November 20th, 2009 at 10:01:04 AM
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In a previous thread, one poster suggested that after 10,000 or so rolls, your results would be "durn close" whether you take the odds bet or not. I disagree, and here is my attempt to explain why.

I'd like to modify the premise slightly from 10,000 rolls to 3,000 Pass Line bets resolved. This is a bit more than the number of expected Pass Line bets resolved over 10,000 rolls, and it is easier to solve.

I found in a Probability text book that for n independent trials with probability of success, p, and probability of failure, q, that the variance is n*p*q. Furthermore, I found that you can sum indpendent variances. The formula assumes a payout of 1 for success and 0 for failure. However, that is not the case in craps, so we need to define another variable "x" as the difference between the payout for winning and losing (e.g., if point is 8, a win is worth +7 and a loss -6, so x=13). When computing the variance we have to square x and multiply it to n*p*q.

For Pass Line bets only (no odds), n=3000, p~0.507071, q~0.492929, x=2 => Var=2,999.4 and standard deviation~54.8.

For the case of making Pass Line bets and taking 3x4x5x odds, the true calculation should be the sum for every possible outcome of the product of the square of the difference between that outcome from the mean and the probability of that outcome. It's hard to write in words and even harder to calculate. So I will instead approximate the variance by summing the variances of the possible outcomes of one trial and using their expected number of occurences in 3000 trials as the value for n.

For come out winners and losers, n=1000, p=8/12, q=4/12, x=2 => Var~888.89

For points 6 and 8, n~833.33, p=5/11, q=6/11, x=13 => Var~34,917.36

For points 5 and 9, n~666.67, p=4/10, q=6/10, x=12 => Var~23,040.00

For points 4 and 10, n=500, p=3/9, q=6/9, x=11 => Var~13,444.44

Summing the variances, we get Var~72,290.69, so the standard deviation for 3000 trials, taking odds ~268.9.

I would suggest that even after 3,000 trials 268.9 is significantly higher than 54.8, but I suppose that is a matter of interpretation.

If anyone knows a shortcut for calculating the true variance for the case of taking odds, or has any corrections to my math, please share.

I'd like to modify the premise slightly from 10,000 rolls to 3,000 Pass Line bets resolved. This is a bit more than the number of expected Pass Line bets resolved over 10,000 rolls, and it is easier to solve.

I found in a Probability text book that for n independent trials with probability of success, p, and probability of failure, q, that the variance is n*p*q. Furthermore, I found that you can sum indpendent variances. The formula assumes a payout of 1 for success and 0 for failure. However, that is not the case in craps, so we need to define another variable "x" as the difference between the payout for winning and losing (e.g., if point is 8, a win is worth +7 and a loss -6, so x=13). When computing the variance we have to square x and multiply it to n*p*q.

For Pass Line bets only (no odds), n=3000, p~0.507071, q~0.492929, x=2 => Var=2,999.4 and standard deviation~54.8.

For the case of making Pass Line bets and taking 3x4x5x odds, the true calculation should be the sum for every possible outcome of the product of the square of the difference between that outcome from the mean and the probability of that outcome. It's hard to write in words and even harder to calculate. So I will instead approximate the variance by summing the variances of the possible outcomes of one trial and using their expected number of occurences in 3000 trials as the value for n.

For come out winners and losers, n=1000, p=8/12, q=4/12, x=2 => Var~888.89

For points 6 and 8, n~833.33, p=5/11, q=6/11, x=13 => Var~34,917.36

For points 5 and 9, n~666.67, p=4/10, q=6/10, x=12 => Var~23,040.00

For points 4 and 10, n=500, p=3/9, q=6/9, x=11 => Var~13,444.44

Summing the variances, we get Var~72,290.69, so the standard deviation for 3000 trials, taking odds ~268.9.

I would suggest that even after 3,000 trials 268.9 is significantly higher than 54.8, but I suppose that is a matter of interpretation.

