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"Optimal strategy in craps is pretty simple. Bet the don't pass and/or don't come and back it up laying the maximum odds."
I've never seen someone play the Dark Side with continuous DC bets and lay 100x odds. The closest thing I've seen though was an old guy betting $100 on the DC every roll and then laying roughly 10x odds.
Even if I could afford it, I don't think I'd ever have the guts to attempt this strategy. You can quickly lose a $50K bankroll at a $10 table if you encounter a hot shooter who rolls several dozen box numbers before sevening out.
I watched from the next table. You weren't aloud to stand around the table.
*not to bet more than you are comfortable with
*that the free odds do not help you win more money
putting those things together, the wizard would only recommend doing max odds with 100x odds in the situation where the player was betting $1000 [or whatever] on the line with no odds, and the assumption he was actually comfortable with betting a grand per whack
Quote: odiousgambitlike the OP, I will not go to the trouble of citing my source, but I can remember the Wizard cautioning plenty of times
*not to bet more than you are comfortable with
*that the free odds do not help you win more money
putting those things together, the wizard would only recommend doing max odds with 100x odds in the situation where the player was betting $1000 [or whatever] on the line with no odds, and the assumption he was actually comfortable with betting a grand per whack
For an individual bet, there are multiple parameters needed for comfort for some players.
The one parameter that a player has less control over is the probability of winning. Of the 12 bets with no house edge on the craps table (even without considering the line bet a significant portion of the discussion) the probabilities of winning are another parameter that many people have a hard time wrapping their heads around.
lay 4/10
lay 5/9
lay 6/8
odds 4/10
odds 5/9
odds 6/8
It's unfortunate that you can't have a free COME or DC bet (lose or win half on the 12 for example) as a spoil for a previous bet. Many people like the idea of having an even-money pay or even the ability to hedge a previous bet (last come is a winner).
This is a mental hangup rather than a math hangup. But I just know too many people that want even-money bets when they bet that big or they want a hedge. Giving up the 0.4% edge per roll to get that is, strictly speaking, a costly path in the long run. And even though it helps you lose more money compared to betting odds, there are multiple different things (besides saving money) that motivates a gambler who enjoys gambling at craps.
Even given the opportunity for 5x free come bets instead of free odds if you never bet odds and you have a pass line, it only helps on the 12, and I think that, just like odds, it would help the casino both work tokes from big winners for the crew and it would help increase the line bets for players watching who don't know the details to increase winnings for the house.
But again, the lack of mental grasp for probabilities on the "awesome free odds bets" are a big part of how the casino beats players out of their money so quickly.
The "max odds" provides a certain mathematical advantage, but speaking practically it becomes very expensive. If I could find a game with $1 table min and $100 odds, yeah I'd go for it. But that game is verrrry rare.
Also, gambling is entertainment. Granted some other games allow for AP, but craps is always a -EV proposition (unless you think you are dice influencing) which means you should be treating it as entertainment.
The DP / max odds is the 'best' available element of risk, ie measuring EV as a percentage of total money bet.
But there are other ways of mathematically assessing the situation.
EV as raw $$ per bet resolved (or per roll).
EV as raw $$ per hour played.
More practically, what is your entertainment value versus the EV of your strategy? This is why I play the pass line. It's slightly worse mathematically but I find it more entertaining, so the net result to me as a person is more positive.
the COME would still have a house edge of 0.50 / 1980 losing 50% of the wager on a roll of 12Quote: AhighIt's unfortunate that you can't have a free COME or DC bet (lose or win half on the 12 for example) <snip>
that means the DC would have a player edge of 0.50 / 1980 (very small and almost 0)
(check me math as i was playing with 2 dogs while at the computer)
and over says
90 rolls per hour that is on average 60 DC bets resolved for
one player, say making $10k flat bets and $0 lay odds for an hour win abouts $151 (rounded down)
a team of 10 players then could rake in $1,515 per hour at one craps table (at Ahigh's Casino & Flower Shop for example)
let us see how long he (Ahigh) lets the free bet for the DC continue
thank you for sharing your opinion, as always
in my opinion, please
make more youtube videos
for entertainment value (and google money)
Mully
Quote: cyrusIf I could find a game with $1 table min and $100 odds, yeah I'd go for it. But that game is verrrry rare.
