Cosmo counts odds bet at Craps

Here is what The Wizard says;

"It is worth noting that while taking the odds lowers the combined house edge, the expected loss remains the same. That is because you are applying a lower average house edge to a higher average bet. Readers often write to me, making this point, as an argument against betting the odds. However, if it is you goal to minimize expected loss as much as possible, then don't play at all! If you are going to gamble anyway, and your goal is to minimize the ratio of losses to amount bet, then you should bet as much as you can on the odds."

Road Trip take a bow, the rest back to school!!!
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Originally posted by: clcjim
I have no idea. I just wanted to say I like your screen name, kp.


This site requires six letters for a screen name so I had to go with the longer version of kp in order to delurk and come on Money's defense.

While I'm fanning the flames I'll add support to another controversial point by Money. Once a point has been established, the pass line bet has a better chance of loosing and a don't pass has a better chance of winning. I think everyone would agree thus far. Adding an odds bet at this point to a pass line bet is capping a loosing bet whereas adding odds to a don't pass is capping a winning bet. Sure, both bets have a good payoff if you win. But would you rather cap a winning bet or a loosing bet?
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Originally posted by: MoneyLA
thank you Kaypea. It's nice to know someone understands reality.

For my math guy friends, please consider this:

For the passline with any odds, the house has an advantage. According to your friend the Wizard of Odds, with 100X odds, the house has an edge of 0.021%.

Tell me, at what point does a negative edge for the player (i.e. a house edge) become a positive game for the player?

I would like to know when a negative becomes a positive? (Hint: it doesn't.)

A lower house edge if continued to be played only means more money going to the house.

Thanks again, Kaypea.


You are like a broken record. No one has said that the game will turn into a positive expected value. Why do you keep bringing that up?



Any money bet on true odds, or free odds, is a long term break even wager in the world of perfect distribution results and zero deviation.

In the world of perfect results, you will win some, and lose some. But the end results will still be exactly on par.

Win 1x2, lose 2 x1 and you break even after 3 decisions for the 4/10 point. True Odds 2-1
Win 2x3, lose 3 x 2 and you break even after 5 decisions for the 4/9 point. True Odds 3-2
Win 5x6, lose 6 x 5 and you break even after 11 decisions for the 6/8 point. True Odds 6-5

The true odds portion of a bet, for the "long term" player is worth exactly zero. No profit, no loss.

IF your bankroll can withstand the deviation of larger wagers, and if you get rated for betting true odds, than you could gain from the increase in comp value.

IF those increased comps are worth anything to you.



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Originally posted by: KayPea
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Originally posted by: RoadTrip
Therefor, our actual theoretical loss for those 1000 hands would be the same $70.50. 1.41%. That house advantage may appear diluted when the full odds are implemented, but if you examine perfect results on the points vs true odds, you will KNOW they cancel each other eventually in the long run. :::shrug:::


Then explain the 1.41 edge for the pass line only vs the .606 edge for the pass line with 2x odds. If the 1.41 is always applied to the line portion of the bet, where does the .606 come in?


That's because some of the money is wagered at a ZERO expected value which would reduce the overall expected value of the total amount wagered. Upon further review, I will concede that wagering more money will result in a greater loss even with a lower expected value. However, if you are wagering the same total amount then it is always better to place as much as possible behind the line. $15 on the pass line is not as beneficial as $5 on the pass and $10 behind the line.

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That's because some of the money is wagered at a ZERO expected value which would reduce the overall expected value of the total amount wagered. Upon further review, I will concede that wagering more money will result in a greater loss even with a lower expected value. ........


This is what I just don't get. :::shrug:::

My expected loss on a $10.00 line bet with no odds is 1.41%, call it 14 cents.

Since true odds cancel themselves out, my expected loss on a $10.00 line bet with 10x odds is still about 14.1 cents.

I can understand how adding the odds will reduce the overall expected percentage value, but just don't get it if you are saying I can than expect to lose more than the original 14.1 cents.

