The hoax that is the 1.41% house advantage on pass line bets
Quote: superrickHere is the point I am trying to make, most players will never only play the pass-line bet. They take the so-called free odds, and they do not take full odds behind! If they do you now have more on the one bet then even you would bet, as you said in your last post!
I'm not sure what your point is there. The passline is a marginally better bet than the place 6/8. It's meaningfully better than the rest of the place bets. Just by itself. The odds bets, which are separate wagers, are unequivocally better than their analogous place bets. Place 4 pays 9-5, odds 4 pays 10-5, etc. There is no situation in which a player would not be better off making an odds bet than placing the same number.
Quote: superrickWhile againI might just bet on the 6 or 8 that would have a lower house edge then your bet that is locked in!
Not just misinterpretation, but blatant misstatement of fact (from wizard's page): House Advantage on 6 and 8 is 0.015152.
Quote: SanchoPanzaNot just misinterpretation, but blatant misstatement of fact (from wizard's page): House Advantage on 6 and 8 is 0.015152.
What he means is that the passline bet that's already gone to a point of 6 or 8, viewed as a put bet, has a 9.09% disadvantage. Of course, that bet had already experienced the comeout roll. This is a typical "place-bet fallacy", comparing the second stage of the pass/come bet to a place bet.
Cheers,
Alan Shank
Woodland, CA
Quote: MathExtremistI'm not sure what your point is there. The passline is a marginally better bet than the place 6/8. It's meaningfully better than the rest of the place bets. Just by itself.
I don't believe in equating "lower HA" with "better"; there's more to a bet than the house advantage. A place bet on six or eight has more variance than a passline bet, as well as the flexibility that superrick seems to value so highly. "Better" must be defined for each individual in light of his/her goal(s).
Quote: MathExtremistThe odds bets, which are separate wagers, are unequivocally better than their analogous place bets. Place 4 pays 9-5, odds 4 pays 10-5, etc. There is no situation in which a player would not be better off making an odds bet than placing the same number.
Yes, but you cannot make an odds bet without having a line bet in play, already on a point, getting paid even money on what is now a losing proposition. If you are standing next to a place bettor, it's not the comeout roll, and you don't have a flat bet, you don't have the option of making that odds bet.
Cheers,
Alan Shank
Woodland, CA
Quote: goatcabinWhat he means is that the passline bet that's already gone to a point of 6 or 8, viewed as a put bet, has a 9.09% disadvantage. Of course, that bet had already experienced the comeout roll. This is a typical "place-bet fallacy", comparing the second stage of the pass/come bet to a place bet.
Oh great. In that case, maybe superrick will allow me and other don't bettors to bray about the great advantages of being behind the line, as we ever so conveniently forget the comeout hurdle we had to overcome to get there.
Oh, I'm a DontBettor and I never forget that comeout hurdle. I'm reminded of it often, particularly at 7 and 11 having been rolled. I do remember it fondly once I'm passed that comeout hurdle though because then its such a sweet sensation to know that a "7" roll, the most likely roll, is going to be good.Quote: SanchoPanzaas we ever so conveniently forget the comeout hurdle we had to overcome to get there.
Quote: FleaStiffOh, I'm a DontBettor and I never forget that comeout hurdle. I'm reminded of it often, particularly at 7 and 11 having been rolled. I do remember it fondly once I'm passed that comeout hurdle though because then its such a sweet sensation to know that a "7" roll, the most likely roll, is going to be good.
That time at Cache Creek when I played the don't come I was very lucky with comeout rolls - very few naturals; actually, very few comeout resolutions at all.
Cheers,
Alan Shank
Woodland, CA
Quote: ahiromuVery aggressive for someone who has absolutely no idea of what's going on. As stated this is a probability... I guarantee you if you did 495 million pass line bets and divided your win/losses by one million you'd come fairly close.
My hand has a habit of influencing the dice what are the numbers on that....
