Calculating Craps Cost Per Hour
I've gone down a rabbit hole and now can't stop overthinking what my EV per hour is for craps. I play $10 pass line with up to 3 $10 come bets backed up by 2x odds always working.
Since the house edge for the odds is 0, am I able to safely exclude it from my EV calculation? I know odds will affect the variance but not the EV.
Assuming 100 rolls per hour, HE of 0.42% per roll, 1 pass and 3 come bets ($40 total) I've worked out a loss of around $16.80 an hour. However, there are times where I won't have all 3 come bets on the table (come out roll for example).
As I understand it, despite having up to 4 bets on the table, is my combined HE is still going to be around 0.42% per roll? Or does adding come bets increase the combined HE?
Any help is appreciated, thank you
For 1 pass bet (no odds) - 100 rolls per hour = -$4.20 per hour. If this was $30 pass it would be about -$12.60 an hour. Is this correct?
Using the combined house edge table: /games/craps/basics/#strategy the house edge per roll for 2X odds is 0.18%.
Therefore a $10 pass bet with $20 odds ($30) assuming 100 rolls per hour costs -$5.40 an hour.
This shows that a $10 pass bet with 2X odds costs less per hour than just $30 pass with no odds. (-$5.40 < -$12.60)
But what I am having trouble understanding is how doing the 2X odds costs more per hour if the house edge on the odds bet is 0? (-$4.20 > -$5.40)
So to summarise, I've come up with 2 different numbers that show the cost per hour of the strategy I want to use and want to know which one is correct?
1. Combined house edge 2X (0.18%) x 100 rolls x 3 (2 extra come bets) x $30 per bet (10 for pass 20 for odds) = -$16.20 per hour
2. Ignore odds bet in EV calculation as HE is 0: Pass bet (0.42%) x 100 rolls x 3 (2 extra come bets) x $10 per bet (ignoring odds) = -$12.60 per hour.
Thank you in advance
https://wizardofodds.com/games/craps/basics/#strategy
yesQuote: ddsdoniI've done some more thinking and came up with a couple of calculations I want to see are correct.
For 1 pass bet (no odds) - 100 rolls per hour = -$4.20 per hour. If this was $30 pass it would be about -$12.60 an hour. Is this correct?
maybe there is an error you've found.Quote:Using the combined house edge table: /games/craps/basics/#strategy the house edge per roll for 2X odds is 0.18%.
Therefore a $10 pass bet with $20 odds ($30) assuming 100 rolls per hour costs -$5.40 an hour.
This shows that a $10 pass bet with 2X odds costs less per hour than just $30 pass with no odds. (-$5.40 < -$12.60)
But what I am having trouble understanding is how doing the 2X odds costs more per hour if the house edge on the odds bet is 0? (-$4.20 > -$5.40)
One error in your thinking is that you can maintain the line bet and 2 come bets like that. Sometimes you will 7-out early or, even better, hit one of your numbers early, and you don't maintain 3 bets. Even if the dice "get" what you want to do and cooperated, you still have to wait for the process to build to 3. So you're loss per hour on average is going to be less than you show. The formulas are only good for playing the pass line one bet at a time, if you want an accurate loss per hour. Even 100 rolls per hour is somewhat of a wild guess at a real craps table.Quote:So to summarise, I've come up with 2 different numbers that show the cost per hour of the strategy I want to use and want to know which one is correct?
1. Combined house edge 2X (0.18%) x 100 rolls x 3 (2 extra come bets) x $30 per bet (10 for pass 20 for odds) = -$16.20 per hour
2. Ignore odds bet in EV calculation as HE is 0: Pass bet (0.42%) x 100 rolls x 3 (2 extra come bets) x $10 per bet (ignoring odds) = -$12.60 per hour.
Thank you in advance
link to original post
Someone here might be able to figure it out. You seem pretty adept, maybe you should try. Maybe you can figure out a possible error that's made. Here is where the Wizard shows his work
https://wizardofodds.com/games/craps/appendix/1/
PS: the Wizard is very touchy about errors. PM him instead of posting in a thread, it if you can find it
Quote: ddsdoniHi everyone,
I've gone down a rabbit hole and now can't stop overthinking what my EV per hour is for craps. I play $10 pass line with up to 3 $10 come bets backed up by 2x odds always working.
Since the house edge for the odds is 0, am I able to safely exclude it from my EV calculation? I know odds will affect the variance but not the EV.
Assuming 100 rolls per hour, HE of 0.42% per roll, 1 pass and 3 come bets ($40 total) I've worked out a loss of around $16.80 an hour. However, there are times where I won't have all 3 come bets on the table (come out roll for example).
As I understand it, despite having up to 4 bets on the table, is my combined HE is still going to be around 0.42% per roll? Or does adding come bets increase the combined HE?
Any help is appreciated, thank you
link to original post
I assume you are thinking of reality play, so any answer should first require: (1) the player buy-in, (2) the target $$ win result desired as a percentage of the buy-in, and (3) the time required to effect that result.
Others here will get into the math weeds, but talking real $$ as a result of real play would probably be a more effective and more helpful approach to the game.
tuttigym
Quote: tuttigymQuote: ddsdoniHi everyone,
I've gone down a rabbit hole and now can't stop overthinking what my EV per hour is for craps. I play $10 pass line with up to 3 $10 come bets backed up by 2x odds always working.
Since the house edge for the odds is 0, am I able to safely exclude it from my EV calculation? I know odds will affect the variance but not the EV.
Assuming 100 rolls per hour, HE of 0.42% per roll, 1 pass and 3 come bets ($40 total) I've worked out a loss of around $16.80 an hour. However, there are times where I won't have all 3 come bets on the table (come out roll for example).
As I understand it, despite having up to 4 bets on the table, is my combined HE is still going to be around 0.42% per roll? Or does adding come bets increase the combined HE?
