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RE: "You" should expect to break even on the odds bets is false! page 2This is a discussion thread · 46 replies Lady Shooter: [nq:1]The casino might PAYOFF the odds bet using TRUE ODDS, but the casino is still more likely to win each and every odds bet because the casino holds the advantage on each and every passline bet. sigh<[/nq]I can't believe I'm jumping into this. Newsguy, the casino "holds the advantage" only on wagers it doesn't pay at true odds. That's what "advantage" means. The casino is not more likely to "win" each and every odds bet. Conceive of a craps game where there are always exactly the same number of players betting the PL as the DP and where equal odds are taken and given accordingly. Three right siders playing minimum line bets with double odds. Three wrong siders playing minimum line bets with double odds. In this game, the casino will always* win on *only the 12. Otherwise, the casino will simply transfer the losers' wagers to pay the winners' wagers. [nq:1]That has been my position, but the math guys say I am wrong. They say the odds bet will always be 50-50,[/nq] Whaaaaaa-a-a-a-t? Prepare for massive incoming, Newsguy. You just set yourself up for a solid wall of napalm. [nq:1]Hence my question: when do the dice know when its time to show a winner?[/nq] They "show a winner" everytime the shooter makes his point or 7s out. You just need to figure out which side to play to reap the benefit. Instead of asking a nonsensical question over and over again, try imagining you're playing the don'ts. Better yet, play them. It will suddenly all become clear to you. Lady Shooter
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alan: Lady Shooter, thanks for your response."Newsguy, the casino "holds the advantage" only on wagers it doesn't pay at true odds. That's what "advantage" means. The casino is not more likely to "win" each and every odds bet." What you wrote in this paragraph is correct and true and I have no dispute with it. But where my brain short-circuits and melts down is when you link the "odds bet" to the flat bet, and the casino has the edge on the flat bet. That "linkage" tells me the casino has the advantage. I do not separate the odds bet from the flat bet my brain will not let me do that. I must be a moron or have a defective brain.
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alan: Mark, I think John is using a fixed universe of decisions to make his point.Let's say his fixed universe of decisions is 5 decisions. The odds bet loses the first three decisions; in that case it is impossible for a 50-50 split. or the odds bet wins the first three decisions; in that case it is impossible for a 50-50 split. But... If you add another decision, or an infinite number of decisions, then it is possible for a 50-50 split. JK if I screwed up the meaning of your post, Im sorry.
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Cymbal Man Freq. [nq:1]"Newsguy, the casino "holds the advantage" only on wagers it doesn't pay at true odds. That's what "advantage" means. The ... flat bet my brain will not let me do that. I must be a moron or have a defective brain.[/nq]It is not like the double down bet in BJ: you can't (or shouldn't) do it unless your hand is favorable (better than 50.000% chance of winning the hand). In BJ you are more likely to win the double down bets on favorable hands than to lose and it pays even money; but they allow it because you are losing badly on other unfavorable hands and the double down feature lowers the overall HA of BJ.
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John Kerr: Re: "You" should expect to break even on the odds bets is false!Group: rec.gambling.craps Date: Mon, Dec 5, 2005, 10:25am (CST-2) From: (Email Removed) (Mark=A0Rafn) It would not be the mode for one individual set of wagers, of which I mention. One individual set of wagers must start with a win or a loss, once the game is underway (reality), the mode is no longer expected to be, "break-even". =3D=3D=3D=3D=3D=3D=3D And yet, it turns out to work anyway! Past results in craps have no effect on future outcomes, so your expectation at any point in time on no-vig bets is to be exactly where you are at the point you're calculating. All individual sets of bets must start with 1, which has no break-even possibility. In reality, the actual mode will never be, exactly break-even for an individual bet set. This is completely wrong, both empirically (via simulation) and mathematically (by calculation). Proof by absurdity: if the mode for an even number of trials of a fair bet is not 0, what is it? No other value makes any sense. Mark Rafn =A0 (Email Removed) =A0 </ =3D=3D=3D=3D=3D=3D=3D=3D=3D Mark, you are absolutely correct "before" the game set is played! But for an actual single game set....it does not hold true. I propose that we have to play the game in order to "prove" your posit! The game will always start out with me being either ahead or behind after the first bet....and for that individual game, or any individual game, my expectation is to be either ahead or behind, not even after the first decision, no matter that decision. Right? JB
