Calculate 1.4% on pass line
Quote: ChumpChangeI'm managing to get a RTP of less than 80% over 4 hours of play on the PL and PB 6/8. People are not making significant points or winning on the 6,8. But the one time I see a player who did lousy the time before so I bet $1 on the hard way 6,8's to wait him out, he hits 8 easy ways that make me lose and no hard way that would pay me. The RTP was down around 70% at the 3 hour mark. Losing 25 times to point 7-outs is what basically happened, although sometimes it took 10 rolls to get the same result.
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Of course, there are a few players who do win. Professional card counters exist; they’re not entirely mythical. It’s just that I know that these professional players are so exceptional, so obsessed, so dedicated, such gluttons for punishment, so terror stricken by the concept of working a nine-to-five job, so few and far between in every sense of few and far between, that, honestly, you are highly unlikely to be one of these human anomalies. And the most honest thing I can say to you, if you tell me that you really want to become a professional blackjack player, is:
You won’t win.
And the reason is: fluctuations.
If you are anything like the masses of humanity, if you like to be rewarded for your efforts within some reasonable time frame, you won’t be able to take the fluctuations. Those negative downswings will be bigger, and harder, and longer lasting, and more upsetting, and more unbelievable, than your level of toleration.
Your losses will tear at your heart, and fill you with emptiness, and leave you in a state of quiet desperation. I hear this from players over and over again. I hear this from players who claim to have studied diligently, and practiced for hours on end, for weeks and months with a singular dream—to beat the casinos.
And they don’t win.
And they ask me why.
And I say, “Oh, it’s just normal standard deviation. A negative fluctuation. It could happen to anyone.”
But it happened to you.
Your money.
Your hours.
Your months of dreaming.
And you didn’t win.
So, over and over again, in my books, and my columns, and my magazine articles, I feel compelled to deliver the message I have been delivering since my very first book in 1980:
You won’t win.
Some card counters will win, but not you. Some card counters will actually experience inordinate positive fluctuations! Wow!
But not you.
You won’t win.
Other card counters will be having champagne parties in their hotel rooms, celebrating that marvelous life of freedom and money and adventure that just seems to come naturally with the lifestyle of a professional gambler. But not for you. You will be among the unfortunate few who, statistically speaking, will be located in the far left tail of the Gaussian curve. Someone has to be there. It will be you.
I have been in that tail; it is a cold and lonely place. I suspect many of those who write about this game have been there, and they know what a cold and lonely place it is. Every professional card counter I know has been there. And if they have played blackjack professionally for many years, they have been there many times. These players have hearts stronger than mine, and I suspect, stronger than yours.
This much I know: it is easier to make a living writing about this game than it is playing it.
In any case, instead of filling an entire chapter of this book with some fifteen articles, written over a period of seventeen years, every one of which simply says, you won’t win, I’ve tossed the whole chapter out in favor of leaving you with just those three words of blackjack wisdom:
You won’t win.
The HE of 1.41% of the PL in craps is a HOAX. I will be specific and invite ALL rebuttals to MY specific points:
1. The Rule of 495 (my term) above has NEVER NEVER been PERFORMED at the tables.
2. It would require the use of two different colored dice, and to my knowledge, no casino in Vegas plays craps with two different colored dice. The only place I know using that format are casinos in California.
3. The PRACTICAL MATH has not happened, but the fantasy MATH via computer SIMULATION can make it occur due to billions of dice rolls and hands played.
4. In order for the Rule of 495 to actually happen, each winning and losing hand must occur ONLY ONCE, not consecutively, in 495 hands or come outs, i.e., one 12 loser in 495 come outs; one 2 loser in 495 CO, one 4-3 and one 3-4 winner in 495 come outs etc. So, the practical questions are (for you truly gifted math people) what are the ODDS of such happening? Or, what are the ODDS of only throwing six winner 7's in 495 come outs?
5. Remember, after each losing hand, the dice must be passed to a different shooter, and that would happen dozens of times during the 495 process.
It is important to state that if any come out winner or loser is duplicated at any time while this process is happening, the 495 processes must begin all over again by definition.
As an editorial comment, this does NOT impugn the Wizard or HIS math by a logical rebuttal. It is reality.
I await responses to my specificities.
tuttigym
Quote: tuttigymI await responses to my specificities.
3. The PRACTICAL MATH has not happened, but the fantasy MATH via computer SIMULATION can make it occur due to billions of dice rolls and hands played.
It is important to state that if any come out winner or loser is duplicated at any time while this process is happening, the 495 processes must begin all over again by definition.
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The "495 processes," as you put them, "begin all over again" with every roll. "Practical math" implies that past results affect future ones, which, for the most part, are not true.
Here's an example: let's use a red die and a blue due. Any particular combination (e.g. red 1, blue 6; red 3, blue 3) has an expected probability of 1/36 of being rolled. "Practical math" seems to mean that I will call the "rule of 36" needs to happen, in that all 36 combinations must be rolled in some order in consecutive rolls. However, in order for this to hold, the 36 rolls beginning with the second roll must also contain all 36 combinations - and since the combination in the first roll is the only one missing in the 36 that begin with the second roll, then it must also be the 37th combination rolled. Similarly, the second roll is the only combination in rolls 3-38 that does not happen in rolls 3-37, so it must be the 38th roll, and so on. Hardly random, if you ask me.