Any help is appreciated, thank you
link to original post
I assume you are thinking of reality play, so any answer should first require: (1) the player buy-in, (2) the target $$ win result desired as a percentage of the buy-in, and (3) the time required to effect that result.
Others here will get into the math weeds, but talking real $$ as a result of real play would probably be a more effective and more helpful approach to the game.
tuttigym
link to original post
Oh yeah? And, how would you come up with an expected loss per hour? Would your only acceptable answer be to physically go with a specific buy-in and play Craps for 10,000 hours and determine what your average loss per hour was?
Further, haven't you denied that the House Edge is even, "Real," in your opinion? And yet, when I asked about a coin flip game where you get paid $1 if you win, but have to pay me $10 if you lose and whether or not the House Edge matters then---crickets.
That, by itself, proves the concept of House Edge and proves that the House wins (long run) by virtue of the games being fundamentally unfair. Some games are simply more unfair than others.
There are some other Forums that you might be interested in checking out that are just chock full of people who post in a fashion that suggests the math is irrelevant. I don't know that I can mention these in an open post, but feel free to PM me and I will give you one or two suggestions of blissful paradises where the math is ignored/disregarded and your apparent fantasies (as I take your posts) of finding ways to beat -EV games are fully embraced.
Quote: Mission146
Oh yeah? And, how would you come up with an expected loss per hour? Would your only acceptable answer be to physically go with a specific buy-in and play Craps for 10,000 hours and determine what your average loss per hour was?
There you go again. Why don't you allow the thread starter to determine if my questions are valid and answer?
Quote: Missionn146Further, haven't you denied that the House Edge is even, "Real," in your opinion?
I have stated a number of times, which you have somehow ignored, that the HE on PL bets greatly EXCEEDS the "establishment's" math touting that 1.41% figure.
When and if you ever decide to start actually playing the game again, maybe you will also acknowledge that fact.
tuttigym
Quote: tuttigymQuote: Mission146
Oh yeah? And, how would you come up with an expected loss per hour? Would your only acceptable answer be to physically go with a specific buy-in and play Craps for 10,000 hours and determine what your average loss per hour was?
There you go again. Why don't you allow the thread starter to determine if my questions are valid and answer?Quote: Missionn146Further, haven't you denied that the House Edge is even, "Real," in your opinion?
I have stated a number of times, which you have somehow ignored, that the HE on PL bets greatly EXCEEDS the "establishment's" math touting that 1.41% figure.
When and if you ever decide to start actually playing the game again, maybe you will also acknowledge that fact.
tuttigym
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I've nary seen anything mathematically valid expressed in your posts, though I have seen your posts totally dismiss anything mathematical---and the OP is asking a highly precise math question, so I figured I'd save him the time and let him know what was up here.
What have I ignored? Feel free to explain to me how the House Edge on the Pass Line exceeds 1.41%...Thanksgiving is coming up, so it will give me a chance to practice my carving.
Quote: Mission146I've nary seen anything mathematically valid expressed in your posts, though I have seen your posts totally dismiss anything mathematical---and the OP is asking a highly precise math question, so I figured I'd save him the time and let him know what was up here.
Yes you have and have ignored such. It is called 4th grade arithmetic, and when Mr. W and others play the "dark side," it is their primary go to calculations. That should be your proof of concept.
My hope is that your carving practices increase your abundant skills and that your Thanksgiving time exceeds all your expectations.
tuttigym
Quote: tuttigymQuote: Mission146I've nary seen anything mathematically valid expressed in your posts, though I have seen your posts totally dismiss anything mathematical---and the OP is asking a highly precise math question, so I figured I'd save him the time and let him know what was up here.
Yes you have and have ignored such. It is called 4th grade arithmetic, and when Mr. W and others play the "dark side," it is their primary go to calculations. That should be your proof of concept.
My hope is that your carving practices increase your abundant skills and that your Thanksgiving time exceeds all your expectations.
tuttigym
link to original post
Feel free to do one of the following:
1.) Link me to the post where this is mathematically demonstrated.
2.) Mathematically demonstrate why the House Edge is higher here in this thread.
OR:
3.) Drop the subject.
Nonsensical blanket statements and references to, "Fourth grade math," prove nothing.
Do you not want the locations of those other Forums? You choosing to take these kind of posts somewhere else would give me something to be thankful for---might even get a mention at the table as we go around the circle.
https://wizardofvegas.com/forum/off-topic/general/34651-craps-3-point-molly/#post768558Quote: ddsdoniHi everyone,
I've gone down a rabbit hole and now can't stop overthinking what my EV per hour is for craps. I play $10 pass line with up to 3 $10 come bets backed up by 2x odds always working.
Since the house edge for the odds is 0, am I able to safely exclude it from my EV calculation? I know odds will affect the variance but not the EV.
Assuming 100 rolls per hour, HE of 0.42% per roll, 1 pass and 3 come bets ($40 total) I've worked out a loss of around $16.80 an hour. However, there are times where I won't have all 3 come bets on the table (come out roll for example).
As I understand it, despite having up to 4 bets on the table, is my combined HE is still going to be around 0.42% per roll? Or does adding come bets increase the combined HE?
Any help is appreciated, thank you
link to original post
Quote: Mission146
Feel free to do one of the following:
1.) Link me to the post where this is mathematically demonstrated.
Link me to a site that actually demonstrates or shows craps play with 495 consecutive CO hands and subsequent play that proves the 1.41% HA for the PL bet.
Quote: Mission1642.) Mathematically demonstrate why the House Edge is higher here in this thread.
Mr. W has posted somewhere that he estimates that 73% of point play, which includes the PL wager, 7-out plus half of CO rolls lose to craps. That is about an 8 to 1 disadvantage for the PL which is far greater than the 1.41% PL HA loser.