Quote: JB Administrator3.5848%, or approximately one "chunk" of 495 Pass Line resolutions out of 28 "chunks" of 495 Pass Line resolutions.
The exact formula is: 495!/(244! × 251!) × (244/495)244 × (251/495)251
To address the notion that the house advantage is a hoax, consider a coin toss. The chance of it landing on heads is 50%, and the chance of it landing on tails is the other 50%. But the probability of exactly 250 heads and 250 tails in 500 coin tosses is 3.5665%, also roughly 1 in 28. Does this mean that the probability of heads being 50% and tails being 50% is a hoax? Of course not. In the short term, anything can happen. But over time, the results will approach expectations (50% heads, 50% tails). It's no different with the Pass Line bet in craps. In the short term, anything can happen, but over time you will win 49.2929% of the time and lose 50.7071% of the time. Just because every sequence of 495 rolls doesn't produce exactly 244 wins and 251 losses does not make it a hoax.
Webmaster - WizardOfVegas.com
JB: I have thought very often about your analysis and formula above. My conclusion is that the 3.5848% is woefully short. Let me state at the outset of this post that it is MY fault that the answer is wrong. I was dwelling on the proposition that the 1.41% HA on PL decisions were based on the outcomes of 495 PL decisions where there are 244 wins vs 251 loses. That assumption is only a small part of the equation. The 495 PL decisions MUST be made only one time each within the body of the totality of the resolutions.
Therefore, I must ask these pertinent questions:
1. What are the odds of having only seven losses in 495 PL decisions?
2. What are the odds of throwing only one 12 or one 2 natural craps come out losers in 495 PL decisions?
3. What are the odds of throwing only six different natural 7s come out winners in 495 PL decisions?
The questions could go on and on in the same mode.
While the 1.41% HA on PL decisions is the conventional wisdom and perhaps mathematically correct, it is absolutely unachieveable and my guess is that the reality of the odds is 1 to the 495th power to one.
What say you?
tuttigym
Quote: tuttigymWhile the 1.41% HA on PL decisions is the conventional wisdom and perhaps mathematically correct, it is absolutely unachieveable and my guess is that the reality of the odds is 1 to the 495th power to one.
What say you?
I say: study limits and asymptotes. What you're getting into is how a binomial distribution of N independent trials of a random variable (in this case, with p=0.492929) approaches its theoretical mean as N grows. In other words, the Law of Large Numbers. Read this:
http://en.wikipedia.org/wiki/Law_of_large_numbers
So why not just humor me and answer the three questions above especially the first question so that our friends here on this forum will, hopefully, be able to realize that the perceived HA is pretty much a fantasy.
Our mutual objective is to win not to be taken to the cleaners by the house's con regarding the PL wager along with the associated FO bet.
One other thing which goes to some simple math, in order for a player to receive excess $$$ over any given Place bet, the player must risk the PL + 5X odds. Anything less increases the house profit. Somehow we almost never hear about that part of the PL/FO equation. Some casinos only offer 4X odds (AC) and cruise ships offer only 2X odds. How is that for a con?
What say you?
tuttigym
Once again, we will try to get second grade math thru your thick skull.Quote: tuttigym1. What are the odds of having only seven losses in 495 PL decisions?
The math does NOT say that there will, on average, be seven losses per 495 decisions. It says that there will, in average, be seven MORE losses THAN wins.
I.E.
251 wins plus 244 losses = 495 decisions.
251 wins minus 244 losses = 7 more losses than wins.
To answer your question: Very small.
However, when phrased correctly: What are the odds of having seven more losses than wins in 495 PL decisions? Very high.
Quote: DJTeddyBearHowever, when phrased correctly: What are the odds of having seven more losses than wins in 495 PL decisions? Very high.
Well, not absolutely. But relatively, that's the most likely outcome.
Similarly, 5 heads and 5 tails in 10 flips is not very likely at 24.6%, but it's the most likely outcome -- all the others possibilities are less likely than that. The moral of the story is that the top of the bell curve doesn't have to be very high, but it's still the top.