Probability is nothing but speculation. Nobody challenges that. That's probably why it's called "probability" and not, say, "certainability."
That being said, the probability of winning a pass line bet, assuming fair dice, is 244/495, and the probability of losing is 251/495. That means that the expected mean loss per pass line bet is 7/495, or 1.4141% of what you bet.
I’m not going to try and help you here. You really don’t want the help.
In 12th grade, that’s 1976, I had to do a ‘math project’. I analyzed craps, to show the house edge, and a few other things. It was 1.4% then, and is now.
Your questions and analysis show you don’t have even a rudimentary understanding of statistics. Where can we start?
Quote: ThatDonGuyQuote: tuttigymI await responses to my specificities.
3. The PRACTICAL MATH has not happened, but the fantasy MATH via computer SIMULATION can make it occur due to billions of dice rolls and hands played.
It is important to state that if any come out winner or loser is duplicated at any time while this process is happening, the 495 processes must begin all over again by definition.
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The "495 processes," as you put them, "begin all over again" with every roll. "Practical math" implies that past results affect future ones, which, for the most part, are not true.
Here's an example: let's use a red die and a blue due. Any particular combination (e.g. red 1, blue 6; red 3, blue 3) has an expected probability of 1/36 of being rolled. "Practical math" seems to mean that I will call the "rule of 36" needs to happen, in that all 36 combinations must be rolled in some order in consecutive rolls. However, in order for this to hold, the 36 rolls beginning with the second roll must also contain all 36 combinations - and since the combination in the first roll is the only one missing in the 36 that begin with the second roll, then it must also be the 37th combination rolled. Similarly, the second roll is the only combination in rolls 3-38 that does not happen in rolls 3-37, so it must be the 38th roll, and so on. Hardly random, if you ask me.
Probability is nothing but speculation. Nobody challenges that. That's probably why it's called "probability" and not, say, "certainability."
That being said, the probability of winning a pass line bet, assuming fair dice, is 244/495, and the probability of losing is 251/495. That means that the expected mean loss per pass line bet is 7/495, or 1.4141% of what you bet.
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Nice spin, but NOT reality.
I truly respect your math abilities for they far exceed my capabilities, but answer my questions about the actual ODDS of a 12 loser happening once in 495 come outs.
You posted it: "probabilities are just speculations." I am talking REALITIES, so get real with me.
PS: Expectations are nothing more than educated guesses which for me, does not cut it as a part of this discussion.
tuttigym
Quote: SOOPOOWelcome back tutti. Hope your health keeps proving.
I’m not going to try and help you here. You really don’t want the help.
In 12th grade, that’s 1976, I had to do a ‘math project’. I analyzed craps, to show the house edge, and a few other things. It was 1.4% then, and is now.
Your questions and analysis show you don’t have even a rudimentary understanding of statistics. Where can we start?
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Thanks for the well wishes.
Really? So, did you set up a table in the classroom? Did you use two different colored dice? How long and how many dice rolls did you do.
If you did NOT do the above, you did not PROVE anything. Using fantasy math like computer simulations for me, is NOT proof.
Allowing oneself to accept, what you call "statistics" shows that you do not allow yourself to understand the realities of practical craps. So, do you want to tell us the ODDS of throwing one 12 loser come out in 495 consecutive hands? So, why not tell me "statistically" how often that can actually happen?
tuttigym
Quote: tuttigymQuote: ThatDonGuyQuote: tuttigymI await responses to my specificities.
3. The PRACTICAL MATH has not happened, but the fantasy MATH via computer SIMULATION can make it occur due to billions of dice rolls and hands played.
It is important to state that if any come out winner or loser is duplicated at any time while this process is happening, the 495 processes must begin all over again by definition.
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The "495 processes," as you put them, "begin all over again" with every roll. "Practical math" implies that past results affect future ones, which, for the most part, are not true.
Here's an example: let's use a red die and a blue due. Any particular combination (e.g. red 1, blue 6; red 3, blue 3) has an expected probability of 1/36 of being rolled. "Practical math" seems to mean that I will call the "rule of 36" needs to happen, in that all 36 combinations must be rolled in some order in consecutive rolls. However, in order for this to hold, the 36 rolls beginning with the second roll must also contain all 36 combinations - and since the combination in the first roll is the only one missing in the 36 that begin with the second roll, then it must also be the 37th combination rolled. Similarly, the second roll is the only combination in rolls 3-38 that does not happen in rolls 3-37, so it must be the 38th roll, and so on. Hardly random, if you ask me.
Probability is nothing but speculation. Nobody challenges that. That's probably why it's called "probability" and not, say, "certainability."
That being said, the probability of winning a pass line bet, assuming fair dice, is 244/495, and the probability of losing is 251/495. That means that the expected mean loss per pass line bet is 7/495, or 1.4141% of what you bet.
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Nice spin, but NOT reality.
I truly respect your math abilities for they far exceed my capabilities, but answer my questions about the actual ODDS of a 12 loser happening once in 495 come outs.