OR:
Quote: Mission1463.) Drop the subject.
When you decide to go to the tables and play and perhaps try to prove your position, I might drop the subject.
Quote: Mission146Nonsensical blanket statements and references to, "Fourth grade math," prove nothing.
4th grade math is easily understood and the foundation of the game, i.e., 36 ways to roll the dice, 6 ways to throw a 7, five ways to throw a 6 or 8, etc., etc. hardly "nonsensical."
tuttigym
Quote: tuttigym
Link me to a site that actually demonstrates or shows craps play with 495 consecutive CO hands and subsequent play that proves the 1.41% HA for the PL bet.
I don't think so. Even if I knew I could, I wouldn't do it. The House Edge on the Pass Line is readily accepted fact in the gambling community, if you want to prove it wrong, then prove it wrong. If you want to spout bull excrement in your posts, then I commend you on your excellent work.
Truthfully, I don't know what your problem with the term, 'House Edge,' is. "House Edge," is a term that the gambling community came up with to describe the sum of the probability of different winning events multiplied by the amount to be won subtracted by the sum of the probabilities of different losing events and the amount to be lost.
What makes Craps Pass Line so easy, in this regard, is that there is only one amount that can be either won or lost and there are no pushes.
In other words, I can't even figure out the basis of your objection because your entire objection is predicated upon the notion that a term that my side were the ones to define doesn't mean what we think it does.
So, to even begin an intelligent argument, you would argue that what happens in your so-called, 'Reality,' does not reflect what we call the House Edge...but not that the House Edge is itself different from what we are saying. We defined the House Edge. IOW, if you do the math problem that results in what we call, 'House Edge,' that is the result you will always get.
Quote:
Mr. W has posted somewhere that he estimates that 73% of point play, which includes the PL wager, 7-out plus half of CO rolls lose to craps. That is about an 8 to 1 disadvantage for the PL which is far greater than the 1.41% PL HA loser.
Find, "Somewhere," and link me to it.
Quote:When you decide to go to the tables and play and perhaps try to prove your position, I might drop the subject.
I have proven my position to my satisfaction. That's what the math is for. If you can prove the concept of House Edge flawed, then just do it. If you demand that I play a sufficient time to be to your liking, then I should remind you that I'm an advantage player. Any game you want---cover the action, we'll discuss my hourly rate on top of that and I'll play it for you. Unlike some people, I have no desire to lose money on Craps.
Quote:
4th grade math is easily understood and the foundation of the game, i.e., 36 ways to roll the dice, 6 ways to throw a 7, five ways to throw a 6 or 8, etc., etc. hardly "nonsensical."
tuttigym
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We agree as to this point and these things are essential to calculating the House Edge.
Your average total amount bet per hour is 100 * 165/557 * $10 * 7061/2925 = $715.10. The house edge is that times 7/495 = $10.11Quote: ddsdoniHi everyone,
I've gone down a rabbit hole and now can't stop overthinking what my EV per hour is for craps. I play $10 pass line with up to 3 $10 come bets backed up by 2x odds always working.
Since the house edge for the odds is 0, am I able to safely exclude it from my EV calculation? I know odds will affect the variance but not the EV.
Assuming 100 rolls per hour, HE of 0.42% per roll, 1 pass and 3 come bets ($40 total) I've worked out a loss of around $16.80 an hour. However, there are times where I won't have all 3 come bets on the table (come out roll for example).
As I understand it, despite having up to 4 bets on the table, is my combined HE is still going to be around 0.42% per roll? Or does adding come bets increase the combined HE?
Any help is appreciated, thank you
link to original post
This calculation ignores any odds bets since there’s no edge on them
Quote: Ace2
Your average total amount bet per hour is 100 * 165/557 * $10 * 7061/2925 = $715.10. The house edge is that times 7/495 = $10.11
This calculation ignores any odds bets since there’s no edge on them
link to original post
Awesome thanks for your help, I ended up opting to ignore odds bets in my calculation as well. My number of $12.60 included the 2 extra come bets all the time just to be safe even though there's not that many out at all times. I read your other thread and it helped me understand the average amount of pass/come bets for a 2-3 point molly player.
Quote: Mission146Quote: tuttigym
Link me to a site that actually demonstrates or shows craps play with 495 consecutive CO hands and subsequent play that proves the 1.41% HA for the PL bet.
I don't think so. Even if I knew I could, I wouldn't do it.
So your steadfast position is that no site exists that show craps play with 495 consecutive CO hands that proves the 1.41% HA for PL bets, AND if one existed you would NOT present it to prove you correct and me wrong because it is a "readily accepted fact"? It would seem to me that if such proof existed, this debate could be ended in a heartbeat, AND I would be the first to jump in your "gambling community" line.
Quote: Mission146Truthfully, I don't know what your problem with the term, 'House Edge,' is. "House Edge," is a term that the gambling community came up with to describe the sum of the probability of different winning events multiplied by the amount to be won subtracted by the sum of the probabilities of different losing events and the amount to be lost.
Truthfully, I have stated several times that I have no problem with the term "House Edge." In fact, I have pronounced the PL HE to be far greater than the "gambling community" number. Yet somehow you have failed to acknowledge such.
Quote: Mission146So, to even begin an intelligent argument, you would argue that what happens in your so-called, 'Reality,' does not reflect what we call the House Edge...but not that the House Edge is itself different from what we are saying. We defined the House Edge. IOW, if you do the math problem that results in what we call, 'House Edge,' that is the result you will always get
"Doing the math problem," and playing the game by performing 495 consecutive CO hands showing the end result of 1.41% are galaxies apart.
Quote: tuttigymMr. W has posted somewhere that he estimates that 73% of point play, which includes the PL wager, 7-out plus half of CO rolls lose to craps. That is about an 8 to 1 disadvantage for the PL which is far greater than the 1.41% PL HA loser.
