22222 = -0.088888
Point of Four or Ten: -.66667 * .16667 = -0.111111
Now, we just total these losses and then subtract the result from our Come Out advantage:
.111111 - (.050506 + .088888 + .111111) = -0.139394 (Error is due to rounding)
Quote truncated. Take the 4/10 calculation where you’ve doubled the HE impact. I think you doubled it twice when you combined 4 and 10 for convenience.
Contribution from the 4 is:
3/36 * (3/9 - 6/9) = -0.0277777
So contribution for 4/10 is double that: -0.05555556
You doubled it again to get to -0.1111111111
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Thank you! This is what happens when I try to do things quickly, plus I had three different text exchanges going on.
Quote: GenoDRPhWith a break even at 1/6.23 and a +EV at 1/6.5 with no odds behind, and a +EV with max odds at 1/6.03, maybe this isn't outlandish after all....
Gene
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Shows how little casinos are actually worried about it. If you could really do 1/6.5 for the sevens, I show a 17% player advantage playing 3/4/5 max odds.
(Note I still don’t think this analysis is accurate with the assumption of ratably increasing the probability of all non-7 numbers. But whatever.)
Quote: unJonQuote: Mission146
Thus, on a bet of one unit, we have an expected gain of 11.1111% looking at the Come Out roll in isolation.
The point being established, we only care about sevens and that point number. With that, we have the following expected losses in the following cases:
Point of Six or Eight: (5/11 - 6/11) * 2 = --0.18182
Point of Five or Nine: (4/10 - 6/10) * 2 = -.4
Point of Four or Ten: (3/9) - (6/9) * 2 = -0.66667
Finally, to get our total expected loss from each situation, we must multiply by the probability of that situation occurring on the CO roll:
Point of Six or Eight: -.18182 * .277778 = -0.050506
Point of Five or Nine: -.4 * .22222 = -0.088888
Point of Four or Ten: -.66667 * .16667 = -0.111111
Now, we just total these losses and then subtract the result from our Come Out advantage:
.111111 - (.050506 + .088888 + .111111) = -0.139394 (Error is due to rounding)
Quote truncated. Take the 4/10 calculation where you’ve doubled the HE impact. I think you doubled it twice when you combined 4 and 10 for convenience.
Contribution from the 4 is:
3/36 * (3/9 - 6/9) = -0.0277777
So contribution for 4/10 is double that: -0.05555556
You doubled it again to get to -0.1111111111
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I wanted to thank you again for finding this and I also like your formula that just combines the whole thing, instead.
To all: Please note that I am in the process of editing my earlier post which will contain the following disclaimer:
Quote:DISCLAIMER: THIS POST HAS BEEN SUBSTANTIALLY EDITED TO REMOVE MISTAKES DUE TO GOING TOO FAST AND ALSO THE FACT THAT I AM AN IDIOT.
Quote: Mission146Quote: GenoDRPhSo as not to hijack an existing thread, here's a hypothetical:
A DI practitioner claims to be able to throw the dice and reduce a probability of a seven rolling from 1 in 6 to 1 in 8.
How does that change the HE on a 1 unit pass line bet with no odds if there is no point established? How does the HE change with a decrease in chance of come out roll wins, but increase in chance of come out roll loses, and increased chance of points being established.
How does that change the HE on a 1 unit pass line bet with no odds if there is a point established? How does the HE change due to a decrease in chance of seven out, an increase in chance of making the point, and an increase in chance of a push?
Feel free to assume either the probability of the other number combinations increase equally, or increase in proportion of their existing probabilities. Ideally, an analyses of both assumptions would be ideal.
Gene
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DISCLAIMER: THIS POST HAS BEEN SUBSTANTIALLY EDITED TO REMOVE MISTAKES DUE TO GOING TOO FAST AND ALSO THE FACT THAT I AM AN IDIOT.
First of all, I apologize for the delay, but my big project took a little longer than anticipated.
Secondly, I'm going to answer this, but I do want to somewhat modify the parameters. The key to being a good snake oil salesman is that the product you are offering must superficially sound viable, with that, I think 1 in 8 sevens is entirely too high. It wouldn't even take a very large sample of tosses to prove it is mathematically unlikely that the claimant does that long-term, so I seriously doubt anyone would even make that claim.
