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Discussion in 'General Craps Discussion' started by RPrinceton, Jul 9, 2019.
Thanks to all for the insightful replies.
The doey-don't player you describe is doing nothing but guessing when he changes between taking and laying odds.
It doesn't reduce the house edge on the bets he was making. He's using the house edge of buy,lay and place bets verses the house edge the way he's playing to make the comparison.
If you can't afford 10x odds or greater and fairly large bets it doesn't make much sense. You still have bankroll tied up in the doey/don't bets that will never earn a penny. Most people don't factor that money into the equation.
How busy was the table and how receptive was the crew to this type of play?
TDVegas has admitted on several occasions that he is not a numbers guy, but his only mistake in this instance is that he, like many non-numbers guys and gals, used expected value (EV) and house advantage (HA) interchangeably. They are different ways of quantifying what you, as a player, are expected to lose: EV expresses this expectation as a dollar amount; HA expresses it as a percentage of bet handle. TDVegas explicitly made the point that a $100 flat bet on one of the lines is expected to lose $1.41 regardless of how much is taken/laid in odds; your point seems to be that the HA can be manipulated by increasing the bet handle without increasing the expected loss.
Alan Shank (aka goatcabin on both this forum and the old Usenet newsgroup) championed the illusory effect of using HA to "prove" that benefit from taking/laying odds because doing so reduces the HA. A decade or so ago Zeke Feinberg used the same sleight of hand to conjure a "synergistic effect" from placing both 6 and 8 vs. just one of them for the same total amount (e.g. $6 each vs. $12 on one). That is the illusion to which I referred.
In any case, you CAN reduce H/A by by betting the odds. This is not the only way to reduce H/A. I've stated it before, but just to refresh. On a $5 table, a player can receive very cheap buy or lay bets, by using the doey/don't, and the ten times odds available at some casinos. In a normalized run of 36 come outs, a player figures to lose one $5 do bet, for which he receives the opportunity to take or lay the 10X odds, on 24 points. Normally these bets would require $2 times 24 bets = $48 in vig, but with the "doey" you get them for about 21 cents each, about 1/10 the normal cost. This is on a vig up front table, which most are, these days. The only sticking point would be, that you have to except the point thrown, as the number you would buy or lay. Nonetheless, a substantial reduction in H/A over normal buy/lay bets. There are also many fine systems you might employ, with these now, very cheap bets.
We’ve been through this before.
The HE is the ratio of the average loss to the initial bet.
A bet is a risk. In this case, it is a risk of a sum of money against an eventual outcome.
A risk that cannot win or lose, is not a bet.
If you cannot win or lose money on the DP, because it is hedged on every roll by a corresponding PL bet, what are you actually betting on the DP?
Zero. Because there is zero risk.
The corresponding PL bet guarantees the DP can only push.
The offsetting PL bet is the only bet that can lose. It loses 1 in 36 rolls, for a HE of 2.77%
WRONG! The two bets are not related to each other. They are settled separately. In the case of the doey/don't, 72 separate bets are placed and settled, with the expectation of a net loss of, one, of those 72 bets. Your assessment, is completely wrong! The H/E on the Doey/don't, is 1.38888888%. This is not my opinion, it is a fact.
Agreed, but only because there is no other bet you can make with the money you are using for odds that carries a lower HA. On one of my favorite threads on the old newsgroup I beat John Patrick into submission with an argument to that effect. (Don't worry, Patrick and Python fans, he got better.)
The cause of the illusion/misunderstanding is not knowing what HA really is.
Which is the better bet, $100 PL that is expected to lose $1.41, or $100 PL w/$100 odds that is expected to lose $1.41?
Which is the better bet, $100 PL that is expected to lose 1.41%, or $100 PL w/$100 odds that is expected to lose 0.8%?
A similar illusion appears when calculating how large a draw game lottery (e.g. Powerball) jackpot must be before the game becomes player positive.
As much as it pains me to side with von duck I must do so in this case if for no other reason than your logic does not pass the smell test. How can a combination of two bets NOT have a HA between their individual HAs?
You claim that the DP bet is not at risk because it can only push, whereas the PL bet is at risk and therefore must be included in the bet handle. The DP is "not at risk" because it is offset by the PL on 35 of the 36 comeouts and by the bar number on the 36th comeout, so why does that method of determining "at risk" not also apply to the PL? On 35 of the 36 comeouts it is offset by the DP, and on the without a bar number is not saved from losing. Why, then, is its bet handle not just that 36th (losing) roll for a HA of 100%?
You are treating a combination of bets as a single bet and must therefore count the total amount of that combination as the bet handle. You are expected to lose half a unit per 36 rolls of one unit each for, as von duck calculated, a HA of 1.3888...%.
What about the 12 ??
And also playing:
The correct answer, of course, is C: the red-head who looks like Lindy Booth.
Mssthis1: Table was full. The crew knew this person and seemed to be impressed with this betting scheme. RPrinceton
When one of them is no longer a bet.
Because, although the PL bet will lose an average of 1 in 36 rolls, it is always at risk of losing on every roll of the dice. The 12 can roll much more than 1 in 36 rolls. The DP is never at risk because it is always offset and can only push against the corresponding PL bet.
There is no ‘combination of bets’. You are thinking the DP as a bet. Once offset by the PL, it is merely a hedge that can only push. The house doesn’t count it as a bet. You will be rated a zero bettor. They are not stupid. Since the DP is 100% guaranteed to push on every roll, there is no risk at all, and consequently, by definition it is no longer a bet.
The only side that can lose is the PL. Once in 36 rolls on average. For a HE of 2.77%. The PL becomes a really bad bet once you give up the possibility of winning 8 in 36 rolls with a 7 or 11.
You also eliminate the possibility of losing the PL bet on 4 of 36 rolls.
Hypothetical situation. Two guys playing as a team, unbeknownst to the house, one on one side of the table, one on the other. They are combining to play a Doey/don't, so how does the house "rate " the action of the don't player? TD.
Those of us old enough to remember
know exactly what Comback's young lady asked just prior to the taking of the picture he posted.
Then you have a single bet, not a combination, and the HA would be that of the bet that remains in action. In the current discussion, if the Don't side of a Doey/Don't is not a bet then it is basically not there and cannot win, which it must be capable of doing in order to offset a loss on the Do side. If it is there then it can also lose. IOW it is in action as is included in the bet handle.
Chip magnet has that part about the don't NOT being a bet, wrong, but this discussion has revealed a few things. Number one, H/A in itself, does not tell you what your chances of winning are. Case in point, even though the doey/don't has a better H/A than a straight pass line bet, the pass line bet, has an infinitely better chance of showing a winner, since the d/d has no chance at all of showing a winner. There is more to betting, then just knowing what the H/A is.