You posted it: "probabilities are just speculations." I am talking REALITIES, so get real with me.
PS: Expectations are nothing more than educated guesses which for me, does not cut it as a part of this discussion.
tuttigym
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Welcome back. Glad your healing is progressing.
Why do you think a 12 out come roll only happens once out of the 495?
Quote: odiousgambitT-gem, I'll grant one thing, you are Consistent
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Thanks OG, I try to keep it real. It is an uphill task, but until I get any answers by way of specific rebuttals to specific questions, and my mind does really remain open, I will maintain this position.
You know, the Wizard will NOT jump in here because I believe he thinks I am insulting his math, his strategies, etc.
Nothing could be further from the truth. This platform, his platform is awesome, and I have personally learned a lot, and I truly respect his accomplishments.
tuttigym
Quote: unJonQuote: tuttigymQuote: ThatDonGuyQuote: tuttigymI await responses to my specificities.
3. The PRACTICAL MATH has not happened, but the fantasy MATH via computer SIMULATION can make it occur due to billions of dice rolls and hands played.
It is important to state that if any come out winner or loser is duplicated at any time while this process is happening, the 495 processes must begin all over again by definition.
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The "495 processes," as you put them, "begin all over again" with every roll. "Practical math" implies that past results affect future ones, which, for the most part, are not true.
Here's an example: let's use a red die and a blue due. Any particular combination (e.g. red 1, blue 6; red 3, blue 3) has an expected probability of 1/36 of being rolled. "Practical math" seems to mean that I will call the "rule of 36" needs to happen, in that all 36 combinations must be rolled in some order in consecutive rolls. However, in order for this to hold, the 36 rolls beginning with the second roll must also contain all 36 combinations - and since the combination in the first roll is the only one missing in the 36 that begin with the second roll, then it must also be the 37th combination rolled. Similarly, the second roll is the only combination in rolls 3-38 that does not happen in rolls 3-37, so it must be the 38th roll, and so on. Hardly random, if you ask me.
Probability is nothing but speculation. Nobody challenges that. That's probably why it's called "probability" and not, say, "certainability."
That being said, the probability of winning a pass line bet, assuming fair dice, is 244/495, and the probability of losing is 251/495. That means that the expected mean loss per pass line bet is 7/495, or 1.4141% of what you bet.
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Nice spin, but NOT reality.
I truly respect your math abilities for they far exceed my capabilities, but answer my questions about the actual ODDS of a 12 loser happening once in 495 come outs.
You posted it: "probabilities are just speculations." I am talking REALITIES, so get real with me.
PS: Expectations are nothing more than educated guesses which for me, does not cut it as a part of this discussion.
tuttigym
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Welcome back. Glad your healing is progressing.
Why do you think a 12 out come roll only happens once out of the 495?
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Thank you
If one looks at the "Rule of 495," each hand in the 1.41% calculation is an individual one-time occurrence in the set. That is why two different colored dice are required to differentiate each possible point converted or lost.
tuttigym
Which I’m guessing will offend your sensibilities even more than it happening once!
How does it happen 0.75 times?
Quantum Dynamics Craps!
Quote: tuttigym
If one looks at the "Rule of 495," each hand in the 1.41% calculation is an individual one-time occurrence in the set. That is why two different colored dice are required to differentiate each possible point converted or lost.
tuttigym
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I have found your Rule of 495, and I don't think anyone else here thinks that it is even remotely true.
Here is the quote from the original post:
Quote: tuttigym
The house advantage (HA) or edge on Pass Line (PL) bets is unequivically stated at 1.41%. The formula for that HA is also readily available, and broken down to its simplest form, states that out of 495 PL plays there will be 244 PL wins against 251 PL losses.
I have simplified the above by calling it the "Rule of 495." If there has been another individual who has previously used that term, my apologies there is no plagerism intended.
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Again, here is why the Rule of 495 is completely meaningless:
Take a run of 1000 pass line plays, and number them 1, 2, 3, ..., 1000 in order.
If the Rule of 495 is accurate, then there are exactly 244 wins in plays 1 through 495.
If play 1 is a winner, then there are exactly 243 wins in plays 2 through 495, which means that, in order for there to be exactly 244 wins in the 495 plays starting with play 2, play 496 must also be a winner - not "possibly" or "with probability 244/495" be a winner; play 496 must be a winner.
Similarly, if play 2 is a loser, then, since play 496 must be a winner, there are exactly 244 wins in plays 3 through 496, so in order for there to be exactly 244 wins in the 495 plays starting with play 3, play 497 must be a loser.
In short, if play N is a winner, then play N + 495 must be a winner, and if play N is a loser, then play N + 495 is also a loser.
Also note that you are implying that each of the 495 "plays" is equally likely. This is untrue. The number is 495 because the probabilities of winning or losing a pass line bet, when reduced to simplest terms, have a denominator of 495. The "play" of rolling a 12 on the comeout has a probability of 1/36, or 13.75/495. Since plays are discrete, there can't be exactly 13.75 of them out of any set of 495 that have a 12 on the comeout.
Quote: unJonThat’s not how I understand it. I would think you get 13.75 twelve outs on the come out roll in the 495.