Quote: Mission146Find, "Somewhere," and link me to it.
Go to odiousgambit's blog entitled "Handy Craps Math" June 6, 2021. The explanation and the link are there.
tuttigym
Quote: DieterHelpful link:
odiousgambit's blog: handy craps math
link to original post
Thank you. I do not know how to do that. Computer illiterate.
tuttigym
Quote: tuttigym
So your steadfast position is that no site exists that show craps play with 495 consecutive CO hands that proves the 1.41% HA for PL bets, AND if one existed you would NOT present it to prove you correct and me wrong because it is a "readily accepted fact"? It would seem to me that if such proof existed, this debate could be ended in a heartbeat, AND I would be the first to jump in your "gambling community" line.
That bears no relationship whatsoever to what I actually said. I'm unsurprised. A sample size of 495 wouldn't prove anything unless you think that the wins and losses in that sample would always be right in tune with expectation.
Again, you seem not to understand that Expectation is what makes gambling, gambling. If it weren't gambling, then it would simply be solving a math problem, there would be no variance and the game would be played by you handing in $100, receiving $98.59 back and getting to play to resolution.
So, the mathematical answer is Expectation, which can be looked at in the context of what we call, "House Edge," but that doesn't mean that plays out to the penny in any sample of any size. If it did, then it wouldn't be gambling. A key component of what makes gambling gambling is uncertainty.
But, I'm certain of what the expected result is or is not.
Quote:
Truthfully, I have stated several times that I have no problem with the term "House Edge." In fact, I have pronounced the PL HE to be far greater than the "gambling community" number. Yet somehow you have failed to acknowledge such.
Demonstrate your position or link me to where you have already done so. The mere act of repeating something does not automatically make it true.
Quote:"Doing the math problem," and playing the game by performing 495 consecutive CO hands showing the end result of 1.41% are galaxies apart.
In a sample of 495 trials, that likely won't be the actual result. That's what makes it expected. If I'm expected to be somewhere at 5:00, that doesn't mean I won't be a few minutes (or even several minutes) early or late.
Quote: tuttigym
Go to odiousgambit's blog entitled "Handy Craps Math" June 6, 2021. The explanation and the link are there.
tuttigym
link to original post
Okay, I'll take a look.
Quote: tuttigymQuote: DieterHelpful link:
odiousgambit's blog: handy craps math
link to original post
Thank you. I do not know how to do that. Computer illiterate.
tuttigym
link to original post
You've got to be messing with me. The sample size of two dice rolled represented in the pie chart was 100 rolls. 100 rolls is a joke.
If you had a set of 100 rolls of two dice with no 12's (this has a roughly 6% probability of happening) would you also conclude that rolling a 12 is impossible?
You want really simple math, here it is:
Pass Line Win, 7 or 11: 8/36
Pass Line Loss, 2, 3, 12: 4/36
Point Established 4 or 10: 6/36
Point Established 5 or 9: 8/36
Point Established: 6 or 8: 10/36
TOTAL: 36/36
Pass Line Win CO Probability: 8/36 = .2222222
Pass Line Point Established 4 or 10 and WIN Probability: (6/36) * (3/9) = 0.05555555555
Pass Line Point Established 5 or 9 and WIN Probability: (8/36) * (4/10) = 0.08888888888
Pass Line Point Established 6 or 10 and WIN Probability: (10/36) * (5/11) = 0.12626262626
Sum of Point Established + WIN: 0.12626262626 + .08888888888 + .05555555555 = 0.27070707069
Sum of ALL WIN Probabilities: 0.27070707069 + .222222222 = 0.49292929289
Pass Line Loss CO Probability: 4/36 = .1111111
Pass Line Point Established 4 or 10 and LOSE Probability: (6/36) * (6/9) = .1111111111
Pass Line Point Established 5 or 9 and LOSE Probability: (8/36) * (6/10) = 0.13333333333
Pass Line Point Established 6 or 8 and LOSE Probability: (10/36) * (6/11) = 0.15151515151
Sum of Point Established + LOSS: 0.15151515151 + .133333333333 + .111111111111 = 0.39595959595
Sum of ALL LOSS Probabilities: 0.39595959595 + .1111111 = 0.50707070706
Difference in Probabilities: 0.50707070706 - 0.49292929289 = 0.01414141417
Ergo, a House Edge of 1.41414141%.
The end. Case closed. That's just how dice probabilities work which is what makes analyzing this bet so trivial. I can't make these terms any simpler above, though I guess I didn't necessarily have to account for equal probability results together and could have done them separately.
1.) Assuming a point has been established, the probability of winning depends on the point in question, but overall, is 27.070707069%. The probability of losing when a point is established is 39.595959595%. As anyone would expect, this lines up with the fact that 24/36 numbers are point numbers.
2.) The probability of winning on the CO is 11.111111% more than of losing on the CO, the difference in probability of winning or losing, a point having been established, is about 12.525252525252% in favor of losing. When you subtract the difference in probability of losing with a point established from the difference in probability of winning without, you will also arrive at the House Edge of roughly 1.4141414141%. Everyone who even understands the basic Rules of Craps would know that is where this comes from.
3.) The House wins approximately 50.70707070--- of ALL hands, regardless of whether or not there is a point established.
4.) The Odds Bet should be viewed as a totally separate bet and has a House Edge of 0% anyway. While you need to have a Point Established to make an Odds bet, making an Odds bet is not required when you have a Point Established.
When you refer to, "Establishment," bettors who have continuous Come Bets or up to a certain number of Come Bets working and you look at all of those losing on a Seven Out, what you are ignoring is that these Come Bets + Odds can also win on the same number multiple times prior to the original Point Established ever being resolved.
