-X > -2X

In this specific case, that 1/36 times you hit a 12 on the come out roll you're screwed.

https://wizardofodds.com/gambling/tencom.html

Don't pass: Negative EV before the point (what you're trying to negate) and positive EV after the point

Pass: Exact opposite

Example:

If a point of 4 comes up, fair odds would be getting paid 2-1 if the 4 hits... but you're getting paid 1-1 because you can't take off that pass line bet. So, yes, you are taking away the (majority) of losses before the point... but you're also taking away the wins AND you're forced to keep a negative EV bet up on the table.

Lay-man's terms: In essence, although you are negating a bet which will lose you money in the long term before points... you're making it so that throughout the entire game you're going to have a bet on the table that will lose you money. This compares to playing just the dark side, which has a negative expected value before the point and a positive expected value after the point (discounting all odds).

Quote:theanswer200Do you decrease your player disadvantage against the house if on your first bet you bet both on the pass line and the don't pass, and then simply bet the don't pass odds,or do you have a bigger advantage simply betting one way and then odds.

the Wizard has covered this HERE about half way down the page.

Quote:DeMangoI'm always amazed at the amount of people who believe that the HA is double when you play the doey don't. One chance out of 72 to lose, no more, no less.

Quote:rudeboyoi1/36 you lose which comes out to 2.77% which is also the sum of the HA on the pass 1.41% plus the HA on the dont pass 1.36%

Quote:RonCHow is it 1/72 when the chances of getting a 12 and losing both bets is 1/36 on each come out roll?

Well, geez. I thought I understood this stuff just a little, but maybe not.

For the moment, ignore the free odds bet. It seems to me that (in terms of long term average expectations) for every 36 come out rolls betting doey don't -- that is, on average for 36 pass bets and 36 simultaneous and equal don't pass bets -- everything should cancel out except for the lose/push when the 12 is rolled. That means 1/36 of the come out rolls would give a loss and 1/72 of the units bet would be lost. Does that mean that both DeMango and rudeboyoi are correct or that both of them are in error or neither of the above? Or maybe we just can't communicate any more?

And the free odds bet (ignored for a moment above), no matter which side it is placed on, reduces the percentage of house advantage but not the dollar value of the house advantage -- so should we now argue about whether or not the free odds bet changes the house advantage?

Geez.

And perhaps RonC loses the don't pass bet on a come out 12, but I never have.

This is a House Advantage of 0.27777 / 20 = 1.3888888% on a TWENTY dollar bet, or 2.77777% on a TEN dollar bet.

So... rudeboyoi is more right. It makes a lot more sense to calculate the HA in this case against the $10, since it's all that is at risk.

If so, consider my (foolishly) simultaneously wagering $1 on each of the 38 numbers on a double-zero roulette table. My expected average loss is $2. My maximum loss is $2. Does that make the house advantage 100%?

My opinion is that the house advantage in double-zero roulette is 1/19 (ignoring the 5-number, really-stupid bet). Placing multiple, opposing bets, similar to doey don't, just affects variance, I think.

The HA on any individual bet is still 1/19, and the house edge against the total amount wagered is 1/19. But the house edge on the money at risk (the working money, if you will) is %100.

Why do you know ahead of time it's foolish to place this combination of bets? Because the combination of bets effectively has a house edge of 100%

No, the combination of bets is for $38 and has a house advantage of 1/19, giving an expected average loss of $2. Similarly, for $20 wagered in a crap game and split doey don't, the overall house advantage is 1/72 or 1.389% of the $20 wagered for an expected average loss of 27.8 cents.

When I made my original post (on the previous page), I thought the disagreement between DeMango and rudeboyoi was basically a matter of clarity of language: percentage of what? One chance out of 72 to lose what? I really intended to mock our own group process of stumbling over the words in a discussion, and saying the same thing at cross purposes. I'm not totally certain that DeMango and rudeboyoi actually disagree, if they could just agree on terminology. Instead of mocking, I think I just accidentally trapped the two of us into doing it even more!

To answer your question, covering the roulette table is a foolish bet because the standard deviation of the outcome is much smaller than the expected value of the loss. In this case, the standard deviation is zero, but a different scenario of equally betting even 35 of the 38 numbers (giving a very small but not-quite-zero standard deviation) would also be a foolish bet, I think, even though it would be possible to come out ahead.

if youre betting $5 on the dontpass, you are losing 1.36% of $5.

if you bet $5 on the passline and $5 on the dontpass, you are losing 2.77% of $5.

if you bet $5 on the passline and $5 on the dontpass, you are losing more than if you just bet $10 on the dont pass and you are losing less than if you just bet $10 on just the passline.

If there were a game where we could place two wagers of $100 each that somehow together had a maximum possible loss of $75 and a expected value of total loss equal to $15, then I suspect we could get this thread to run another couple of pages with at least some of us disagreeing on what the percentage house advantage was.

I still haven’t figured out how RonC was going to lose both pass and don’t pass on a come out 12, but I suspect that was just a slip.

Quote:rudeboyoiif you bet $5 on the passline and $5 on the dontpass, you are losing more than if you just bet $10 on the dont pass and you are losing less than if you just bet $10 on just the passline.

That's interesting.

What they need to let you do is split your bet, $2.50 on each [vbg]

Quote:rudeboyoiif youre betting $5 on the passline, you are losing 1.41% of $5.

if youre betting $5 on the dontpass, you are losing 1.36% of $5.

if you bet $5 on the passline and $5 on the dontpass, you are losing 2.77% of $5.

if you bet $5 on the passline and $5 on the dontpass, you are losing more than if you just bet $10 on the dont pass and you are losing less than if you just bet $10 on just the passline.

Nope, you are losing one half bet of $10. EVERYTHING ELSE PUSHES. You have to look at the total bet. So on the 12 you lose $5 out of $10 or 1/2 / 36, again -1.389

1. $5 don't with 2x lay no matter the point for 2 numbers(6 for 5, 3 for 2, 2 for 1)

2. always do $2 c/e on come out and for second point

3. if I get a 7 after 2nd number is established, for the next come out roll I go to $10 with $3 c/e twice, then $15 with $4...

If I lose I go back to $5 with $2 c/e for 2 numbers.

This has been working for me the last 6 months. I have been walking away with money, sometimes a lot when the table goes cold with a lot of shooters. Have I been lucky or have I hit on something? In some casinos my comps have dropped off to nothing.

Thanks