Try this. Bet $50 DP, $51 Pass. Once a point is established, bet $100 (double odds).
Quote: FatGeezusTry this. Bet $50 DP, $51 Pass. Once a point is established, bet $100 (double odds).
?
Am I missing something here? You're not trying to say that's the same as a $1 wager on pass or don't pass, are you? What happens if the "push" number comes up (2 or 12 depending on the casino)? Sound like you lose one of them in their entirety instead of just $1. Please explain as I must not be reading your comment correctly.
Quote: FatGeezusTry this. Bet $50 DP, $51 Pass. Once a point is established, bet $100 (double odds).
The EV of your bet will be the HE calculated against $101, not $1. The odds will not affect the EV.
Each bet has a negative EV. You cannot cancel the EV of one bet by making another in negative expectation. This is why smart pit bosses pay no attention to hedging and such.
It's true that you are exposed to rolling a 12 on the come-out roll and losing $51.
On a per bet resolved basis that is EV of 1/36 * -51 = -$1.42
Compared to a flat pass line $51 with EV of -0.0141 * $51 = -$0.71
Hmm on second thought, in the long run it's worse than doing just a flat bet. Although, if you can just avoid that pesky 12 you are in good shape.
So, as long as you roll the numbers that win, and avoid the numbers that lose, it's a great bet!
But if a hedge can lower the -EV significantly, it can prove worthwhile.Quote: odiousgambitYou cannot cancel the EV of one bet by making another in negative expectation.
Quote: SanchoPanzaBut if a hedge can lower the -EV significantly, it can prove worthwhile.
In negative expectation, the EV can only accumulate. Of course, if you can make a bet with positive expectation ... An example of that I believe is in BJ, where you must make your initial bet that is -EV, then see you were dealt 11 vs 6. Now you are allowed to make another bet, the double down, that is +EV. But we don't call this a hedge either.
Quote: SanchoPanzaBut if a hedge can lower the -EV significantly, it can prove worthwhile.
Any examples?
Betting 50 on the DP and 51 on the PL has an HE of 1.41%, which is the same as betting the PL alone.
Taking 2X odds on the PL in this example only reduces the HE to .84.
Taking 2X odds on the PL when there is no DP has an HE of .6
The DP 'hedge' actually increased the HE in this example.
I don't see any advantage in hedging.
Quote: Tanko
Betting 50 on the DP and 51 on the PL has an HE of 1.41%, which is the same as betting the PL alone.
Still not seeing how it could be the same when you are at risk for losing the "push" on the two or twelve.
Quote: bodyforlifeStill not seeing how it could be the same when you are at risk for losing the "push" on the two or twelve.
Depends on what is meant. It's roughly 1.4% on $101 in the example. $101 on the pass line is about the same in EV. Tanko is wrong if he meant about 1.4% on $50 or $51 in total, but I did not take that as his meaning.
OP of the example I am pretty sure is thinking 1.4% on $1
Quote: bodyforlifeStill not seeing how it could be the same when you are at risk for losing the "push" on the two or twelve.
Only at risk for losing the push on the 12, not the 2.
The DP will push when the 12 rolls, so the player is risking the $51 PL on every roll.
HE is the ratio of the average loss to the initial bet.
Since the 12 is expected once in every 36 rolls, the bettor is expected to lose an average of $51 for every 36 rolls.
(-51 * 1/36)/101=1.40% HE
If the PL was $50, the HE would be 1.39%.
The $1 hedge added to the PL bet only increases the HE.
Quote: drjohnnyAccording to the Wizard...
"Optimal strategy in craps is pretty simple. Bet the don't pass and/or don't come and back it up laying the maximum odds."
I've never seen someone play the Dark Side with continuous DC bets and lay 100x odds. The closest thing I've seen though was an old guy betting $100 on the DC every roll and then laying roughly 10x odds.