Exactly, that is why the Rule of 495 is a hoax. It is only supposed to occur ONCE in 495 CO.
Quote: unjonWhich I’m guessing will offend your sensibilities even more than it happening once!
How does it happen 0.75 times?
No, you misunderstand. My point is that it cannot happen 0.75 times in 495 CO and therefore, by definition, the "equation" is a fraud or hoax. BTW I do NOT get offended. You are the first to actually answer the simple question posed, now postulate that 0.75 occurrence 495 times and tell me what the ODDS of that happening.
Quote: unjonQuantum Dynamics Craps!
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Third grade arithmetic
tuttigym
Quote: ThatDonGuyI have found your Rule of 495, and I don't think anyone else here thinks that it is even remotely true.Quote: tuttigym
If one looks at the "Rule of 495," each hand in the 1.41% calculation is an individual one-time occurrence in the set. That is why two different colored dice are required to differentiate each possible point converted or lost.
tuttigym
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That just it, Don, everybody thinks it is true because they use the 495 equation posted of 244 PL wins vs 251 PL losses everywhere. Just look at all of the push back. The prevailing thought is: The PL bet in craps is one of the best bets in the casino because the HE is only 1.41%. Where does that come from? It is just not true.
Here is the quote from the original post:
Quote: tuttigym
The house advantage (HA) or edge on Pass Line (PL) bets is unequivically stated at 1.41%. The formula for that HA is also readily available, and broken down to its simplest form, states that out of 495 PL plays there will be 244 PL wins against 251 PL losses.
I have simplified the above by calling it the "Rule of 495." If there has been another individual who has previously used that term, my apologies there is no plagerism intended.
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Quote: ThatDonGuyAgain, here is why the Rule of 495 is completely meaningless:
Take a run of 1000 pass line plays, and number them 1, 2, 3, ..., 1000 in order.
If the Rule of 495 is accurate, then there are exactly 244 wins in plays 1 through 495.
If play 1 is a winner, then there are exactly 243 wins in plays 2 through 495, which means that, in order for there to be exactly 244 wins in the 495 plays starting with play 2, play 496 must also be a winner - not "possibly" or "with probability 244/495" be a winner; play 496 must be a winner.
Similarly, if play 2 is a loser, then, since play 496 must be a winner, there are exactly 244 wins in plays 3 through 496, so in order for there to be exactly 244 wins in the 495 plays starting with play 3, play 497 must be a loser.
In short, if play N is a winner, then play N + 495 must be a winner, and if play N is a loser, then play N + 495 is also a loser.
Also note that you are implying that each of the 495 "plays" is equally likely. This is untrue. The number is 495 because the probabilities of winning or losing a pass line bet, when reduced to simplest terms, have a denominator of 495. The "play" of rolling a 12 on the comeout has a probability of 1/36, or 13.75/495. Since plays are discrete, there can't be exactly 13.75 of them out of any set of 495 that have a 12 on the comeout.
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Thank you Don, it is great to know that you believe the Rule of 495 is "meaningless."
One cannot "take a run of 1000 pass line plays" and just go after it. That is not the equation.
In order for one to accept your explanation above, I suggest you go to a craps table at home with $495 and play exactly as you have posed betting $1 on each CO. Again, you are talking "probabilities" which for me have no place in the realities of casino play. Do not forget to use two different colored dice.
Remember this equation is set in stone throughout the gambling world. I dissent because it has yet to be PERFORMED.
tuttigym
Quote: tuttigymRemember this equation is set in stone throughout the gambling world. I dissent because it has yet to be PERFORMED.
tuttigym
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Excuse me? Set in stone by whom?
Here is the problem I have: you use the fact that the Rule of 495 is a hoax to claim that the house edge of 1.4141% on a pass line bet is also a hoax. They have little to do with each other.
The 1.4141% house edge is valid. In fact, it disproves the Rule of 495 as it is valid on every pass line bet, which would not be the case if the Rule of 495 was valid, as the number of upcoming wins in the next 494 pass line bets, and thus the probability that the next one wins, would have to be different once a pass line bet won.
You know what this reminds me of? Somebody came here with a book that claimed that he had beaten baccarat, and included the computer code that proved it. His theory was based on something called the Great Universal Theory, which claims that, in roulette, after 37 spins, 1/3 of the numbers will not come up at all, 1/3 will come up exactlly once, and 1/3 will come up more than once. He used the same fallacy that the Rule of 495 does - that a probability of a single event means that a set of upcoming events will happen, similar to thinking that if you toss a fair coin 45 times and it comes up heads 25 times, then the next five are all expected to be tails as the coin "should" come up heads 25 times in 50 tosses.
Quote: ThatDonGuy
That being said, the probability of winning a pass line bet, assuming fair dice, is 244/495, and the probability of losing is 251/495. That means that the expected mean loss per pass line bet is 7/495, or 1.4141% of what you bet.
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So at a $10 table if I bring $4,950 and bet $10 per come-out roll, the house advantage will be $70 over 495 come-out rolls. Winning more than $70 ahead of the House Advantage will put me in profit. So how many rolls does it take to get to 495 come-out rolls? 1980? I guess I wanted to add odds with a bigger bankroll.