Of course, this is in my world of perfect distribution and expectation with zero deviation.



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Originally posted by: RoadTrip
I can understand how adding the odds will reduce the overall expected percentage value, but just don't get it if you are saying I can than expect to lose more than the original 14.1 cents.

Can you or anyone show me an example formula how the 1.41 (no odds edge) and the .606 (2x odds edge) come out with the same expected loss?

For example:

$10 pass line with no odds bet at 1.41 edge = (10 x 1.41 / 100) = $.141 loss

??? pass line with 2x odds bet of ??? at .606 edge = (? x .606 / 100) = $.141 loss
For Alex:

Look. My whole point about the odds betting is that odds will not turn craps into a positive expectation game. Thats all Ive been saying. Ive also said that odds, while not turning craps into a positive expectation game, increases the chances that you will lose.

Im afraid, Alex, that you came in between a discussion between me and shlomo, who says you have to be crazy (my words) not to bet full odds. I was just trying to show him and explain to him, that betting full odds would not turn the negative expectation of craps into a positive expectation game... so, why bother?

True: betting full odds might minimize losses, but on the other hand betting full odds can maximize losses. The bottom line is that the odds bet has no player advantage and no house advantage. That has been my point the entire time.

The reason that odds can lower the house edge is that when you do win on the odds it can cancel out some losses of the odds money.

Still craps is a NEGATIVE EXPECTATION GAME. Betting odds will not make you a winner. Why is that so difficult? And why must shlomo maintain that if youre not betting full odds youre an idiot?

Now, if something got lost in the translation and through reading statements on the internet, Im the first to say Im sorry.

edited to add:

And a note to Demango who quoted the Wizard who wrote: "It is worth noting that while taking the odds lowers the combined house edge, the expected loss remains the same."

Do you know what that means? That means that craps is a negative expectation game, and even though the house edge might be reduced with betting the odds it is still a negative expectation game. Enough already kids. Enough.
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Originally posted by: KayPea
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Originally posted by: RoadTrip
I can understand how adding the odds will reduce the overall expected percentage value, but just don't get it if you are saying I can than expect to lose more than the original 14.1 cents.

Can you or anyone show me an example formula how the 1.41 (no odds edge) and the .606 (2x odds edge) come out with the same expected loss?

For example:

$10 pass line with no odds bet at 1.41 edge = (10 x 1.41 / 100) = $.141 loss

??? pass line with 2x odds bet of ??? at .606 edge = (? x .606 / 100) = $.141 loss


I seem to recall seeing that question addressed somewhere, probably on one of the wizard sites.

Perhaps you'll search for it, and than come back and enlighten us all.

I've spent more than enough time on this thread already, and getting tired of it. .


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Originally posted by: KayPea

a) $5 line bet x 1000 hands = $5000 wagered x 1.41 edge = $70.50 loss
b 2x) $5 line bet + $10 odds = $15 x 1000 hands = $15000 wagered x .606 edge = $90.90 loss
b 100x) $5 line bet + $500 odds = $505 x 1000 hands = $505,000 wagered x .021 edge = $106.05 loss


Fixed my math:

a) $5 line bet x 1000 hands = $5000 wagered x 1.41 edge = $70.50 loss
b 2x) $5 line bet + $10 odds on 24/36 hands = $11.67 x 1000 hands = $11,666 wagered x .606 edge = $70.70 loss
b 100x) $5 line bet + $500 odds on 24/36 hands = $338 x 1000 hands = $338,333 wagered x .021 edge = $71.04 loss

The flaw in the earlier calculations is the assumption that you always have an odds bet, when in reality you only place an odds bet when a point is established (24 out of 36 possible come out rolls). These new calculations show that the actual loss is always the same (with some round off errors I presume).

The primary difference is that to play the odds requires a much larger bankroll. If you're willing and able to risk the extra $$$ short term to play for comps, then the Cosmo deal is a good one. I believe they will come out ahead as people are generally better about walking away at a stop loss than a stop win.
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