Even if I could afford it, I don't think I'd ever have the guts to attempt this strategy. You can quickly lose a $50K bankroll at a $10 table if you encounter a hot shooter who rolls several dozen box numbers before sevening out.
when the wiz and (the several) others make that statement they are being factually correct...however to be better as a practical matter they might insert "affordable" or "comfortable" after the word "maximum"...this does violence to the meaning of "optimal" you might say?...but, no, i would disagree as no strategy could be deemed "optimal" if you cannot afford it or if it is otherwise unrealistic: indeed it might make it "stupid" more than optimal...anyway this suggested modification of the statement would satisfy dr j's valid objection, give it more of a generally good meaning, still imply that maximum is best and sacrifice nothing factual...tom "home runs are sometimes boring" p
Lay the point and go down on it as soon as the DC moves behind the line. That vastly improves your chances from being a 1-to-2 or 2-to-3 underdog.Quote: TankoAny examples?
Quote: SanchoPanzaLay the point and go down on it as soon as the DC moves behind the line. That vastly improves your chances from being a 1-to-2 or 2-to-3 underdog.
The lay bet wins more often than it loses, but it loses more money than it wins.
The combined losses with the lay bet as a hedge, are higher than the combined losses without it.
Adding the lay as a hedge, only increases the overall HE.
Krigman on Hedging
Quote: SanchoPanzaLay the point and go down on it as soon as the DC moves behind the line. That vastly improves your chances from being a 1-to-2 or 2-to-3 underdog.
John Patrick calls it 'The Richochet' method of betting.
The important difference being the taking down of the bet. It does not stay up there. None of the mathematicians who advocate staying away from hedging address the extremely limited use of the lay technique.Quote: TankoThe lay bet wins more often than it loses, but it loses more money than it wins. The combined losses with the lay bet as a hedge, are higher than the combined losses without it. Adding the lay as a hedge, only increases the overall HE. Krigman on Hedging
Is that name really allowed to be mentioned here?Quote: FatGeezusJohn Patrick calls it 'The Richochet' method of betting.
Quote: SanchoPanzaThe important difference being the taking down of the bet. It does not stay up there. None of the mathematicians who advocate staying away from hedging address the extremely limited use of the lay technique.
I am aware that you are taking down the lay bet after the DC goes to a number.
The presence of the lay bet, even for one roll, increases the HE.
Whether one roll, or every roll, the lay will lose more money than it wins.
I simulated it as you described using Wincraps Pro.
The EV for the combined Pl and lay is 2.4%
Without the lay, it is 1.41%
With the DC added in, the overall EV is around 1.7%.
It would be 1.4% without the lay as a hedge bet.
The worst part playing the Don't is the Come Out obviously, after that odds are in your favor, correct? But if you are playing the DC continuous on every roll, don't you mitigate that significantly? Come out you have 7/11 against, 2/3 for everything else is point or wash. But if point is established, on your next roll, 7 becomes a wash (lose on the DC, win against the point), 2/3 still win, 11 lose, and now the point lose. Having the point against instead of the 7 against, seems quite advantageous, and this is compounded the more number's you are don't on. (Apologies if terminology is off, like I said I'm new).
Has anyone calculated the house edges playing this way? Am I right in that it decreases by playing continuous DC or is it the same but you just win or lose much faster?
Quote: SanchoPanzaIs that name really allowed to be mentioned here?
Few people know that John Patrick is the pen name used by , no, I can nor say it. Gee Wiz
Quote: buurmjHas anyone calculated the house edges playing this way? Am I right in that it decreases by playing continuous DC or is it the same but you just win or lose much faster?
The HE stays exactly the same if you play continuous DC or if you just play one hand at a time.
Your total action goes up, thus the negative EV also increases. I learned the hard way you need to limit your action, even if you are in the position of, say, having the 4,5, and the 10 to resolve ... good numbers darkside. Getting excited about that a while ago, I figured why not get all 6 numbers up to resolve darkside if possible? sweet! But it is putting too much in action. If having one or two numbers, maybe 3, to resolve is a good fit for your rightside betting, then it is also right for your darkside betting. Trust me on this.