THAT is the difference. The Point Established number on the original Pass Line bet can ONLY win or lose once in that hand. Other Point Numbers (assuming continuous Come + Odds betting) can win multiple times before the original Pass Line bet has been resolved.
For example, consider the following hand:
8, 5, 4, 5....
Okay, so now you have a five that has been rolled twice and the Come Bet + Odds has already won once on the Five, though the eight is unresolved. Prior to the five being rolled, the player could have lost the equivalent of three PL + Odds bet (two of these are COME + Odds) had a seven been the next number. The third Come Bet would have won.
So, it is true that the seven can cause multiple bets to lose. However, a seven cannot cause the same number to lose twice...but during Continuous Come Bet action, the same number can win multiple times in a hand and can only lose once in a hand.
Quote: Mission146In a sample of 495 trials, that likely won't be the actual result. That's what makes it expected. If I'm expected to be somewhere at 5:00, that doesn't mean I won't be a few minutes (or even several minutes) early or late.
You can perform the math, but here is what you or anyone else cannot PERFORM.
495 unique one time only consecutive COME OUT rolls with 495 unique one time only results of ANY sample size.
That is what you are missing.
Perhaps you can calculate that problem. Try it and let me know.
tuttigym
Flip a fair coin and record the results.
After 100 flips you’ll have a 95% chance of 40% to 60% heads
After 10000 flips you’ll have a 95% chance of 49% to 51% heads
After 1000000 flips you’ll have a 95% chance of 49.9% to 50.1% heads
No matter how many times you flip, you’ll never be 100% certain of hitting the expected 50% heads, but you can get very close with enough flips
Quote: tuttigymQuote: Mission146In a sample of 495 trials, that likely won't be the actual result. That's what makes it expected. If I'm expected to be somewhere at 5:00, that doesn't mean I won't be a few minutes (or even several minutes) early or late.
You can perform the math, but here is what you or anyone else cannot PERFORM.
495 unique one time only consecutive COME OUT rolls with 495 unique one time only results of ANY sample size.
That is what you are missing.
Perhaps you can calculate that problem. Try it and let me know.
tuttigym
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I don't even understand what you mean in your second sentence. The only way I could perform 495 consecutive Come Out rolls (with no intervening rolls) is if I rolled 495 consecutive results of either 2, 3, 7, 11 or 12. I have to say that such a series is incredibly unlikely.
If you're simply asking that I perform and log the results of 495 hands of Craps, sure, no problem. I'll calculate my expected loss (Pass Line only) of such rolls, estimate how long I think it will take for me to do that and give you a price based on my hourly + the expected loss on my action.
Also, where is your fourth grade math that purports to disprove anything I said above?
Quote: TuttigymMr. O: Further down in this blog, I believe your calculations show an overall point conversion rate at 27+%. Does that mean the House wins approximately 73% of all hands where there is a point established? I am using simple math here. Are my assumptions correct? If so, how does the House's 73% winning advantage on point establishments square with a 1.41% HA.? Can you tell me, approximately, the percentage of outcomes of Come Outs vs Point establishments?
https://wizardofvegas.com/member/odiousgambit/blog/2/#post2286
(Quote clipped, relevance)
That is the dumbest Craps-related question that I have ever seen asked.
The Point Conversion Probability of 0.27070707069 does NOT mean that the House wins approximately 73% of all hands where a point is established.
Start with the probability of the House winning on a Point of either Four or Ten is (6/9) = 66.666666% against (3/9) = 33.333333% probability of losing. This is the second-worst case scenario for the Pass Line with the only thing worse being to lose immediately.
Now, let's look at tis section of what I said:
Quote:1.) Assuming a point has been established, the probability of winning depends on the point in question, but overall, is 27.070707069%. The probability of losing when a point is established is 39.595959595%. As anyone would expect, this lines up with the fact that 24/36 numbers are point numbers.
Okay, so what you want to do is simply index all of these results to Points Established. That's easy, and this is how you would do it:
.27070707069/(.27070707069+.39595959595) = 0.40606060605 or 40.6061%
.39595959595/(.27070707069+.39595959595) = 0.59393939394 or 59.3939%
Assuming a point is established, irrespective of what the individual point actually is, above are the probabilities of winning and losing.
Did it escape your attention that .22222222 + .27070707 = .4929292929?
Does fourth grade math not get that far?
Quote: ChumpChangeWell, I was kind of stuck on all points only being hit 27%, but with the posts here, I'll revise it to a 40% average, or the the odds of winning a put bet.
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Personally, it seems like the sort of argument that a person would make if the goal was to make the topic of calculating the House Edge as convoluted as possible. I'm not saying that is what happened here, but if it was my goal to talk someone off of the belief that the House Edge is simply what the House Edge is (which would be a weird goal to have) that is the sort of seemingly mathematical argument that I would advance.
Nothing wrong with being stuck on something like that---it's the phrasing of the question that seems to obfuscate the issue. I had to read it through a few times to even figure out what it was supposed to be asking.
Quote: JimRockfordTuttigym has been arguing against the fundamental probability of a six sided die for over A DECADE. No matter how elegant the proofs, no matter how concisely or creatively it is explained he just comes back for more. For some reason that I can’t explain, I find it entertaining.
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Here's the thing: I'm more than happy to answer questions like this. If certain people could quit dropping terms such as, "Fourth Grade Math," and, "Simple Arithmetic," then I would even be happy not to post in a snarky tone.
When people come on here and ask about systems, for example, from a position of intellectual curiosity as to, "Why doesn't this system work? It seems like it should work. Here is the way I am looking at it and I can't find the problem," then I'm only too happy to try to find the problem with it.