Superficially, it looks like just a 4.1% change of outcome, but look at it over 100 rolls with 12.5 (actual) compared to 16.6 (normally expected) and that example would be saying that it effectively reduces the frequency of sevens by about 25%.
What I am going to do is reduce this to a more reasonable 1 in 6.5, which would change the expected percentage from 16.6667% to about 15.3846%.
The second modification I am going to make is that we are just going to assume that the CO roll is thrown such as to not change anything vis-a-vis the long-term probabilities. After all, a PL bet enjoys its only advantage (in terms of probability of winning as opposed to losing) on the CO anyway. Now, one might argue, "But, I try to roll more sevens on the CO, but I don't care," since this is all hypothetical anyway.
The one thing that I am keeping is that we will treat the increase to other combinations as equal. I get that dice influencers say they are trying to hit certain other numbers with greater frequency, but in the non-hypothetical world, we have yet to see any actual mathematical evidence to back the assertion that they can even reduce sevens, much less increase the probability of other numbers.
With that, the probabilities on the CO are:
Snake Eyes: 2.7778%
Three: 5.5556%
Four: 8.3333%
Five: 11.1111%
Six: 13.8889%
Seven: 16.6667%
Eight: 13.8889%
Nine: 11.1111%
Ten: 8.3333%
Yo: 5.5556%
Midnight: 2.7778%
Okay, so what we have are the following:
Immediate Loss: 11.1111%
Immediate Win: 22.2222%
Point Established (4, 10): 16.6667%
Points Established (5,9): 22.2222%
Point Established (6, 8) 27.7778%
Any Point Established: 66.6667%
Therefore, our expectation on the Pass Line bet (taken alone) is simply:
.222222 - .111111 = .111111
Thus, on a bet of one unit, we have an expected gain of 11.1111% looking at the Come Out roll in isolation.
The point being established, we only care about sevens and that point number. With that, we have the following expected losses based on the probability of arriving at that Point Number (via the CO) to begin with and then comparing to sevens:
Point of Four or Ten: 3/36 * (3/9 - 6/9) = -0.02777777777 * 2 = -0.05555555555
Point of Five or Nine: 4/36 * (4/10 - 6/10) = -0.02222222222 * 2 = -.00444444444
Point of Six or Eight: 5/36 * (5/11 - 6/11) = -0.01262626262 * 2 = -0.02525252525
Now, we just total these losses and then subtract the result from our Come Out advantage:
.111111 - (.0555555555555 + .0444444444444 + .02525252525) = -0.01414152524
Thus, reflecting the House Edge of the Pass Line of about 1.414%.
That's the Pass Line bet in a nutshell.
Okay, so what we want to do is reduce the odds of a seven to 1 in 6.5 rather than 1 in 6, which will result in the following probability shift:
(1/6) - (1/6.5) = 0.01282051282
Okay, so we want sevens to decrease by that and everything else to increase proportionately. It's important to note that, given the intended impact of dice setting, these things are not meant to increase proportionately to one another, but there's really no way to measure the success rate of something that has not been demonstrated to be successful in the first place. For that reason, we can't know the exact probabilities of what should shift away from sevens to other numbers.
Also, we have taken away from sevens, so we do not want to add to sevens. For that reason, what we will do is take this change in percentage and multiply it by the fact that the sevens must also be something else and then add those together.
0.01282051282 + (0.01282051282 * 1/6) = 0.01495726495
With that out of the way, we simply multiply our other results' normal probabilities by the above, then add what the probability would normally be into that, and hopefully, the sum of all probabilities after we have done so will be something very close to 1.
Snake Eyes: 2.7778%---> (1/36 * 0.01495726495) + 1/36 = 0.02819325735
Three: 5.5556%---> (2/36 * 0.01495726495) + 2/36 = 0.05638651471
Four: 8.3333%---> (3/36 * 0.01495726495) + 3/36 = 0.08457977207
Five: 11.1111%---> (4/36 * 0.01495726495) + 4/36 = 0.11277302943
Six: 13.8889%---> (5/36 * 0.01495726495) + 5/36 = 0.14096628679
Seven: 16.6667--->0.15384615384
Eight: 13.8889%---> 0.14096628679
Nine: 11.1111%---> 0.11277302943
Ten: 8.3333%---> 0.08457977207
Yo: 5.5556%---> 0.05638651471
Midnight: 2.7778%---> 0.02819325735
SUM: 0.02819325735 + 0.02819325735 + 0.05638651471 + 0.05638651471 + 0.08457977207 + 0.08457977207 + 0.11277302943 + 0.11277302943 + 0.14096628679 + 0.14096628679 + 0.15384615384 = 0.99964387454
I'd assume this is slightly off of one due to rounding. It's certainly a hell of a lot closer than it was last time.