Quote: ChumpChangeQuote: ThatDonGuy
That being said, the probability of winning a pass line bet, assuming fair dice, is 244/495, and the probability of losing is 251/495. That means that the expected mean loss per pass line bet is 7/495, or 1.4141% of what you bet.
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So at a $10 table if I bring $4,950 and bet $10 per come-out roll, the house advantage will be $70 over 495 come-out rolls. Winning more than $70 ahead of the House Advantage will put me in profit. So how many rolls does it take to get to 495 come-out rolls? 1980? I guess I wanted to add odds with a bigger bankroll.
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The mean number of rolls in 495 comeouts is 1971.
Of course, "the mean number" and "the actual number" will almost certainly be two different things. Remember that the mean number of heads in 100 tosses of a coin is 50, but in any 100 tosses, you will get exactly 50 heads only about 1/13 of the time.
Quote: ThatDonGuyQuote: tuttigymRemember this equation is set in stone throughout the gambling world. I dissent because it has yet to be PERFORMED.
tuttigym
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Excuse me? Set in stone by whom?
Here is the problem I have: you use the fact that the Rule of 495 is a hoax to claim that the house edge of 1.4141% on a pass line bet is also a hoax. They have little to do with each other.
[q-ThatDonGuy]The 1.4141% house edge is valid. In fact, it disproves the Rule of 495 as it is valid on every pass line bet, which would not be the case if the Rule of 495 was valid,
I am sorry Don. There is a stretch here. The Rule of 495 is the mathematical BASIS of the 1.41% HE, I do not believe that in practice one can show that HE "on every pass line bet" as you claim.
Quote: ThatDonGuyas the number of upcoming wins in the next 494 pass line bets, and thus the probability that the next one wins, would have to be different once a pass line bet won.
You know what this reminds me of? Somebody came here with a book that claimed that he had beaten baccarat, and included the computer code that proved it. His theory was based on something called the Great Universal Theory, which claims that, in roulette, after 37 spins, 1/3 of the numbers will not come up at all, 1/3 will come up exactlly once, and 1/3 will come up more than once. He used the same fallacy that the Rule of 495 does - that a probability of a single event means that a set of upcoming events will happen, similar to thinking that if you toss a fair coin 45 times and it comes up heads 25 times, then the next five are all expected to be tails as the coin "should" come up heads 25 times in 50 tosses.
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I have NEVER posted that the Rule of 495 is a hoax. The Rule of 495 is the basic equation that results in the 1.41% HE. I acknowledge that fact. Your postings of trying to prove some sort of validity meander around events that are PROBABILITIES. You completely ignore that my position is that the Rule of 495 cannot/has not been performed at the tables thus making the perceived HE of 1.41% invalid in a practical sense.
I find it difficult to understand why anyone would embrace an axiom of math, even though it is correct on its face but cannot, as a matter of practice cannot be produced in the real world. You posted in a previous post that the 1.41% HE is "meaningless" and, I believe this is correct, that nobody actually believes in the 1.41% HE. If I got this last part incorrect, then I apologize for the erroneous quote.
You posted that (paraphrase) you don't believe that anyone believes that the Rule of 495 is true. Most everyone who entered this discussion uses the 251/244 CO as their basis for accepting the 1.41% HE.
tuttigym
Quote: tuttigymQuote: unJonThat’s not how I understand it. I would think you get 13.75 twelve outs on the come out roll in the 495.
Exactly, that is why the Rule of 495 is a hoax. It is only supposed to occur ONCE in 495 CO.Quote: unjonWhich I’m guessing will offend your sensibilities even more than it happening once!
How does it happen 0.75 times?
No, you misunderstand. My point is that it cannot happen 0.75 times in 495 CO and therefore, by definition, the "equation" is a fraud or hoax. BTW I do NOT get offended. You are the first to actually answer the simple question posed, now postulate that 0.75 occurrence 495 times and tell me what the ODDS of that happening.Quote: unjonQuantum Dynamics Craps!
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Third grade arithmetic
tuttigym
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I want to re-visit this post as I mis-spoke or mis-posted in my zeal to further the conversation.
First, the Rule of 495 is NOT a hoax. It just cannot be performed at the table
Second, the fact that the accepted equation cannot be performed actually shows that HE on this bet is basically and totally unreliable for the player, and should not be a part of any PL conversation IMHO.
tuttigym
I agree that if you bet $100 on the line, there isn't a way to walk away from the table with $98.59.
The 1.4% can still be useful as a measure of how much more likely you are to lose than win.
Best of luck!
Quote: tuttigymQuote: ThatDonGuyThe 1.4141% house edge is valid. In fact, it disproves the Rule of 495 as it is valid on every pass line bet, which would not be the case if the Rule of 495 was valid,
I am sorry Don. There is a stretch here. The Rule of 495 is the mathematical BASIS of the 1.41% HE, I do not believe that in practice one can show that HE "on every pass line bet" as you claim.
You posted that (paraphrase) you don't believe that anyone believes that the Rule of 495 is true. Most everyone who entered this discussion uses the 251/244 CO as their basis for accepting the 1.41% HE.
tuttigym
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Then the Rule of 495 is not "out of 495 PL plays there will be 244 PL wins against 251 PL losses" as you stated in an earlier post.
