December 17th, 2011 at 12:54:52 PM
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What would be the odds or house edge on single roll betting all numbers?

December 18th, 2011 at 4:08:30 AM
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Quote:captmjrussWhat would be the odds or house edge on single roll betting all numbers?

usually you decide whether you want to do the Horn bet [12.5%] or the Whirl [13.33%]. I don't think you can bet 4,5,6,8,9,10 individually on a single roll basis, but maybe I'm wrong. There's the field though at 2.78% that covers some while duplicating others if you also do the Horn or Whirl.

IMO all single roll bets are murder for the bankroll in short order.

the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell! She is, after all, stone deaf. ... Arnold Snyder

January 15th, 2012 at 10:54:47 PM
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Of course you can bet single roll proposition bet on all numbers. It's called a hop bet. It has a terribly high House Edge, tell you that in sec.

The house edge of the field bet is 5.56% on the 2x Fields for 2 and 12. Won't see too many 3x pays on 2and12 on the Field unless you're in Vegas. 3x on the field 2/12 drops the edge to 2.78 like you mentioned.

Hop Bets Odds: 17-1 odds for any hop bet, but the casino will only pay 15-1. Knowing that the casino doesn't pay true odds on proposition bets, they should be avoided.

Hop Bets House Edge:

11.11% individually.

So for every $100 wagered on the hop, you will lose $11 approx. Bet on all box numbers, on the hop. You are wagering $30 on $5 minimum. And for every $100 wagered, expect to lose $66. Yuck.

Yikes. And for those of you that don't know, the Whirl is the 2,3,7,11,12. Please don't make these bets.

But it is your money....

The house edge of the field bet is 5.56% on the 2x Fields for 2 and 12. Won't see too many 3x pays on 2and12 on the Field unless you're in Vegas. 3x on the field 2/12 drops the edge to 2.78 like you mentioned.

Hop Bets Odds: 17-1 odds for any hop bet, but the casino will only pay 15-1. Knowing that the casino doesn't pay true odds on proposition bets, they should be avoided.

Hop Bets House Edge:

11.11% individually.

So for every $100 wagered on the hop, you will lose $11 approx. Bet on all box numbers, on the hop. You are wagering $30 on $5 minimum. And for every $100 wagered, expect to lose $66. Yuck.

Yikes. And for those of you that don't know, the Whirl is the 2,3,7,11,12. Please don't make these bets.

But it is your money....

January 16th, 2012 at 1:38:52 AM
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when you say betting all numbers...

wouldnt you accomplish this by having a bet on all possible outcomes of the dice: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 and have them all turned "on" during the comeout? or any particular roll of the dice?

I'm no good at math, but would it be some blend of the house edge on each of those individual bets?

wouldnt you accomplish this by having a bet on all possible outcomes of the dice: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 and have them all turned "on" during the comeout? or any particular roll of the dice?

I'm no good at math, but would it be some blend of the house edge on each of those individual bets?

January 16th, 2012 at 2:22:37 AM
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Quote:YoDiceRoll11Of course you can bet single roll proposition bet on all numbers. It's called a hop bet.

didnt think of that. Not all places take them though, right?

Quote:AlanMendelsonwouldnt you accomplish this by having a bet on all possible outcomes of the dice

what would you say to the dealer in order to place these bets?

the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell! She is, after all, stone deaf. ... Arnold Snyder

January 16th, 2012 at 3:36:05 AM
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Quote:odiousgambitusually you decide whether you want to do the Horn bet (12.5%) or the Whirl (13.33%).

You may wish to bet the C&E and play the Seven as three separate $1 Hop (single combination) bets, as this strategy will return all but 11.1%, so that you gain 2.2% more of your $5 betting this way instead of plopping your nickel on that big, conspicuous, convenient "Whirl" zone!

Quote:AlanMendelsonwouldnt you accomplish this by having a bet on all possible outcomes of the dice?

what would you say to the dealer in order to place these bets?