Where I get annoyed is when we have posts that are written in a tone of assured intellectual superiority that are just flatly wrong. I'm going to be happy to rebut a snarky tone with a similarly snarky tone and enjoy the added benefit of actually being right. If it would please Tuttigym to present questions in a neutral tone, then it would please me to answer them similarly.
Quote: ChumpChangeIf I'm laying odds on the Don't Pass, is it better to lay odds of $12, $15, & $20 on the points, or to lay odds of $12 on any point?
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Makes no difference EV-wise. I guess Laying $12 Odds on a Point of 5 or 9 would be pretty weird, so there's a chance that they would screw up and pay you incorrectly if they haven't seen that specific lay before and aren't good at mental math.
tuttigym
You just don’t get it and apparently never willQuote: tuttigymMr. Aee2: With all due respect, IMHO there is no parallel. Coin flips are basically 50% events. There are no variables that would influence the 50/50 result. PL bet results vary greatly from an almost 2 to 1 CO natural favoring the player for one roll with the exception of the 12, to a point resolution, which according to most, is skewed toward the House. Plus with coin flipping, the performance is easy, short term, and reliable kinda like 4th grade arithmetic.
tuttigym
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A PL bet is very similar to a coin flip in terms of winning percentage and variance. Flip a coin 990 times and you will see heads 495 times (50%) on average. Play the pass line 990 times and you will win 488 (49.3%) times on average. Standard deviation is 15.7 for both cases
The PL is essentially a coin flip with a 1.4% house edge
Quote: Mission146I don't even understand what you mean in your second sentence.
Let me apologize for not making myself clear.
The statement should read 495 consecutive hands including 495 consecutive Come Out rolls.
Quote: Mission146If you're simply asking that I perform and log the results of 495 hands of Craps, sure, no problem. I'll calculate my expected loss (Pass Line only) of such rolls, estimate how long I think it will take for me to do that and give you a price based on my hourly + the expected loss on my action.
I am not. A CO roll can NOT be duplicated. For example: if a 12 is rolled twice at CO, the series ends and must be started all over again from the beginning. Every hand is played consecutively and no CO or result of those 495 hands can be duplicated. If any duplication occurs, the series is void and must be started over again. That is what the Rule of 495 is. That is why PERFORMING the math is easy and readily embraced while PERFORMING that math at the table is virtually impossible and disqualifies the 1.41% HA.
Quote: Mission146Also, where is your fourth grade math that purports to disprove anything I said above?
My fourth grade math does NOT seek to disprove anything you have written; it is there to help any player simplify the approach to the game. Are you aware that Mr. W. uses 4th grade math when he plays as does most all who know the game and play?
tuttigym
tuttigym
We can agree that you disagree with very basic mathQuote: tuttigymI just guess we will have to agree to disagree.
tuttigym
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Quote: tuttigym
Let me apologize for not making myself clear.
The statement should read 495 consecutive hands including 495 consecutive Come Out rolls.
I am not. A CO roll can NOT be duplicated. For example: if a 12 is rolled twice at CO, the series ends and must be started all over again from the beginning. Every hand is played consecutively and no CO or result of those 495 hands can be duplicated. If any duplication occurs, the series is void and must be started over again. That is what the Rule of 495 is. That is why PERFORMING the math is easy and readily embraced while PERFORMING that math at the table is virtually impossible and disqualifies the 1.41% HA.
See, I thought it was disqualified because you said the probability of the House winning once a point is established is something like 73%, or, asked if that was the case anyway. Of course, I easily proved that wasn't the case despite the probability of a winning hand with a Point Established actually being just north of 27%.
This argument is fantastic. Is all math on Video Poker disqualified until someone can see all 2,598,960 starting hands...not to mention all of the things that can potentially happen on the draw...without any repeating? Or, would 311,875,200 be required because you decide that you also care about card positions on the deal?
How does it work for Roulette? Someone would have to go and personally witness all 36 numbers and both zeroes come up without any repeating, and only then, is the House Edge of Roulette proven?
Your argument boils down to math doesn't exist, only results exist.
I know...I know...you can demonstrate that 2 - 1 = 1 pretty readily. If I have two bananas in a room, but then I open up the window and throw one of the bananas out of the window for no reason, then there is demonstrably one banana remaining in the room---that's not fourth grade math, though, it's preschool math and even preschoolers shouldn't throw bananas.
I get it. My suggestion to anyone who thinks that math only exists if it can immediately be visually represented by something is to never gamble.
Besides, I could visually demonstrate what you are asking, if it wouldn't be a total waste of my time. I could get hundreds of pairs of physical dice and line them up in such a way that every possible series of outcomes (ignoring results that have no impact on the bet) is covered. You can visually demonstrate the effects of all 495 total possible results for a Pass Line bet; it would just be pointless to do that.
Quote:
My fourth grade math does NOT seek to disprove anything you have written; it is there to help any player simplify the approach to the game. Are you aware that Mr. W. uses 4th grade math when he plays as does most all who know the game and play?
tuttigym
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I don't know what you mean by, "Approach to the game." Here are the possible bets and here are the house edges of each. Multiply the house edge by the amount you are betting and that will be the expected loss of that bet. That aside, make any combination of bets that you wish---why would I care?
tuttigym
Quote: tuttigymI do NOT find your tone to be "snarky." You have actually reinforced my stance by using the word "perform." In describing the math, you said you "performed" the math which brought me to the realization that in order to validate the "performed" math one would have to "perform" those 495 consecutive hands and CO's with no duplications of results and CO tosses. So thank you for that and bringing me even more clarity to my position.
tuttigym
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What about one roll bets?
An extension of your...I'm going to be charitable and call the contents of your post, 'Logic,' would be that every possible result of a one-roll bet would have to take place without any repeating to prove the House Edge of those. Okay, so if you bet, "Any Seven," are you telling me that you would want to see all 36 possible combinations of rolls without any repeating in order to believe that the probability of a seven is 6/36 and the probability of anything else is 30/36?