Okay, so now what we will do is look at our situations the same way, so we only care about those numbers and seven.
Four or Ten + Seven Combined Probability:
0.08457977207 + 0.15384615384 = 0.23842592591
Four and Ten Solved:
(0.08457977207/0.23842592591) - (0.15384615384/0.23842592591) = -0.29051530996 * 6/36 = -0.04841921832
Five and Nine Combined Probability:
0.15384615384 + 0.11277302943 = 0.26661918327
Five and Nine Solved:
(0.11277302943/.26661918327) - (0.15384615384/.26661918327) = -0.15405164739 * 8/36 = -0.03423369942
Six and Eight Combined Probability:
0.15384615384 + 0.14096628679 = 0.29481244063
Six and Eight Solved:
(.14096628679/.29481244063) - (.15384615384/.29481244063) = -0.04368834307 * 10/36 = -0.01213565085
As we can see, the disadvantage of these situations has been reduced due to the reduced frequency of sevens thus necessitating an increased frequency of these other numbers. Will it change our results overall to an expected positive, let's find out:
.111111 - 0.04841921832 - 0.03423369942 - 0.0121356085 = 0.01632247376
The result is a player advantage of 1.632247376%, which would be even more insurmountable than the, 'Normal,' disadvantage on the Pass Line bet is over the long-term.
Also, keep in mind we made the following other assumptions:
1.) We roll, 'Normally,' on the Come Out.
2.) While we reduce the frequency of sevens to 1 in 6.5, we are not disproportionately increasing the frequency of other numbers. This is important to note because people who purport to be dice influencers will often state that they are trying for sixes and eights. Since nobody has ever demonstrated that they can reduce the frequency of sevens to 1 in 6.5, much less demonstrated what doing so will increase their expected sixes and eights to, I really don't care.
3.) This is Pass Line only. If you want to know what this does to Odds Bets, it would be trivial to incorporate the Odds into what I have already done. Alternatively, I can always do so later.*
*Obviously, given the above assumptions, Odds bets would be made at an advantage if you could reduce the frequency of sevens (without reducing the frequency of the desired number) at all.
DISCLAIMER: THIS POST HAS BEEN SUBSTANTIALLY EDITED TO REMOVE MISTAKES DUE TO GOING TOO FAST AND ALSO THE FACT THAT I AM AN IDIOT.
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Quote: unJonI get a player edge of 1.6459% on a flat passline if you roll sevens 1/6 of the time on come out and 1/6.5 of the time after. Assuming the extra probability is spread among the other outcomes ratably.*
Breakeven on no odds passline I get 1/6.23 of the time 7.
Mission I don’t know where you messed up (or if I messed up, but I just double checked my Excel).
* As I said in my first (possibly unintelligible) post in this thread, I don’t think raising the probability of rolling other numbers is the right way to analyze the issue. For instance, if the DI in theory was setting so to get a 1 and 6 not to show one of the die, and was semi-successful in that, then the chance of a 2 or 12 would of course fall also, in addition to the chance of 7. And the chance of other numbers would increase more than ratably.
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I'm getting a Player Edge of 1.632247376%, but it could just be rounding, at this point. Thank you again for all of the help! In trying to go fast, I made the equations needlessly complicated and ended up just confusing myself.
With the Pass Line out of the way and our methodology (I hope) down, we can calculate the Player Edge or remaining House Edge for Odds.