Please tell me what, exactly, as opposed to "in its simplest form," the Rule of 495 is.
I submit that the probability of winning a particular pass line bet in craps is 244 / 495. That, and only that, is the basis for the statement that the house edge on the bet (rounded to the nearest 0.01%) is 1.41%.
And are you certain that, in any set of 495 consecutive pass line bets, there have never been exactly 244 wins? The probability is only about 1 / 28.
a) If PUSH is a valid round, rolls required/round = 3/36 + 1/36 + 8/36 + 15/36 + 15/36 + 92/180 + 92/180 + 235/396 + 235/396 = 557/165 = 3.3757576
b) If PUSH is NOT a valid round, rolls required/round = 3/35 + 1/35 + 8/35 + 15/35 + 15/35 + 92/175 + 92/175 + 235/385 + 235/385 = 6684/1925 = 3.4722078
The rolls made during the PUSH round will be considered valid rolls in both case a) and case b).
Don't Pass/Don't Come
The probability of winning on the come out roll is pr(2)+pr(3) = 1/36 + 2/36 = 3/36.
The probability of pushing on the come out roll is pr(12) = 1/36.
The probability of establishing a point and then winning is pr(4)×pr(7 before 4) + pr(5)×pr(7 before 5) + pr(6)×pr(7 before 6) + pr(8)×pr(7 before 8) + pr(9)×pr(7 before 9) + pr(10)×pr(7 before 10) =
(3/36)×(6/9) + (4/36)×(6/10) + (5/36)×(6/11) + (5/36)×(6/11) + (4/36)×(6/10) + (3/36)×(6/9) =
(2/36) × (18/9 + 24/10 + 30/11) =
(2/36) × (1980/990 + 2376/990 + 2700/990) =
(2/36) × (7056/990) = 14112/35640
The total probability of winning is 3/36 + 14112/35640 = 17082/35640 = 2847/5940
The probability of losing is 1-(2847/5940 + 1/36) = 1-(3012/5940) = 2928/5940
The expected return is 2847/5940×(+1) + 2928/5940×(-1) = -81/5940 = -3/220 ≈ 1.364%
Most other sources on craps will claim that the house edge on the don't pass bet is 1.403%. The source of the discrepancy lies is whether or not to count ties. I prefer to count ties as money bet and others do not. I'm not saying that one side is right or wrong, just that I prefer counting them. If you don't count ties as money bet then you should divide by figure above by the probability that the bet will be resolved in a win or loss (35/36). So 1.364%/(35/36) ≈ -1.403%. This is the house edge assuming that the player never rolls a 12 on the come out roll.
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I just thought it was 1.3889% or 1/72 for the bar 12, but I haven't been paying attention.
Quote: ThatDonGuyQuote: tuttigymQuote: ThatDonGuyThe 1.4141% house edge is valid. In fact, it disproves the Rule of 495 as it is valid on every pass line bet, which would not be the case if the Rule of 495 was valid,
I am sorry Don. There is a stretch here. The Rule of 495 is the mathematical BASIS of the 1.41% HE, I do not believe that in practice one can show that HE "on every pass line bet" as you claim.
You posted that (paraphrase) you don't believe that anyone believes that the Rule of 495 is true. Most everyone who entered this discussion uses the 251/244 CO as their basis for accepting the 1.41% HE.
tuttigym
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Then the Rule of 495 is not "out of 495 PL plays there will be 244 PL wins against 251 PL losses" as you stated in an earlier post.
Please tell me what, exactly, as opposed to "in its simplest form," the Rule of 495 is.
I submit that the probability of winning a particular pass line bet in craps is 244 / 495. That, and only that, is the basis for the statement that the house edge on the bet (rounded to the nearest 0.01%) is 1.41%.Quote: ThatDonGuyAnd are you certain that, in any set of 495 consecutive pass line bets, there have never been exactly 244 wins? The probability is only about 1 / 28.
YES. Can I prove it? NO, but unless there is documented table play proof. (REMEMBER TWO DIFFERENT COLORED DICE), I will hold at that position.
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OK. That is a number I can work with: a probability 1/28 or a 99.65% against the 1.41% HE occurring at the table. That is a microscopic virtually unattainable event. So, as you posted before = "meaningless."
Allow me to take it one step further. IMHO the PL bet is a terrible wager because it is a (usually) table MINIMUM, even money loser. With that in mind, as a player myself, I do not understand why, in this instance, the HE for PL bets is in any ways important. Apparently, the "establishment" would have one believe that because one can tack on a true odds bet to it it becomes something special.
tuttigym
Quote: DieterWelcome back, tuttigym. Glad to hear you're recovering well.
I agree that if you bet $100 on the line, there isn't a way to walk away from the table with $98.59.
The 1.4% can still be useful as a measure of how much more likely you are to lose than win.
Best of luck!
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Thanks for the shoutout Dieter.
tuttigym
Quote: KevinAAIf the house edge is not 1.41%, then what is it?
