craps bet

help me destroy this logic. $25 don't bet with $1 on the 2,3,12, and $2 on the 11 = $30 total bet. on 36 bets a 7 rolls 6 times. = $180 loss. 2 rolls once plus $51. 3 rolls twice $36 dollars times two= plus $72. 12 rolls once plus $26. 11 rolls twice plus $2. total don'ts = plus $151. That leaves 24 bets left where you are betting $30 to win $25 but with better odds than that on 14 rolls. trying to make up the $29 dollar difference. 0n the 10 rolls on the 6&8 true odds. This seems to work on a regular basis but has to be negative ev. Seems like you could get good comp ratings also. Help Thanks

strings of place numbers (4,5,6,8,9,10) will cause you to lose the don't as well as the single roll bets. You may get the "perfect" expected distribution in 36 rolls, but it could come with the seven rolling six times in the first six rolls...

I don't expect the perfect roll. I know there will be runs where your bank will go up and down. A $500 dollar bank has so far handled these distribution differences.

Quote: Wizard's Craps Appendix - Don't Pass

Don't Pass/Don't Come
The probability of winning on the come out roll is pr(2)+pr(3) = 1/36 + 2/36 = 3/36.
The probability of pushing on the come out roll is pr(12) = 1/36.
The probability of establishing a point and then winning is pr(4)×pr(7 before 4) + pr(5)×pr(7 before 5) + pr(6)×pr(7 before 6) + pr(8)×pr(7 before 8) + pr(9)×pr(7 before 9) + pr(10)×pr(7 before 10) =
(3/36)×(6/9) + (4/36)×(6/10) + (5/36)×(6/11) + (5/36)×(6/11) + (4/36)×(6/10) + (3/36)×(6/9) =
(2/36) × (18/9 + 24/10 + 30/11) =
(2/36) × (1980/990 + 2376/990 + 2700/990) =
(2/36) × (7056/990) = 14112/35640
The total probability of winning is 3/36 + 14112/35640 = 17082/35640 = 2847/5940
The probability of losing is 1-(2847/5940 + 1/36) = 1-(3012/5940) = 2928/5940
The expected return is 2847/5940×(+1) + 2928/5940×(-1) = -81/5940 = -3/220 =~ 1.364%

Most other sources on craps will claim that the house edge on the don't pass bet is 1.403%. The source of the discrepancy lies is whether or not to count ties. I prefer to count ties as money bet and others do not. I'm not saying that one side is right or wrong, just that I prefer counting them. If you don't count ties as money bet then you should divide by figure above by the probability that the bet will be resolved in a win or loss (35/36). So 1.364%/(35/36) =~ -1.403%. This is the house edge assuming that the player never rolls a 12 on the come out roll.

Quote: Wizard's Craps Appendix - Horn

The probability of rolling either a 2 or 12 is 1/36 + 1/36 = 2/36.
The probability of rolling either a 3 or 11 is 2/36 + 2/36 = 4/36.
The probability of roling anything else is 1-2/36-4/36 = 30/36.
Remember that the horn bet is like all four craps bets in one. Even if one wins the other three still lose.

Wizard's calculation for the Horn bet with adjustment for betting high on the Yo (aka, eleven):
The house edge is:
[(2/36)×26 + (2/36)×11 + (2/36)x27 + (30/36)×(-5)]/4 = (-18/36)/4 = -15%

You can't add two negative expectations and get a positive outcome. The weakness appears to be the horn bet. Of the 36 rolls, you can only win on six combinations, and then only part of your original wager.

To minimize the House edge, Try the same $25 Don't Pass, and use the "hedge fund" to lay free odds.

On 6 rolls you lose on the come out. (7) on 6 rolls you win on the come out. (2,3,11,12). $180 loss vs a $151 win. on 24 rolls you have an advantage bet. This seems to limit the variance. When the table is cold which it usually seems to be, you almost always can get out with a win. not 100% but seems to work pretty well, that's why I'm having trouble getting a handle on this. Been on the Wizards original site a long time so know all the probabilities.

I think it still boils down your having a DontPass Bet ... and then you are diluting its benefit if it wins under the guise of "protecting yourself" from the effects of it losing. All you really accomplish is increasing the "price" of your DontPass bet. You are wagering 30 dollars when you would be better off wagering 25.00 and accepting the consequences of being wrong.

By lowering the variance and not taking odds to start with, if you get down because a lot of numbers are being made, I've found that I can then add odds on my don't bet and almost always recoup the losses. If a lot of 2's,3's and 12's hit you also catch up very fast.

I must play at a cheat table then...
I played $25 Dont Pass the other night, and I was down $500 on a COLD table! (Six people went 7-point-7), which only made me EVEN on those props, then someone shot SIX consecutive passes. That plus some 7/11's on the come-outs...