ODDS BETS
This is easy as it just requires changing one part of our formulas to reflects the Odds Payout:
Odds for Four and Ten:
((0.08457977207/0.23842592591)*2) - (0.15384615384/0.23842592591) = 0.06422703504
The above reflects the advantage on the Odds Bet taken alone. For its influence on a, "Per Come Out," basis, simply multiply the above by the overall probability of occurrence based on the CO, which is 6/36:
0.06422703504 * 6/36 = 0.01070450584
That is the advantage per Come Out roll given the potential for being able to make the Odds Bet. As we have previously established, the player is at an overall advantage anyway. The new Expected Value of a Point of Four Established that we determined is -0.04841921832, so our expectation on Single Odds is insufficient to offset that, 3x/4x/5x odds also wouldn't, because this would be the 3x, but it becomes offset and overall profitable EVEN WITH THIS POINT HAVING BEEN ESTABLISHED at 5x Odds.
Odds for Five and Nine:
((0.11277302943/.26661918327)*3/2) - (0.15384615384/.26661918327) = 0.05743544075
The above reflects the advantage on the Odds Bet taken alone. For its influence on a, "Per Come Out," basis, simply multiply the above by the overall probability of occurrence based on the CO, which is 8/36:
0.05743544075 * 8/36 = 0.01276343127
As we can see, our expectation for Single Odds is slightly greater in this scenario, which makes sense, because trying to Make a Point of 5 or 9 is not as bad of a situation relative to 4/10 to begin with. With regard to the Pass Line bet, -0.03423369942 is the new expected loss of a Point of Five or Nine established given the adjusted probabilities. With that, any Odds greater than 2x Odds would turn a point of five or nine established profitable.
Odds for Six and Eight:
((.14096628679/.29481244063)*6/5) - (.15384615384/.29481244063) = 0.05194282261
With that, we multiply that by the probability of even being the Point Number and get:
0.05194282261 * 10/36 = 0.01442856183
Six and Eight becomes positive even with single odds as our new expected loss of these points established is -0.01213565085, which is less than our positive expected outcome from the Odds Bet. Therefore, in addition to just winning on the Come Out roll, having a Come Out roll of six or eight has positive expected value if we are able to take any Odds whatsoever.
The reason that 6/8 becomes better (with odds) than 5/9 followed by 5/9 being better than 4/10 is because of these changed probabilities in the first place.
Once again, dice influencing proponents will argue that sixes and eights will increase out of proportion to other numbers given the reduced amount of sevens, but once again, nobody has ever demonstrated that they can knock the frequency of sevens down to 1 in 6.5 rolls, so I really don't care.
However, when we assume that the frequency of sevens has dropped so, then because numbers such as six and eight are more likely to begin with, they ultimately end up with a greater percentage of the roll frequency that was taken away from sevens. For that reason, single odds would produce an advantage on 6/8, at least 3x odds to produce an advantage on 5/9 and starting at 5x Odds for Points of four and ten.
With that all considered, if you were at a table upon which you could take 5x Odds, 10x Odds, or something even more than that...and again assuming you could roll sevens with a frequency of 1 in 6.5, or less often, then every single roll of the dice would see you at an advantage. Recall, just rolling the dice, 'Randomly,' you have an advantage on the Come Out roll to begin with.
Quote: unJonQuote: GenoDRPhWith a break even at 1/6.23 and a +EV at 1/6.5 with no odds behind, and a +EV with max odds at 1/6.03, maybe this isn't outlandish after all....
Gene
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Shows how little casinos are actually worried about it. If you could really do 1/6.5 for the sevens, I show a 17% player advantage playing 3/4/5 max odds.
(Note I still don’t think this analysis is accurate with the assumption of ratably increasing the probability of all non-7 numbers. But whatever.)
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Would you be willing to show your work?
Gene
Yes but will be about a week. I’m traveling and want to check it. I see Mission posted Odds calculations above. Haven’t reviewed if it’s consistent with mine.Quote: GenoDRPhQuote: unJonQuote: GenoDRPhWith a break even at 1/6.23 and a +EV at 1/6.5 with no odds behind, and a +EV with max odds at 1/6.03, maybe this isn't outlandish after all....
Gene
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Shows how little casinos are actually worried about it. If you could really do 1/6.5 for the sevens, I show a 17% player advantage playing 3/4/5 max odds.
(Note I still don’t think this analysis is accurate with the assumption of ratably increasing the probability of all non-7 numbers. But whatever.)
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Would you be willing to show your work?
Gene
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ETA Also just occurred to me the 17% was based on the $1 pass line bet. That’s not really right as you have more at risk once you put your odds out there. So it will come down a bunch.