Quote: tuttigymThanks for the offering, however, if the "large number" is beyond comprehension, then the bet as stated by that conventional wisdom becomes a con for the house.
There's nothing beyond comprehension about the number 495. There's also nothing beyond comprehension about the LLN, but for those who haven't taken the time to study it yet. Your lack of study doesn't imply that the "HA is a fantasy". It just means you don't understand it yet.
Upon further reflection, it would seem that you don't believe in the concept of a probability. In order to make any progress on this topic, you have to get the fundamentals down first. For example, do you believe that over many flips of a fair coin, the ratio of heads to flips will converge upon 1/2? If you don't, you can't really begin to dissect the rest.
You are a smart man, the So-Called Free Odds was developed as a marketing ploy to get the players to bet more money...!
I try to point this out to everybody that ask about the PL bet, all the Math guys tell me I am an idiot for not betting on the PL and taking the so-called free odds, I guess it's their new math, that I can't comprehend...!
Their way of thinking is a $10 PL bet on the 10 with $20 in odds is the bet to make, their bet pays them $10 on their flat PL bet and $40 on their so-called free odds total of $50 fantastic....!
My old way of doing math and betting $30 placed bet and to make things simple we will say you had a free buy on the 10, my bet just paid me $60 dumb old me, I could take the bet down at any time and turn it off, I didn't have to wait for the point to be made if the point was something else. If I bet on the 10 because it was the come-out point and I wanted to bet on that number, when the point was hit I didn't have to replace the bet again, if I wanted it to bet the point again! Or I could have regressed the bet down.
I never have a PL bet unless I am paying rent on the table to shoot the dice, I don't care about all the come-out rolls that you all say are the best ways of making money at a craps table.
So please explain why I won more money then you, with your PL bet and your so-called free odds...!
My math and your's must be different, free odds are a joke, you are paying for them with your PL bet, and unless you are taking max odds on the bet they are doing nothing for you!
Most players will never take full odds if they are playing on a 10 x odds table or more, this is a fact, just watch how they play, sure there are a lot of smart players that have the bankroll to take advantage, but the average Joe Blow will not bet full odds!
I never worry what the odds are on a table, and if I forget where I am playing and they have Strip Odds, I just place the point for what ever I want to bet!
If there are any question about playing rent, it's only when you have the dice, and you need a PL bet to shoot the dice, other wise you never have a PL bet!
,,,
Quote: superricktuttigym
You are a smart man, the So-Called Free Odds was developed as a marketing ploy to get the players to bet more money...!
I try to point this out to everybody that ask about the PL bet, all the Math guys tell me I am an idiot for not betting on the PL and taking the so-called free odds, I guess it's their new math, that I can't comprehend...!
Their way of thinking is a $10 PL bet on the 10 with $20 in odds is the bet to make, their bet pays them $10 on their flat PL bet and $40 on their so-called free odds total of $50 fantastic....!
My old way of doing math and betting $30 placed bet and to make things simple we will say you had a free buy on the 10,
If you have a free buy bet, by all means make it, and forget about the pass line w/odds. But, in fact, you are going to pay at least $1 vig on that bet. With the best conditions, a buy 4 or 10 can have a house edge of just 1.11% (vig for $30 bet is $1, collected only on wins), lower than the "hoax" of 1.41% on the pass line. However, that $10 bet on the pass line has an ev of just -$.14, not a whole dollar, and the odds bet has an ev of zero.
Quote: superrickmy bet just paid me $60 dumb old me, I could take the bet down at any time and turn it off, I didn't have to wait for the point to be made if the point was something else. If I bet on the 10 because it was the come-out point and I wanted to bet on that number, when the point was hit I didn't have to replace the bet again, if I wanted it to bet the point again! Or I could have regressed the bet down.
All true. Place and buy bets have flexibility, and many people like that. Of course, odds bets have exactly the same flexibility -- you can call them off, increase/decrease them or take them down any time. You just can't do it with the flat part (except DP/DC).