That's why I switched to adding the 2,3,11,12 on the come out. Instead of 8 losses on the come out vs 3 wins (7,11 vs 2,3) It is 6 losses vs 6 wins (7 vs 2,3,11,12). Then if I get beat because a lot of points are made, so be it.

Quote: dihaig

help me destroy this logic. $25 don't bet with $1 on the 2,3,12, and $2 on the 11 = $30 total bet. on 36 bets a 7 rolls 6 times. = $180 loss. 2 rolls once plus $51. 3 rolls twice $36 dollars times two= plus $72. 12 rolls once plus $26. 11 rolls twice plus $2. total don'ts = plus $151. That leaves 24 bets left where you are betting $30 to win $25 but with better odds than that on 14 rolls. trying to make up the $29 dollar difference. 0n the 10 rolls on the 6&8 true odds. This seems to work on a regular basis but has to be negative ev. Seems like you could get good comp ratings also. Help Thanks

Just do the math like this.......by the numbers on your $30 bet assuming you are playing on a 15 to 1 and 30 to 1 table and not a "FOR" table. On 36 come outs!

2 is rolled......You win 25+30-4x1 net +51
3 is rolled......you win 25+15-4x2 net +72
4 is rolled......you lose 5x3...........net - 15
5 is rolled......you lose 5x4...........net - 20
6 is rolled......you lose 5x5...........net - 25
7 is rolled......you lose 25+5x6.....net - 180
8 is rolled......you lose 5x5...........net - 25
9 is rolled......you lose 5x4...........net - 20
10 is rolled....you lose 5x3...........net - 15
11 is rolled....you win 30-25x2.....net + 10
12 is rolled....you win 30-4x1.......net + 26

So before any points are passed or not you are net -141

24 points will be established......25x24 = $600 total in play on points established
40% of points will be passed, so you will win 60% of the time
360 won on points missed
240 lost on points made
net gain on points established +120

Total on 36 come out rolls net -21

Of course you cant win 360 making $25 Don't pass bets. that's just how the math came out using 60%/40% on 24 points 14.4 points will be missed. so if you want to figure close with round numbers

of 24 points
10 made - 250
14 missed + 350
net 100-141 net -41

or

of 24 points
9 made - 225
15 missed + 375
net 125-141net -16

Any way you slice it even if you round 14.4 points missed to 15 points missed you are a net loser. By making the prop bets you are taking out some volatility but giving yourself a larger expected lose.

EDIT POST....sorry changed numbers as I forgot to -4 on the 12 on the come out before so it is only +26 NOT +30

Perfect. Thank you.

First of all, I will assume that 2 and 12 pay 30-1, and 3 and 11 pay 15-1.

I'll put it another way...
If the come-out is:
2 (1 roll): you gain 30 on the 2, lose 4 on the other numbers, and gain 25 on the don't = +51
3 (2 rolls): you gain 30 on the 3, lose 4 on the other numbers, and gain 25 on the don't = +51 x 2 = +102
4 (3 rolls): you lose 5 on the numbers; 1/3 of the time, you make the point, and lose another 25 on the don't, while 2/3 of the time, you seven out and gain 25, so that's a total of (-5 + (1/3 x -25 + 2/3 x +25)) x 3 = +10
5 (4 rolls): you lose 5 on the numbers; 2/5 of the time, you make the point, and lose another 25 on the don't, while 3/5 of the time, you seven out and gain 25, so that's a total of (-5 + (2/5 x -25 + 3/5 x +25)) x 4 = 0
6 (5 rolls): you lose 5 on the numbers; 5/11 of the time, you make the point, and lose another 25 on the don't while 6/11 of the time, you seven out and gain 25, so that's a total of (-5 + (5/11 x -25 + 6/11 x +25)) x 5 = -13.6364.
7 (6 rolls): you lose 5 on the numbers and 25 on the don't = -30 x 6 = 180.
8 (5 rolls): same as 6 = -13.63634.
9 (4 rolls): same as 5 = 0
10 (3 rolls): same as 4 = +10.
11 (2 rolls): you gain 30 on the 11, lose 3 on the other numbers, and lose 25 on the don't = +2 x 2 = +4
12 (1 roll): you gain 30 on the 12, lose 4 on the other numbers, and push on the don't = +26
Add those up: over 36 rolls, your expected gain is -4.2727.

The reason it may seem to "work on a regular basis" is, if, in those 36 rolls, you replace a 7 with anything else, that -30 changes to at worst -2.2727, and your 36-roll expected gain is now at least +23.4545.

It's sort of like saying, "I'll cover 35 numbers on a roulette wheel"; you still have an expected loss, but "it works on a regular basis."