# I need a math guy

Discussion in 'Advanced Craps' started by Night Attack, Sep 9, 2020.

1. Night Attack, Sep 9, 2020

### Night Attack Member

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We all know that the house edge on placing a 6 or 8 is 1.52%.

If you place them both, then each one has a house edge of 1.52% but since the seven out will wipe out both bets, is the house edge higher?

I guess what I'm asking is how do you calculate the house edge on placing multiple numbers?

Appreciate it if anyone knows.

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2. Settingcanthurt, Sep 10, 2020

### Settingcanthurt Member

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Would not hitting either wipe out the other as well??

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3. von duck, Sep 10, 2020

### von duck Member

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NO!!!

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4. Settingcanthurt, Sep 10, 2020

### Settingcanthurt Member

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placed

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5. von duck, Sep 10, 2020

### von duck Member

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OK. One way to figure the H/E for this is to sort of consider it one bet, even though it's multiple numbers. For simplicity let's say you are placing each number for \$6 (American) you have 10 winning combinations which pay \$7, that's \$70 won. You have 6 losing combinations which lose \$12 each, for a loss of \$72. So for 12 dollars bet 16 times, 12X16= \$192.
\$2 lost for \$192 wagered, 2/192= 1.041%.
For the same amount on one number \$7 won five times, \$35. \$6 lost six times = \$36. \$6 bet 11times, with \$1 lost, 1/66= 1.515%.
It looks like you come out a little bit better % wise, if you place both numbers. I'll have to think about this one for a while, to figure out why. This is turning out to be a better question than I thought it was.

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Last edited: Sep 10, 2020
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6. von duck, Sep 10, 2020

### von duck Member

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That's correct. If you have 6&8 (placed) hitting one, does nothing to the other. Agreed?

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7. von duck, Sep 11, 2020

### von duck Member

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On thinking about this, where you have one bet (sixnieght) vs (sebben) where you have a 10/16 chance of winning, with a 7 to 12 payoff, you wind up with the same \$ amount of loss, but more money in action.
Just for fun, let's run the numbers for an inside place. 6-8-5-9 = 18 winning combinations, with 6 losing combinations. \$22 wagered 24 times, = \$528. 6 X \$22 = \$132 lost, 18 x \$7 won = \$126 Won, for a net loss of \$6 . 6/528 = 1.136%. Not too bad. If you bet it at a level that allows you to "buy" the five and nine, your % loss would be even smaller, especially with vig-after. (.631%). \$20 lost, for \$3168 put in action. Bets would be \$36 on 6&8, \$30 on 5&9. Better odds than you would think.

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8. Night Attack, Sep 11, 2020

### Night Attack Member

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While I can see your calculations and they logically follow the method used to determine the house edge on singular placed numbers, I'm finding it difficult to get it in my head that we can reduce the house edge by placing more numbers.

I may need to send an SOS out to the midnight skulker to see if he can corroborate any of this. I don't think I can go through another thread of whether the passline bet is better than placing numbers.

Thanks for providing these numbers so far.

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9. Mssthis1, Sep 11, 2020

### Mssthis1 Member

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House edge stays the same for each number no matter how many bets you make. Betting more numbers WILL increase the house hold. House hold is, \$ bet x HA x number of resolved bets = expected house hold over time.

Betting across worsens the house edge for you since it is 4% on the 5/9 and over 6% on the 4/10. If you can find a game where you can buy them vig after the pain s reduced but you also usually have to bet greater amounts.

If HA is your only concern, stick with the place six/eight or the best bet of all as far as HA, the line bets.

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10. von duck, Sep 11, 2020

### von duck Member

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I was also surprised, but the numbers aren't lying, or maybe they are. "House edge", is not a completely accurate reflection of ones chances. A very good example would be the doey-don't bet. The "house edge" for that bet is only 1.39%, but, you could never win a dime betting it. Any time a bet has a posible "push" decision, the H/E number will be somewhat skewed. In the case of place betting, the money bet on the numbers NOT hit in the decision, keeps getting counted because it is part of the overall wager. The actual dollar (US) amount lost does not change, but you increase the total action counted. What happens is, the numbers not hit, become pushes, when there is a hit on one of your other numbers.
In conclusion, H/E, is not all there is to it. My numbers are accurate.

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11. von duck, Sep 11, 2020

### von duck Member

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Horseshit! We did not say anything about , "house hold" that number is completely different, then H/E, but it is true, the hold number, would be greater, with each added number. It is a similar situation to the odds bet, which lowers the H/E, but not the house hold.