In Lost Wages... I mean Las Vegas, most boxmen will book these Hop bets if you make them verbally - "call" bets - as many as you like, even all 21 of them. In Atlantic City, where the law requires explicit zoning to mark all bookable bets, as call bets are illegal, you can bet $1 right on the table, for each possible combination separately.

As for the fiscal expectation for a "blackout" bet, in 36 rolls, each of the fifteen X&Y ("easy" or "soft") combos will hit twice, winning $15 each time, while each of the six "hard" X&X combos will strike once, winning $30 each. Each roll will lose all but one Hop bet. So, assuming $21 bet per roll ($21 x 36 = $756 in all), thirty rolls will each see a net loss of $15-$20=-$5, and six will each net you $30-$20=$10. Overall, you will therefore lose $90 of your original $756 investment, which is a house edge of 11.90% against you.

January 16th, 2012 at 4:03:25 AM
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I dont think this is the right place to post this but I am going to reply anyways. I have my bachelors degree in statistics from emporia state university. I very recently brought stanford wongs casino tournament strategy book. Because I was trying to think if the math for it was right for what I was thinking as far as a blackjack method that did not involve counting but consistently beat the house. I also bought blackjack software and tested it for 2,000,000,789 hands. My results for this came up stunning I developed what I thought is a a variation of deep stack level progression non going broke betting strategy based on the laws of probability over time. This produces a positive gain for the player and let me explain how it works by explaining what it is not.

We know the math behind martingale doesnt work because you eventually go broke and you are betting to get one unit. What if you took a betting progression strategy and gave yourself aq bankroll of 200 times that betting progression and never went beyond that level you would then becoming up with something that beats the math behind most games because you never progress beyond that barrier of that which you have 200 times the total amount of the progresssion.

In even money games like blackjack the odds of winning one hand without doubling and splittling is close to 48% the odds of winning one out of two hands in blackjack without doubling or splittling is 73% and the odds of winning one out of three hands without doing the same is 86%. So here are the progressions I ran in my computer giving each player 200 X the total amount of each progression to see if we could yield a positive gain running each trial a minimum of 2,000,000,789 hands to account for the proper adjustments, I did 6 progression levels of each 200 deep for the total amount of progression so the players would not go broke if they lost one progression each player resorts back to the minimum bet after winning one progression bet

For the 1x then 5 X progression 200 deep my profit results yielded a constant growth of 12.7% percent over the long term

For the 1X 3X 15X progression 200 deep my profit results yielded a constant growth rate of 11.3% over the long term in simulation

For the 1X 2X 6X 18X progression 200 deep my profit results yielded a constant growth rate of 10.4% over the long term

For the 1x2X 6x 18X 72X progression 200 deep my profit resutls yielded a constant growth rate 9.6% over the long term

For the 1x 2x 6X 18x 54X 216X progression 200 deep my profit results yielded a constant growth rate of 8.7%

For the 1x 2x 6x 18x 54x 162X 1000X 200 deep my profit results yielded a constant growth rate of 6.5%

These trial progressions were ran 2,000,000,789 hands for each progression again the only reason this worked was because each player was sitting 200 deep of the total progression that way if a progression lost the player did not go broke. Cost is broken down as the following this is assuming a 5 dollar min wager

Bankroll for player A 5 + 25 = 30 * 200 = 6000 dollars

Bankroll for player b 5 + 15 + 75 * 200 = 19000 dollars

Bankroll for player c 5 + 10+ 30 + 90 * 200 = 27000 dollars

Bankroll for Player d 5 + 10 + 30 + 90 + 360 * 200 = 99000 dollars

Bankroll for Player e 5 + 10 + 30 + 90 + 270 + 1080 * 200 297000 dollars

Bankroll for Player F 5 = 10 + 30 + 90 + 270 + 810 + 5000 * 200 1,245,000 dollars

The whole premise behind this is to make sure you have 200 X of the total progression. Then the math does it self. Without the 200 X the strategy is no different then a glamoured up martingale. Without the 200 x the math says that this strategy will fail eventually. I would like someone else to do the math for me on this as well because maybe I have overlooked something, or maybe 2,000,000,789 is not enough hands. Yes that is right over 2 billion hands. Anyways get back to me on this please thank u

Jeremy Noble

We know the math behind martingale doesnt work because you eventually go broke and you are betting to get one unit. What if you took a betting progression strategy and gave yourself aq bankroll of 200 times that betting progression and never went beyond that level you would then becoming up with something that beats the math behind most games because you never progress beyond that barrier of that which you have 200 times the total amount of the progresssion.