Quote: superrickI never have a PL bet unless I am paying rent on the table to shoot the dice, I don't care about all the come-out rolls that you all say are the best ways of making money at a craps table.
So, when you DO shoot and bet the pass line, you refuse to accept any comeout wins, right? We don't need no stinkin' comeout wins! We refuse to consider the part of the passline bets where the player has the advantage; otherwise our argument falls apart.
Quote: superrickSo please explain why I won more money then you, with your PL bet and your so-called free odds...!
My math and your's must be different, free odds are a joke, you are paying for them with your PL bet, and unless you are taking max odds on the bet they are doing nothing for you!
You *are* paying for the odds bet with the flat bet, but you are only paying 14 cents, regardless of how much you bet in odds. Your insistence on comparing place/buy bets with only the post-point two-thirds of the passline bet is truly ignorant. On average, for every 36 $10 passline bets made, you net $40 on the 12 resolved on the comeout. But I guess you don't care about that.
Quote: superrickMost players will never take full odds if they are playing on a 10 x odds table or more, this is a fact, just watch how they play, sure there are a lot of smart players that have the bankroll to take advantage, but the average Joe Blow will not bet full odds!
So, what's your point? You might just as well state that place bettors never bet the table max. A player's betting level is presumably based on his/her risk of ruin. Place bets, buy bets and odds bets all have the same probabilities of winning, less than .5, so you need to size your bets so that you can accept some losses without busting your bankroll, else you will be out of the game.
If you bet $30 buy 4, $1 vig collected only on wins 60 times, the ev is -$20, the standard deviation (SD) $325. If you bet $10 pass and take $20 odds on all points 60 times, the ev is -$8, the SD $221, so you'll also less likely to bust. If you take 3, 4, 5X odds the ev is the same, but the SD is now $381. The odds multiple can be chosen so that the risk of ruin is low, but if you have a large bankroll you can increase the variance without increasing the expected loss; you cannot do that with place or buy bets.
Cheers,
Alan Shank
Woodland, CA
Quote: superrickI never have a PL bet unless I am paying rent on the table to shoot the dice, I don't care about all the come-out rolls that you all say are the best ways of making money at a craps table.
"Best" is too subjective. What's true, however, is that you're giving up more to the house on the place bets than a pass bettor is on the line bets, assuming equal wager sizes. You can make subjective arguments about why you like the place bets better, and that's totally fine, but it's simply incorrect that the place bets have a lower house edge than the passline.
What you seem to favor is the immediacy with which you can make (and remove) a place bet, vs. the restrictions placed on line bets. That's a valid procedural complaint, but it's not relevant to the mathematics. If you're at a table with 10x odds, for example, and that table allows put bets, you're always going to be better off making a put bet + odds vs. a place bet. $50 place 5 pays $70, for example, compared to $5 put + $44 odds ($49 total wager) pays $5 + $66 = $71, for $1 less on the bet and $1 more on the win. There's no arguing that the put/odds approach pays better. It turns out that the pass/odds approach has even a lower house edge, but if you prefer not to wait for your point to roll, that's your call. There's nothing wrong with being impatient at the dice table - it'll just cost more money.
Quote: DJTeddyBearOnce again, we will try to get second grade math thru your thick skull.
The math does NOT say that there will, on average, be seven losses per 495 decisions. It says that there will, in average, be seven MORE losses THAN wins.
I.E.
251 wins plus 244 losses = 495 decisions.
251 wins minus 244 losses = 7 more losses than wins.
To answer your question: Very small.
However, when phrased correctly: What are the odds of having seven more losses than wins in 495 PL decisions? Very high.