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So, to put this all to bed, the HE IS 1.41% based on the Rule of 495 equation which has a likelyhood of being performed AT THE TABLE of 99.65% AGAINST which for me, is a hoax or "meaningless."
tuttigym
Quote: tuttigymQuote: KevinAAIf the house edge is not 1.41%, then what is it?
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So, to put this all to bed, the HE IS 1.41% based on the Rule of 495 equation which has a likelyhood of being performed AT THE TABLE of 99.65% AGAINST which for me, is a hoax or "meaningless."
tuttigym
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Except that the house edge is not determined to be 1.41% merely "based on the Rule of 495 equation."
The 1.41% HE is based on the fact that the probability of winning a pass line bet is 244/495 and the probability of losing is 251/495.
If you were to bet 495 on a pass line bet, the "expected loss" would be 7.
The house edge is therefore 7/495, or 1.4141%.
Any mention of the "Rule of 495" has nothing whatsoever to do with this fact. The number 495 is used because it just so happens that the probability of winning a pass line bet is a rational number whose denominator in lowest terms is 495.
Let me put it to you another way:
Toss a coin 1000 times. Do you expect it to come up 500 heads and 500 tails? The 1000-toss equivalent to the "Rule of 495" says that it will...despite the fact that it will happen just once in about every 40 attempts. Does this mean the probability of a coin toss coming up heads is not 1/2, or the "house edge" of an even-money bet on a coin toss is zero?
Quote: ThatDonGuyQuote: tuttigymQuote: KevinAAIf the house edge is not 1.41%, then what is it?
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So, to put this all to bed, the HE IS 1.41% based on the Rule of 495 equation which has a likelyhood of being performed AT THE TABLE of 99.65% AGAINST which for me, is a hoax or "meaningless."
tuttigym
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Except that the house edge is not determined to be 1.41% merely "based on the Rule of 495 equation."
The 1.41% HE is based on the fact that the probability of winning a pass line bet is 244/495 and the probability of losing is 251/495.
If you were to bet 495 on a pass line bet, the "expected loss" would be 7.
The house edge is therefore 7/495, or 1.4141%.
Any mention of the "Rule of 495" has nothing whatsoever to do with this fact. The number 495 is used because it just so happens that the probability of winning a pass line bet is a rational number whose denominator in lowest terms is 495.
Let me put it to you another way:
Toss a coin 1000 times. Do you expect it to come up 500 heads and 500 tails? The 1000-toss equivalent to the "Rule of 495" says that it will...despite the fact that it will happen just once in about every 40 attempts. Does this mean the probability of a coin toss coming up heads is not 1/2, or the "house edge" of an even-money bet on a coin toss is zero?
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The coin toss analogy, for me, is not even in the ballpark simply because it is a 50/50 event every time. When a point is established, which is about 2/3's of the time, the advantage goes to the house overwhelmingly. The other portion of the PL bet is the odds tacked on which increases the monetary risk. No such risk is associated with the coin toss simply because the toss ends in a win or lose situation.
I believe, in past postings you have indicated that you do not play craps or maybe that you do not gamble at all, if that is true, your postings on probabilities as stated above just ring hollow because there may be such fear of loss. You really understand that the House's advantage far exceeds the supposed published HE.
tuttigym
Quote: tuttigymQuote: ThatDonGuyExcept that the house edge is not determined to be 1.41% merely "based on the Rule of 495 equation."
The 1.41% HE is based on the fact that the probability of winning a pass line bet is 244/495 and the probability of losing is 251/495.
If you were to bet 495 on a pass line bet, the "expected loss" would be 7.
The house edge is therefore 7/495, or 1.4141%.
Any mention of the "Rule of 495" has nothing whatsoever to do with this fact. The number 495 is used because it just so happens that the probability of winning a pass line bet is a rational number whose denominator in lowest terms is 495.
Let me put it to you another way:
Toss a coin 1000 times. Do you expect it to come up 500 heads and 500 tails? The 1000-toss equivalent to the "Rule of 495" says that it will...despite the fact that it will happen just once in about every 40 attempts. Does this mean the probability of a coin toss coming up heads is not 1/2, or the "house edge" of an even-money bet on a coin toss is zero?
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The coin toss analogy, for me, is not even in the ballpark simply because it is a 50/50 event every time.
And the pass line bet is a 244/495 to 251/495 event every time.
Quote: tuttigymI believe, in past postings you have indicated that you do not play craps or maybe that you do not gamble at all, if that is true, your postings on probabilities as stated above just ring hollow because there may be such fear of loss. You really understand that the House's advantage far exceeds the supposed published HE.
I play craps quite a bit when I am in Nevada. I don't in California because all they have is the kind of craps where you have to convert the numbers on the dice to values on cards, which gets too confusing to make the game worth playing.
The house's advantage does not "far exceed the supposed published HE." I can think of reasons why it may appear to do so - for example, players will press losses, so they expose more money to the HE, and if a player's bankroll is X, they may keep playing if they win X more, but if they lose X, they do not have the option to "win it back." However, this is confusing house edge with risk of ruin.
Quote: ThatDonGuyQuote: tuttigymQuote: ThatDonGuyExcept that the house edge is not determined to be 1.41% merely "based on the Rule of 495 equation."
The 1.41% HE is based on the fact that the probability of winning a pass line bet is 244/495 and the probability of losing is 251/495.