Follow Up Question: Would you want the dice to be different colors, or marked in some way, to conclusively prove these combinations so that results of 5-2, 2-5 weren't simply one or the other happening twice, and therefore, repeating?
Quote: Mission146What about one roll bets?
An extension of your...I'm going to be charitable and call the contents of your post, 'Logic,' would be that every possible result of a one-roll bet would have to take place without any repeating to prove the House Edge of those. Okay, so if you bet, "Any Seven," are you telling me that you would want to see all 36 possible combinations of rolls without any repeating in order to believe that the probability of a seven is 6/36 and the probability of anything else is 30/36?
First, I do not know or care what the HE is on any given bet. It does not factor, for me, into my play. I am sure that does not surprise you, but I am fairly confident watching a lot of play, some by recreational players and some by "knowledgeable" players, that they also do not know or care about any perceived HE. All I know is that by my 4th grade arithmetic, the Any 7 bet has a 16.67% chance of winning. I am sure that with your math "magic" you could come up with some low HE number, but for me and anyone I might explain it to, the HA/HE is a whopping 83%. That is a real number most would understand and probably embrace.
Quote: Mission146Follow Up Question: Would you want the dice to be different colors, or marked in some way, to conclusively prove these combinations so that results of 5-2, 2-5 weren't simply one or the other happening twice, and therefore, repeating?
That is a very very intelligent question!! I would be willing to bet that probably 80+% of the folks on this forum would not have asked that question or know that the dice need to be different colors. (So WELL DONE) Absolutely YES two different colored dice is a must requirement.
With that in mind, do you have any guess at the odds of completing such a task? I believe it would be stratospheric.
tuttigym
Quote: tuttigym
First, I do not know or care what the HE is on any given bet. It does not factor, for me, into my play. I am sure that does not surprise you, but I am fairly confident watching a lot of play, some by recreational players and some by "knowledgeable" players, that they also do not know or care about any perceived HE. All I know is that by my 4th grade arithmetic, the Any 7 bet has a 16.67% chance of winning. I am sure that with your math "magic" you could come up with some low HE number, but for me and anyone I might explain it to, the HA/HE is a whopping 83%. That is a real number most would understand and probably embrace.
Other than to make posts that annoy me, why are you here?
Quote:
That is a very very intelligent question!! I would be willing to bet that probably 80+% of the folks on this forum would not have asked that question or know that the dice need to be different colors. (So WELL DONE) Absolutely YES two different colored dice is a must requirement.
With that in mind, do you have any guess at the odds of completing such a task? I believe it would be stratospheric.
tuttigym
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You believe correctly. I know how to determine the odds, but why would you care? You don't even believe in how we define, "House edge." Furthermore, you do not believe in anything unless that thing can be demonstrated physically.
You seem to have ignored my questions about Roulette and Video Poker. Expected.
How do you know that the "Any 7" bet has a 16.67% chance of winning? Have you seen every possible way it can win and every way it can lose consecutively and without interruption, or are you deliberately selective about which probabilities you accept as true based on math and which ones you don't?
Your posts are a waste of this forum's time, my time, your time and the time of everyone who has ever had the misfortune of reading them. I'm sure you're having fun, but I have better things to do in life than read and respond to your posts. I don't know whether or not you have better things to do than to read and respond to mine, but the good news is, you can still do that if you want to; it just won't be a back and forth. See ya.
Here’s how I think about a pass line bet.
There are 36 CO rolls. 4 are losers (2, 3, 3, 12). 8 are winners (7, 7, 7, 7, 7, 7, 11, 11). And 24 are continue playing (point established).
Of the point established:
6 of them are 4/10 and those win 1/3 of the time. So 2 winners and 4 losers.
8 of them are 5/9 and those win 2/5 of the time. So 3.2 winners and 4.8 losers.
10 of them are 6/8 and they win 5/11 of the time. So 4.545 winners and 5.455 losers.
So in total I have:
Losers: 4 + 4 + 4.8 + 5.455 = 18.255
Winners: 8 + 2 + 3.2 + 4.545 = 17.745
(Note total winners plus losers adds up to the 36 possible CO rolls).
All good so far?
So house edge is simply the difference between winners and losers divided by the sum of winners plus losers.
Difference is 18.255 - 17.745 = -0.51
Total is 36
In other words you lose about half a unit every 36 times you make the pass line bet.
-0.51/36 = 1.416% (note the rounding issue because I rounded the 6/8 to three digits).
Hopefully this was helpful to someone in this thread. I fell like it was my penance for enjoying the conversation so much.
Quote: unJonI find it endless entertaining how tuttygym can drag Mission into this same discussion in every craps thread.
Here’s how I think about a pass line bet.
There are 36 CO rolls. 4 are losers (2, 3, 3, 12). 8 are winners (7, 7, 7, 7, 7, 7, 11, 11). And 24 are continue playing (point established).
Of the point established:
6 of them are 4/10 and those win 1/3 of the time. So 2 winners and 4 losers.
8 of them are 5/9 and those win 2/5 of the time. So 3.2 winners and 4.8 losers.
10 of them are 6/8 and they win 5/11 of the time. So 4.545 winners and 5.455 losers.
So in total I have:
Losers: 4 + 4 + 4.8 + 5.455 = 18.255
Winners: 8 + 2 + 3.2 + 4.545 = 17.745
(Note total winners plus losers adds up to the 36 possible CO rolls).
All good so far?
So house edge is simply the difference between winners and losers divided by the sum of winners plus losers.
Difference is 18.255 - 17.745 = -0.51
Total is 36
In other words you lose about half a unit every 36 times you make the pass line bet.
-0.51/36 = 1.416% (note the rounding issue because I rounded the 6/8 to three digits).