You are wrong again DJ. The 495 decisions are specific to the the way the PL decisions can be made in the total set of PL wagers involving everyway one can win or lose from the natural come out winners to the natural come out losers (craps) to each point resolution. Therefore your I.E. above is absolutely wrong in that it assumes that ANY 495 decisions would produce the so-called HA. So go back to school and get your thicker skull drilled and drained of whatever elements are blocking your ability to see the truth.
tuttigym
Quote: MathExtremistQuote: DJTeddyBearHowever, when phrased correctly: What are the odds of having seven more losses than wins in 495 PL decisions? Very high.
Well, not absolutely. But relatively, that's the most likely outcome.
Similarly, 5 heads and 5 tails in 10 flips is not very likely at 24.6%, but it's the most likely outcome -- all the others possibilities are less likely than that. The moral of the story is that the top of the bell curve doesn't have to be very high, but it's still the top.
I am always puzzeled why smart math people revert to coin flipping which has only two possible outcomes to compare the "odds" and "probabilities" to PL outcomes with 495 possibilities. For me, it is comparing an amoeba to a gorilla. Perhaps you could flip a 495 sided coin and have it land on each side just once. There might be a parallel available there.
tuttigym
p.s. I love the Cardano quote
Quote: tuttigymYou are wrong again DJ.
Nope. Nothing DJ said in his post was incorrect. He said that on average, the pass line will experience seven more losses than wins per 495 decisions. That's a true statement. That doesn't imply there *will be* seven more losses than wins, or even that it's is a likely outcome. but in any collection of 495 pass line outcomes, -7 is the mean (average) outcome. For precisely the same reason, and with precisely the same math, in any collection of 3 coin flips, 1.5 is the mean (average) number of heads.
It would appear that you don't understand what the "mean of a distribution" means, and I suggest you learn it rather than insulting other forum members. It's impossible to have a conversation based on mathematics when the parties don't agree on the validity thereof.
Well, I think that's one of the first things you got right.Quote: tuttigymTherefore your I.E. above is absolutely wrong in that it assumes that ANY 495 decisions would produce the so-called HA.
But let me ask you this about the question I was responding to:
Where did you get "seven" and "495" ?Quote: tuttigym1. What are the odds of having only seven losses in 495 PL decisions?
Quote: tuttigymI am always puzzeled why smart math people revert to coin flipping which has only two possible outcomes to compare the "odds" and "probabilities" to PL outcomes with 495 possibilities. For me, it is comparing an amoeba to a gorilla. Perhaps you could flip a 495 sided coin and have it land on each side just once. There might be a parallel available there.
It's a matter of scale, not of quality. The passline bet is the mathematical equivalent of flipping a fair 495-sided coin with 244 heads and 251 tails, where heads is a winner and tails is a loser (fair means that each side is equally-likely). If you accept that equivalence (do you?) then calculating the house edge for that game is trivial. If you don't accept that equivalence, I'd like to hear why.
Quote: MathExtremist"Best" is too subjective. What's true, however, is that you're giving up more to the house on the place bets than a pass bettor is on the line bets, assuming equal wager sizes. You can make subjective arguments about why you like the place bets better, and that's totally fine, but it's simply incorrect that the place bets have a lower house edge than the passline.
What you seem to favor is the immediacy with which you can make (and remove) a place bet, vs. the restrictions placed on line bets. That's a valid procedural complaint, but it's not relevant to the mathematics. If you're at a table with 10x odds, for example, and that table allows put bets, you're always going to be better off making a put bet + odds vs. a place bet. $50 place 5 pays $70, for example, compared to $5 put + $44 odds ($49 total wager) pays $5 + $66 = $71, for $1 less on the bet and $1 more on the win. There's no arguing that the put/odds approach pays better. It turns out that the pass/odds approach has even a lower house edge, but if you prefer not to wait for your point to roll, that's your call. There's nothing wrong with being impatient at the dice table - it'll just cost more money.
I really cannot believe you used the above example to make the case for the PL/FO "pays better" scenario against the Place betting. If you look really close, that $44 odds bet is slightly in excess of 5X the PL wager. Now lets do an apples to whatever and make that put bet $29 ($5 + $24) and compare it to the $30 Place bet.