If you were to bet 495 on a pass line bet, the "expected loss" would be 7.
The house edge is therefore 7/495, or 1.4141%.
Any mention of the "Rule of 495" has nothing whatsoever to do with this fact. The number 495 is used because it just so happens that the probability of winning a pass line bet is a rational number whose denominator in lowest terms is 495.
Let me put it to you another way:
Toss a coin 1000 times. Do you expect it to come up 500 heads and 500 tails? The 1000-toss equivalent to the "Rule of 495" says that it will...despite the fact that it will happen just once in about every 40 attempts. Does this mean the probability of a coin toss coming up heads is not 1/2, or the "house edge" of an even-money bet on a coin toss is zero?
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The coin toss analogy, for me, is not even in the ballpark simply because it is a 50/50 event every time.
And the pass line bet is a 244/495 to 251/495 event every time.Quote: tuttigymI believe, in past postings you have indicated that you do not play craps or maybe that you do not gamble at all, if that is true, your postings on probabilities as stated above just ring hollow because there may be such fear of loss. You really understand that the House's advantage far exceeds the supposed published HE.
I play craps quite a bit when I am in Nevada. I don't in California because all they have is the kind of craps where you have to convert the numbers on the dice to values on cards, which gets too confusing to make the game worth playing.
The house's advantage does not "far exceed the supposed published HE." I can think of reasons why it may appear to do so - for example, players will press losses, so they expose more money to the HE, and if a player's bankroll is X, they may keep playing if they win X more, but if they lose X, they do not have the option to "win it back." However, this is confusing house edge with risk of ruin.
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But that is the real HE, i.e. chasing loses.
California craps confusing? Just an extra step to determine the individual roll. Should not change the HE or does it.?
tuttigym
Quote: ThatDonGuyI play craps quite a bit when I am in Nevada. I don't in California because all they have is the kind of craps where you have to convert the numbers on the dice to values on cards, which gets too confusing to make the game worth playing.
First, I want to apologize for my misstatement about your gambling. Thanks for straightening me out.
Second, as a true math savant, I was wondering if you have developed any kind of different math strategy devoted to craps, or do you play the "establishment" way, i.e., PL + odds + 3 pt. Molly or some variation? Or do you follow the Wizard's lead and play the Don't (dark side)?
Just curious.
tuttigym
Quote: ThatDonGuyI am pretty much strictly a Pass Line & Odds player - if it's a $10 or 15 minimum, I usually bet 30/40/50 odds (i.e. whatever returns 60 plus the PL bet) if I win.
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And tutti, I’m sure TDG would be the first to tell you that he is playing a game he is (slightly) more likely to lose than win! Just his strategy only exposes his initial bet to that wascally 1.4% house edge. All his odds bets are 0 EV affairs.
Quote: ThatDonGuyI am pretty much strictly a Pass Line & Odds player - if it's a $10 or 15 minimum, I usually bet 30/40/50 odds (i.e. whatever returns 60 plus the PL bet) if I win.
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Thanks for the response. I would suspect -- lifetime you might be down, and that is OK knowing you have the discipline to not try to catch up on a choppy or cold table. The highs come with that "hot" shooter and embracing the excitement.
I wish you well. If you are interested, I will PM you with might approach which almost always give me a mathematical edge over the House on any given roll of the dice.
tuttigym
Quote: SOOPOOQuote: ThatDonGuyI am pretty much strictly a Pass Line & Odds player - if it's a $10 or 15 minimum, I usually bet 30/40/50 odds (i.e. whatever returns 60 plus the PL bet) if I win.
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And tutti, I’m sure TDG would be the first to tell you that he is playing a game he is (slightly) more likely to lose than win! Just his strategy only exposes his initial bet to that wascally 1.4% house edge. All his odds bets are 0 EV affairs.
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The supposed 0 EV with odds bets really suffers on a cold or choppy table. There is almost no way recoup multiple consecutive 7 outs with an occasional point winner. The "establishment "right" side player almost always seem to be trying to catch up and that's how the HE really works from my experience.
p.s. So, does a 0 EV carry any HE? Watching multiple players who play that way leave the table bankrupt or a lot lighter in their chip count should make one wonder, right?
tuttigym
Quote: tuttigymQuote: SOOPOOQuote: ThatDonGuyI am pretty much strictly a Pass Line & Odds player - if it's a $10 or 15 minimum, I usually bet 30/40/50 odds (i.e. whatever returns 60 plus the PL bet) if I win.
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And tutti, I’m sure TDG would be the first to tell you that he is playing a game he is (slightly) more likely to lose than win! Just his strategy only exposes his initial bet to that wascally 1.4% house edge. All his odds bets are 0 EV affairs.
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The supposed 0 EV with odds bets really suffers on a cold or choppy table. There is almost no way recoup multiple consecutive 7 outs with an occasional point winner. The "establishment "right" side player almost always seem to be trying to catch up and that's how the HE really works from my experience.
p.s. So, does a 0 EV carry any HE? Watching multiple players who play that way leave the table bankrupt or a lot lighter in their chip count should make one wonder, right?
tuttigym
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SOOPOO: Crickets??