Hopefully this was helpful to someone in this thread. I fell like it was my penance for enjoying the conversation so much.
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tuttygym, you may find the above “Rule of 36” a better way to visualize “house edge” than your Rule of 495. 495 is not really a special number other than it happens to be what the house edge fraction reduces too.
Quote: unJon
6 of them are 4/10 and those win 1/3 of the time. So 2 winners and 4 losers.
All good so far?
Simple question: The point is 10. There are three ways to win, i.e. 5/5; 4/6; and 6/4. Good so far? There are 6 ways to lose, i.e. 1/6; 6/1; 2/5; 5/2; 3/4; and 4/3. Is that correct? I will await your answer, thank you.
tuttigym
Quote: tuttigymQuote: unJon
6 of them are 4/10 and those win 1/3 of the time. So 2 winners and 4 losers.
All good so far?
Simple question: The point is 10. There are three ways to win, i.e. 5/5; 4/6; and 6/4. Good so far? There are 6 ways to lose, i.e. 1/6; 6/1; 2/5; 5/2; 3/4; and 4/3. Is that correct? I will await your answer, thank you.
tuttigym
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Yes I agree. Point of 10 there are 3 ways to win and 6 ways to lose. So you win 1/3 of the time.
Quote: Mission146Responses to Various Sections of Blog Post (See Blog Post Linked Above)
1.) Assuming a point has been established, the probability of winning depends on the point in question, but overall, is 27.070707069%. The probability of losing when a point is established is 39.595959595%. As anyone would expect, this lines up with the fact that 24/36 numbers are point numbers.
The great math scam. Dismissing "probability" and embracing possible "expectation." Whatever makes you happy, makes me happy too.
Quote: Mission1462.) The probability of winning on the CO is 11.111111% more than of losing on the CO,
8 ways to win and 4 ways to lose, right? 2 to 1 advantage to the player on that one CO, so what's with 11.1% vs 22.2%? They are the same, right?
Quote: Mission146the difference in probability of winning or losing a point having been established, is about 12.525252525252% in favor of losing.
You wrote above that the overall probability of losing is 27+%, so now it is 12.5+%. I guess to fit a certain narrative one might be able to do that.
Quote: Mission1463.) The House wins approximately 50.70707070--- of ALL hands, regardless of whether or not there is a point established.
Sorry to disappoint you, but that is flat out false and most anyone who plays the game knows it including Mr. W who is an admitted "dark side" player. Perhaps you can explain that.
Quote: Mission1464.) The Odds Bet should be viewed as a totally separate bet and has a House Edge of 0% anyway. While you need to have a Point Established to make an Odds bet, making an Odds bet is not required when you have a Point Established.
When you refer to, "Establishment," bettors who have continuous Come Bets or up to a certain number of Come Bets working and you look at all of those losing on a Seven Out, what you are ignoring is that these Come Bets + Odds can also win on the same number multiple times prior to the original Point Established ever being resolved.
THAT is the difference. The Point Established number on the original Pass Line bet can ONLY win or lose once in that hand. Other Point Numbers (assuming continuous Come + Odds betting) can win multiple times before the original Pass Line bet has been resolved.
For example, consider the following hand:
8, 5, 4, 5....
Okay, so now you have a five that has been rolled twice and the Come Bet + Odds has already won once on the Five, though the eight is unresolved. Prior to the five being rolled, the player could have lost the equivalent of three PL + Odds bet (two of these are COME + Odds) had a seven been the next number. The third Come Bet would have won.
So, it is true that the seven can cause multiple bets to lose. However, a seven cannot cause the same number to lose twice...but during Continuous Come Bet action, the same number can win multiple times in a hand and can only lose once in a hand.
The above has nothing to do with this basic discussion, and as you have done previously, you seek to distract, divert, and obscure the discussion away from the original subject. As a further example, your invoking a question about video poker previously in regards to the performance of the game was a diversion.
tuttigym
Quote: unJonQuote: tuttigymQuote: unJon
6 of them are 4/10 and those win 1/3 of the time. So 2 winners and 4 losers.
All good so far?
Simple question: The point is 10. There are three ways to win, i.e. 5/5; 4/6; and 6/4. Good so far? There are 6 ways to lose, i.e. 1/6; 6/1; 2/5; 5/2; 3/4; and 4/3. Is that correct? I will await your answer, thank you.
tuttigym
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Yes I agree. Point of 10 there are 3 ways to win and 6 ways to lose. So you win 1/3 of the time.
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Excuse me? 3/6 = 1/2 or half the time. I'll bite; where do you get 1/3 of the time?
tuttigym
Quote: tuttigymQuote: unJonQuote: tuttigymQuote: unJon
6 of them are 4/10 and those win 1/3 of the time. So 2 winners and 4 losers.
All good so far?
Simple question: The point is 10. There are three ways to win, i.e. 5/5; 4/6; and 6/4. Good so far? There are 6 ways to lose, i.e. 1/6; 6/1; 2/5; 5/2; 3/4; and 4/3. Is that correct? I will await your answer, thank you.
tuttigym
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Yes I agree. Point of 10 there are 3 ways to win and 6 ways to lose. So you win 1/3 of the time.
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Excuse me? 3/6 = 1/2 or half the time. I'll bite; where do you get 1/3 of the time?
tuttigym
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3 ways to win out of 9 total ways (3+6). Means you win 1/3 of the time and lose 2/3 of the time.
Do you think you win 1/2 or 50% of the time after a point of 4 or 10 is established?
You win 1/3 of the time, so the fair odds are 2:1.
Let’s use your method on a coin flip. Heads wins and tails loses. One way to win and one way to lose. Your method 1/1 means you win 100%!!! No. One way to win divided by two total ways (1+1) means 1/2 chance of winning. So fair odds are 1:1.