That 5X put bet yields $41 while the $30 Place bet nets $42. Come on M E you are better than that.
tuttigym
Quote: tuttigymI really cannot believe you used the above example to make the case for the PL/FO "pays better" scenario against the Place betting. If you look really close, that $44 odds bet is slightly in excess of 5X the PL wager. Now lets do an apples to whatever and make that put bet $29 ($5 + $24) and compare it to the $30 Place bet.
That 5X put bet yields $41 while the $30 Place bet nets $42. Come on M E you are better than that.
Well, you should first re-read my post where I made no mention whatsoever of a 5x table. But your example is illustrative: you compared betting 29 to win 41 vs. 30 to win 42. That's a win of $12 in both cases, but in the put bet case, you're losing one less dollar when you lose. The put bet + odds is still a better wager.
Quote: MathExtremistIt's a matter of scale, not of quality. The passline bet is the mathematical equivalent of flipping a fair 495-sided coin with 244 heads and 251 tails, where heads is a winner and tails is a loser (fair means that each side is equally-likely). If you accept that equivalence (do you?) then calculating the house edge for that game is trivial. If you don't accept that equivalence, I'd like to hear why.
The 495 sided coin is not a "fair" sided coin in that some of the sides are larger than the others to determine a win or loss just as converting the 10 is less likely to win than converting the 6.
There are folks out there who really believe that if one were to bet $1 for 495 times just on the PL, they would only lose $7. They site the outcome as a statistical "expectation." That is nonsense and promotes real ignorance and huge losses for the any player who buys into the 1.41% HA on PL wagers.
Winning using PL/FO is totally reliant on the "hot" shooter(s) who convert points and/or throws lots of numbers. Those "hot" shooters are few and far between which lead to winning % in the 1 -20% range. For me, winning only 20% of the time is totally unacceptable.
tuttigym
Quote: MathExtremistWell, you should first re-read my post where I made no mention whatsoever of a 5x table. But your example is illustrative: you compared betting 29 to win 41 vs. 30 to win 42. That's a win of $12 in both cases, but in the put bet case, you're losing one less dollar when you lose. The put bet + odds is still a better wager.
You are absolutely correct and that is why the 5X odds is the plateau one must start from in order to get that "pays better" scenario. If you were to go back and re-read my post regarding the "pays better" PL/FO you could note that I was highly critical of players and "experts" touting the PL/FO at less than 4X.
tuttigym
Quote: tuttigymYou are absolutely correct and that is why the 5X odds is the plateau one must start from in order to get that "pays better" scenario. If you were to go back and re-read my post regarding the "pays better" PL/FO you could note that I was highly critical of players and "experts" touting the PL/FO at less than 4X.
This is also incorrect. You have conflated the pass line with a put bet, but those are not equivalent at all. I was discussing put bets + odds in comparison to place bets. Those are equivalent wagers with different winning/losing amounts, so it makes it easy to compare. The pass line does not win/lose with the same outcomes as any place bet, so it's less straightforward to compare, but using a financial metric, the pass line *always* has a lower house advantage, in dollar terms, than any place bet (with equal wagers).
Quote: tuttigymSo go back to school and get your thicker skull drilled and drained of whatever elements are blocking your ability to see the truth.
tuttigym
You DO realize that by arguing with people who are very knowledgable on the subject, and posing unsupported theories, you're pretty much like Elmer Fudd trying to tell the captain of a 767 how to fwy the pwane, er, plane?
I'm still waiting for an answer to this question:Quote: tuttigym1. What are the odds of having only seven losses in 495 PL decisions?
Where did you get "seven" and "495" ?
PLEASE SAY IT AIN'T SO
and now all I have done is add 1 to the count. grrrrrrrrrrr
DJ Thanks for straightening me out on the seven losers. We now have a 495 sided coin that can land only once on each of its sides in 495 flips. (Each side represents one way of winning or losing such as a 12 come out loser and a 1-6 (7) natural come out winner)
tuttigym
comeout win: 110
comeout loss: 55
point win: 134
point loss: 196
----------------------
********495
None of this is going to shift tuttigym off his rock, in any case.