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12. ### FredP Member

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All of your numbers after betting \$12 which can win \$70 and lose \$72 suck.
The answer is 2/12=.166666 or 1.6666%

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13. random_roller, Sep 11, 2020

### random_roller Member

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Thought I'd stop in and say hi. I hope everyone is making the best of the current situation. My mom passed away a few months ago. She was an awesome mother and special person. It's been very tough. I have been focusing my time & effort on other things and will continue to do so; thus, I probably will only drop by once in a while, time permitting.

Place 6 and Place 8 bets, even if made at the same time, are independent. Following the come out roll, if a 6 is rolled before a 7, that has no win/loss impact on the Place 8 wager. Same holds true if an 8 is rolled (before a 7) instead -- no win/loss effect on the Place 6 bet. In both cases, the winning place bet gets paid while the other place bet is unaffected. Well, other than the disgust the bettor might feel for having split place bets equally on 2 complimentary numbers instead of doubling the bet on the number that had hit. Been there, done that.

If a bettor places the 6 & 8 for \$6 each (\$12 total bet/at risk) rather than only one (1) of those bets for \$6, the player's edge remains unchanged at -1.52% (per resolved bet) on the \$12 action. Mathematically, there is no difference in total expected return or player's edge than had the bettor placed either 6 or 8 for \$12.

By making the Place 6 & 8 bets for \$6 each (\$12 total bet/at risk), the bettor is doubling the number of possible winning number combinations from 5 to 10 while also doubling the amount he/she has bet/at risk. But the payout (on a single hit) is *not* doubling. Nor will betting more numbers change the theoretical frequency of a 6 or 8 (or 7) being rolled. Only a DI (or biased dice) might have those elusive powers. Hence, for each place bet the expected return \$(-0.0909) and player's edge remain unchanged (still -1.52%). What it will change is the total expected return (\$), which is 'up' to \$(-0.1818) based upon the increased total bet/at risk (\$12 versus \$6).

With all that being said, we've all been at tables where 1 number seemingly hits early and often while its complement is MIA. In the short term -- and realistically each craps session is just that -- almost anything can happen. Expected return is a theoretical calculation based upon long-term probabilities. It is not the be-all-end-all forgone conclusion of what *will* happen each and every time.

Take care everyone & stay safe. And good luck at the tables if you decide to visit a casino.

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Last edited: Sep 11, 2020
14. random_roller, Sep 11, 2020

### random_roller Member

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IMO, the player's edge (-1.39%) on the doey-don't bet (equal amount wagered on each) is a fair representation. It is a negative number, indicating the house has the edge. Furthermore, I am not aware of any casino making a claim that using such a betting approach (with no other bets, including taking or laying odds) can result in the player coming out ahead (excluding any promotion/comps). However, bettors deploying a doey-don't system often have chips on a bonus bet (i.e., Fire, ATS, Sharpshooter, etc.); also, they're trying to hedge their line bet against a craps number on the come out roll because they often don't have many chips left in the rack.

The only conclusion I will draw is the more money a bettor has at risk per roll and the longer he/she plays, the better the chances the casino will have because of its vastly larger bank roll, table betting limits, and less than true odds payouts on winning bets. Oh, and player greed.

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15. von duck, Sep 11, 2020

### von duck Member

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WRONG!!! Ma'am. You totally fked this up. 2/12 is NOT 1.6666%, it's 16.666%. You should stick to English, and leave the math, to somebody that's a lot better at it than you. To express any number as a %, move your decimal point, two places to the right. For example, the number 10, expressed as a percentage, would be 1000%. DIG? Ma'am. You're out of your element Ma'am.

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16. von duck, Sep 11, 2020

### von duck Member

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Well, your best chance of doubling your bankroll, would be to is let it all ride on one even money bet. I actually knew a guy that played that way, when he went to Vegas, he was a little strange in other ways also.
The best use of the "doey" that I've seen, is to play it at a ten X odds table, and run a D'Pyrimid, or small martingale on the odds bet on the don't side.

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17. random_roller, Sep 11, 2020

### random_roller Member

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Outside of sports betting (specifically college football, especially during the 2nd full week of a normal season), I would agree with you about making 1 big even money bet and leave either way up or cleaned out as being the optimal approach. But most people -- myself included -- enjoy the action. Making 1 bet and then leaving just doesn't cut it for me.

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18. von duck, Sep 11, 2020

### von duck Member

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It doesn't for me either, that's why I indicated that that guy, was a little strange.

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19. ### FredP Member

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Oops. I stand corrected.
But you’re wrong about my math knowledge. I have forgotten more mathematics than you’ll ever know.

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20. Night Attack, Sep 12, 2020

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