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John Kerr: Re: "You" should expect to break even on the odds bets is false!Group: rec.gambling.craps Date: Mon, Dec 5, 2005, 10:47am (CST-2) From: (Email Removed) (alan) DocDice... you have touched on my position. The casino might PAYOFF the odds bet using TRUE ODDS, but the casino is still more likely to win each and every odds bet because the casino holds the advantage on each and every passline bet. That has been my position, but the math guys say I am wrong. They say the odds bet will always be 50-50, even if the 4 and 10 are 2 to 1 underdogs, the 5 and 9 are 3 to 2 underdogs, and the 6 and 8 are 6 to 5 underdogs. Hence my question: when do the dice know when its time to show a winner? Oh well. The math guys say I am a moron. I guess Im a moron. == alan, be careful what ya "guess"... JB
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Alan Shank: [nq:2]I must be a moron or have a defective brain.[/nq]I vote for both! Cheers, Alan Shank
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Alan Shank: [nq:1]Mark, you are absolutely correct "before" the game set is played! But for an actual single game set..it does not hold true.[/nq]Wrong, wrong, wrong, as I have conclusively proven in two other threads. [nq:1]I propose that we have to play the game in order to "prove" your posit! The game will always start ... my expectation is to be either ahead or behind, not even after the first decision, no matter that decision. Right?[/nq] Once the first result is seen, if you then extend the set by one, the expectation, and most likely outcome (of all possible individual outcomes) is to be ahead by one. For example: 20 fair-coin flips1st is heads At this point, if you do 20 MORE flips, the most likely result is for heads to be "up one," since the expectation for these next 20 is 10 of each. However, for the remaining 19 of the ORIGINAL 20, you cannot get 9.5 head or tails, and therein lies the flaw in your reasoning. Well,I have detailed all that relative to coin flips and odds bet on the 4/10 in other posts. Cheers, Alan Shank
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John Kerr: Mark, this is really bothering me...I can't believe I'm that dense!  I know you are right about the theoretical mode. But why do I see the actual as being something other than that?I go to a fair coin flip game with you, I ask you what I should expect if I play a 20 decision session. You answer, you should expect to break-even. I make my first bet, and I lose, I turn to you and ask, where do I stand now...you answer, you can expect to be behnd after this session is over. Or, I win the first bet, and I turn to you and ask, where do I stand now...you answer, you can expect to be ahead after this session is over. No matter what happens on the first bet..I am never expected to break-even according to your answers to my question...except, before I decide to play! Alan set me straight on my poor use of the English language when I first stated that being even was the least likely thing to happen...and I accept that I didn't say what I intended, and he was right in pointing that out. I should have said it was the least likely out of the three scenarios...being ahead, being behind, or being exactly even. He, and you, and I would suppose all knowledgable mathematicians would say that being even is the most likely of all individual possibilities..and I agree with that as a theoretical posit! So, why the hell am I having problems with an actual individual real session? JB
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John Kerr: Re: "You" should expect to break even on the odds bets is false!Group: rec.gambling.craps Date: Mon, Dec 5, 2005, 12:07pm (CST-2) From: (Email Removed) (Alan=A0Shank) Mark, you are absolutely correct "before" the game set is played! But for an actual single game set..it does not hold true. Wrong, wrong, wrong, as I have conclusively proven in two other threads. I propose that we have to play the game in order to "prove" your posit! =3D=3D=3D=3D=3D=3D The game will always start out with me being either ahead or behind after the first bet..and for that individual game, or any individual game, my expectation is to be either ahead or behind, not even after the first decision, no matter that decision. Right? =3D=3D=3D=3D=3D=3D Once the first result is seen, if you then extend the set by one, the expectation, and most likely outcome (of all possible individual outcomes) is to be ahead by one. For example: 20 fair-coin flips1st is heads At this point, if you do 20 MORE flips, the most likely result is for heads to be "up one," since the expectation for these next 20 is 10 of each. However, for the remaining 19 of the ORIGINAL 20, you cannot get 9.5 head or tails, and therein lies the flaw in your reasoning. Well, Ihave detailed all that relative to coin flips and odds bet on the 4/10 in other posts. Cheers, Alan Shank =3D=3D=3D=3D=3D=3D=3D I think the light is starting to get brighter! I think I am confusing the set, and the individual bets. If we just make a bet as to how the set will turn out..then we can expect to break even no matter that the first bet is won or lost. But, if we consider each bet individually, it makes all the difference how the first bet turns out! You would never give me an automatic win on the first bet, then take your chances on the next 19 that you would break even over all. I had to break it down to just two decisions...after two decisions I can in fact expect to be even, no matter the outcome of the first decision. My problem was seeing the only possibilities as being even, and being behind, and being ahead after two decisions...I overlooked the, being even twice scenario, as opposed to either being ahead 2 units, or behind 2 units. I was getting a total of less than 1 for the 4 possibleoutcomes. O.k. I feel kinda dumb! JB
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