Cheers,
Alan Shank
Woodland, CA
Quote: tuttigymWhat are the odds of throwing one 12 natural come out loser in 495 PL outcomes? The questions would continue from there regarding the one time occurence of each possible PL outcome.
tuttigym
What you need to understand, and I think you may, and from what goatcabin (Alan) has tried to explain is the "495" is a rounded down number from 1980 so when you do the calculations that you have asked for you end up with whole numbers instead of rounded numbers.
Example. 1980 pass line bets, 55 will "go down" from a 12 craps that rolls.
In 495 pass line bets, 13.75 (55/4) will "go down" from a 12 craps that rolls.
So does one round up or down that 13.75?
Better to use the 1980.
If you would, start a new thread as your question is a good one and deserves it's own uncluttered thread.
Here is a formula from a friend doing the math for the pass line using a "pass line tree". I think it is overkill but there are no errors and the final answer before rounding down is a good one!
It is interesting as it does cover every "branch" and the result is a kick!
=(8/36)+(2*(((24/36)*(3/24)*(3/9))+((24/36)*(4/24)*(4/10))+((24/36)*(5/24)*(5/11))))
=(8/36)+(2*((216/7776)+(384/8640)+(600/9504))) now we must find a common denominator so we can add all the fractions which is 855,360.
=(8/36)+(2*(115776/855360))
=(8/36)+(231552/855360)
=(190080/855360)+(231552/855360)
final answer = (421632/855360)
now we see why we do "round down"
Quote: guido111What you need to understand, and I think you may, and from what goatcabin (Alan) has tried to explain is the "495" is a rounded down number from 1980 so when you do the calculations that you have asked for you end up with whole numbers instead of rounded numbers.
Example. 1980 pass line bets, 55 will "go down" from a 12 craps that rolls.
In 495 pass line bets, 13.75 (55/4) will "go down" from a 12 craps that rolls.
Careful here, Guido; you are playing into tuttigym's hands when you say, "55 will 'go down'", etc. Better to say something like, "In a very large number of pass decisions, the percentage of comeout 12's will be very close to 55/1980, or 2.78%. Of course, this is nothing but 1/36, the probability of rolling a 12 on any roll.
Cheers,
Alan Shank
Woodland, CA
Quote: MathExtremist"Best" is too subjective. What's true, however, is that you're giving up more to the house on the place bets than a pass bettor is on the line bets, assuming equal wager sizes. You can make subjective arguments about why you like the place bets better, and that's totally fine, but it's simply incorrect that the place bets have a lower house edge than the passline.
What you seem to favor is the immediacy with which you can make (and remove) a place bet, vs. the restrictions placed on line bets. That's a valid procedural complaint, but it's not relevant to the mathematics. If you're at a table with 10x odds, for example, and that table allows put bets, you're always going to be better off making a put bet + odds vs. a place bet. $50 place 5 pays $70, for example, compared to $5 put + $44 odds ($49 total wager) pays $5 + $66 = $71, for $1 less on the bet and $1 more on the win. There's no arguing that the put/odds approach pays better. It turns out that the pass/odds approach has even a lower house edge, but if you prefer not to wait for your point to roll, that's your call. There's nothing wrong with being impatient at the dice table - it'll just cost more money.
Exactly Harrahs near my house has $5 normally and $2 tables 2 days a week with 100X ODDS and they allow Put bets.So i dont do place bets at Harrahs anymore all Put bets.Lowering the HA when you can Put $2 with $200
Which Harrah's is that?Quote: sammy55Exactly Harrahs near my house has $5 normally and $2 tables 2 days a week with 100X ODDS and they allow Put bets.So i dont do place bets at Harrahs anymore all Put bets.Lowering the HA when you can Put $2 with $200
Quote:darnits: 495 PL decisions in a day for 30 yrs?