In even money games like blackjack the odds of winning one hand without doubling and splittling is close to 48% the odds of winning one out of two hands in blackjack without doubling or splittling is 73% and the odds of winning one out of three hands without doing the same is 86%. So here are the progressions I ran in my computer giving each player 200 X the total amount of each progression to see if we could yield a positive gain running each trial a minimum of 2,000,000,789 hands to account for the proper adjustments, I did 6 progression levels of each 200 deep for the total amount of progression so the players would not go broke if they lost one progression each player resorts back to the minimum bet after winning one progression bet

For the 1x then 5 X progression 200 deep my profit results yielded a constant growth of 12.7% percent over the long term

For the 1X 3X 15X progression 200 deep my profit results yielded a constant growth rate of 11.3% over the long term in simulation

For the 1X 2X 6X 18X progression 200 deep my profit results yielded a constant growth rate of 10.4% over the long term

For the 1x2X 6x 18X 72X progression 200 deep my profit resutls yielded a constant growth rate 9.6% over the long term

For the 1x 2x 6X 18x 54X 216X progression 200 deep my profit results yielded a constant growth rate of 8.7%

For the 1x 2x 6x 18x 54x 162X 1000X 200 deep my profit results yielded a constant growth rate of 6.5%

These trial progressions were ran 2,000,000,789 hands for each progression again the only reason this worked was because each player was sitting 200 deep of the total progression that way if a progression lost the player did not go broke. Cost is broken down as the following this is assuming a 5 dollar min wager

Bankroll for player A 5 + 25 = 30 * 200 = 6000 dollars

Bankroll for player b 5 + 15 + 75 * 200 = 19000 dollars

Bankroll for player c 5 + 10+ 30 + 90 * 200 = 27000 dollars

Bankroll for Player d 5 + 10 + 30 + 90 + 360 * 200 = 99000 dollars

Bankroll for Player e 5 + 10 + 30 + 90 + 270 + 1080 * 200 297000 dollars

Bankroll for Player F 5 = 10 + 30 + 90 + 270 + 810 + 5000 * 200 1,245,000 dollars

The whole premise behind this is to make sure you have 200 X of the total progression. Then the math does it self. Without the 200 X the strategy is no different then a glamoured up martingale. Without the 200 x the math says that this strategy will fail eventually. I would like someone else to do the math for me on this as well because maybe I have overlooked something, or maybe 2,000,000,789 is not enough hands. Yes that is right over 2 billion hands. Anyways get back to me on this please thank u

Jeremy Noble

January 16th, 2012 at 7:18:32 AM
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Jeremy- during your statistics study, how would you answer this question?....

Professor, is there any way to sum up a bunch of negative expectation bets so that they will have a positive expectation?

Since you KNOW the answer is NO.... all your subsequent analysis is worth...... zippo....

Professor, is there any way to sum up a bunch of negative expectation bets so that they will have a positive expectation?

Since you KNOW the answer is NO.... all your subsequent analysis is worth...... zippo....

January 16th, 2012 at 7:34:51 AM
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Well you have a degree in statistics. What does the degree in statistics tell you? Find me some equations that show a positive expectation from a negative expectation game.

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You want the truth! You can't handle the truth!

January 16th, 2012 at 7:44:12 AM
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That he doesn't know enough not to post that identical dribble in 11 different places? Hell, he even posted it in at least one blog reply!Quote:boymimboWell you have a degree in statistics. What does the degree in statistics tell you?

I invented a few casino games. Info:
http://www.DaveMillerGaming.com/
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