First, while we are speculating because your set-up is wrong, the table has ten players which is very possible. All random shooters; so the SRR is 6.0. If there was between 45 seconds and one minute between rolls, it would take approximately 38 hours of continuous 24/7 play to complete the 495 PL outcomes.
Second, how many diapers will be soiled? And what is to prevent a premature death at the table?
And are you the only shooter or do the dice rotate between players?
Third, JB, in general terms, what is the liklihood that any given set of 495 PL outcomes will have wins in excess of 251 - very unlikely; somewhat unlikely; likely; somewhat likely; or
very likely??
Fourth, JB, do you know WHO and WHEN this 1.41% HA on PL outcomes was first produced??
Mosca seems to think it was over 3000 years or so ago,and that I am trying to change all that has come before kinda like when the earth was thought to be flat.
tuttigym
Quote: tuttigymDJ Thanks for straightening me out on the seven losers. We now have a 495 sided coin that can land only once on each of its sides in 495 flips. (Each side represents one way of winning or losing such as a 12 come out loser and a 1-6 (7) natural come out winner)tuttigym
tuttigym: 495 sided coin?
First, while we are speculating because your set-up is wrong, a 495 sided coin would be almost impossible to mint in a size small enough to fit into the average human's pocket. Also, with an odd number of sides, it would likely be very difficult to create a roll to package the coins in for banking purposes.
Second, how would a coin like that fit into a video game machine? And what would you put on the faces of this coin?
Third, what would be the denomination of this coin? And would we have to redesign cash register trays, vending machines, etc?
I think we need more thought on this matter before moving forward.
-bruski
the coin is chocolate and yummy and I ate it anyway.Quote: bruskituttigym: 495 sided coin?
First, while we are speculating because your set-up is wrong, a 495 sided coin would be almost impossible to mint in a size small enough to fit into the average human's pocket. Also, with an odd number of sides, it would likely be very difficult to create a roll to package the coins in for banking purposes.
Second, how would a coin like that fit into a video game machine? And what would you put on the faces of this coin?
Third, what would be the denomination of this coin? And would we have to redesign cash register trays, vending machines, etc?
I think we need more thought on this matter before moving forward.
-bruski
The "Coin" was just a euphemism. One of many tools we used to drill thru his thick skull. None were successful.Quote: bruski495 sided coin?
Thankfully, he went away. Unfortunately, the thread still pops up from time to time....
Quote: NowTheSerpentIf we considered every 36,000,000 rolls of 2d6, we would be more correct in saying that 1,000,000 rolls would be '2' than when we say that in 36 rolls, one roll will be '2', speaking in CONCRETE terms.
This to me shows the misunderstanding of the Law of Large Numbers.
The Law deals with ratios or percentages and NOT absolute values.
My college stat professor always stressed that point.
In 36 million craps dice rolls the percentages of rolls that were "2" would be closer to 1/36 than at 36 rolls or even 360,000 rolls.
At the same time the actual number of 2s in 36M tosses would more than likely be farther away from 1 million number due to the Arcsine Law.
This would make a good thread by itself. After the weekend I just may do that.
This is a very difficult thread to read. Some real good posts by member Goatcabin and a lot of fillers.
It is Friday night, time to hit the casinos and party.
Quote: mustangsallyIn 36 million craps dice rolls the percentages of rolls that were "2" would be closer to 1/36 than at 36 rolls or even 360,000 rolls.
At the same time the actual number of 2s in 36M tosses would more than likely be farther away from 1 million number due to the Arcsine Law.
This would make a good thread by itself.
I beg to differ. This is the Wizard of Vegas forum, not the Wizard of Stats.
The only relevant number here from a gambling point of view is the probability of rolling a 2 on the next roll - in this case, p = 1/36 = 0.0277777778.
Why so many players persist in comparing various data samples is beyond me.