If anyone knows a shortcut for calculating the true variance for the case of taking odds, or has any corrections to my math, please share.

The ratio of people to cake is too big.

November 20th, 2009 at 10:41:27 AM
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I just realized I left out a major piece of the puzzle that will help compare the variances over different numbers of trials--Expected Value.

The EV for either betting strategy is the same and is dependent on the number of trials. Specifically EV=~.01414*n.

EV (n=3,000) ~ -42.42

STDEV (Pass Line only, n=3,000) ~ 54.77

STDEV (Pass Line + odds, n=3,000) ~ 268.87

EV (n=30,000) ~ 424.24

STDEV (Pass Line only, n=30,000) ~ 173.19

STDEV (Pass Line + odds, n=30,000) ~ 850.24

EV (n=3,000,000) ~ -42,424.24

STDEV (Pass Line only, n=3,000,000) ~ 1,731.88

STDEV (Pass Line + odds, n=3,000,000) ~ 8,502.39

For 3,000 trials, breaking even is within 1 standard deviation for both strategies.

For 30,000 trials, breaking even is still within 1 standard deviation for the odds bettor, but not for the Pass Line only strategist.

For 3,000,000 trials, breaking even is several standard deviations away from EV, so that even the luckiest player will almost certainly lose over this number of trials.

The EV for either betting strategy is the same and is dependent on the number of trials. Specifically EV=~.01414*n.

EV (n=3,000) ~ -42.42

STDEV (Pass Line only, n=3,000) ~ 54.77

STDEV (Pass Line + odds, n=3,000) ~ 268.87

EV (n=30,000) ~ 424.24

STDEV (Pass Line only, n=30,000) ~ 173.19

STDEV (Pass Line + odds, n=30,000) ~ 850.24

EV (n=3,000,000) ~ -42,424.24

STDEV (Pass Line only, n=3,000,000) ~ 1,731.88

STDEV (Pass Line + odds, n=3,000,000) ~ 8,502.39

For 3,000 trials, breaking even is within 1 standard deviation for both strategies.

For 30,000 trials, breaking even is still within 1 standard deviation for the odds bettor, but not for the Pass Line only strategist.

For 3,000,000 trials, breaking even is several standard deviations away from EV, so that even the luckiest player will almost certainly lose over this number of trials.

The ratio of people to cake is too big.

November 21st, 2009 at 3:54:09 AM
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Again, my shortcomings in math are quite regrettable, but one thing has become clear: 10,000 rolls [come-out rolls?] can mean only 3,000 trials where free odds are involved. Now what you have said makes more sense to me.

At something like 30 come-out rolls per hour, there must be quite a few craps players who would see 10,000 rolls in a reasonable span or certainly in a lifetime.

At something like 30 come-out rolls per hour, there must be quite a few craps players who would see 10,000 rolls in a reasonable span or certainly in a lifetime.

"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." Mark Twain ....................
So it's like love? I don't think craps is like love. I think craps is a fun random game, sometimes exhilarating, sometimes infuriating. If you try to control it, chances are it'll drive you nuts and you'll end up losing a lot of money.
... wait, craps *IS* like love! MathExtremist

December 8th, 2009 at 4:21:37 PM
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In my previous posts to this thread, I had the right idea, but the wrong math. The table below shows my calculations with (hopefully) the right math.

A trial is defined as a completed Pass Line bet.

Expected Value is the same for all cases and is solely dependent on the number of trials.

The first Standard Deviation and Win % columns assume $1 bet on the pass line.

The second Standard Deviation and Win % columns assume $1 bet on the pass line plus 3x4x5x odds.

The third Standard Deviation and Win % columns assume $1 bet on the pass line plus 100x odds.

The Win % is based on the normal curve and is the probability that after the given number of trials, the bettor will be ahead or break even overall. Obviously the Win % should be the same after 1 trial for all betting systems, but the error is due to using the normal curve as an approximation of the possible outcomes. This approximation becomes more valid as the number of trials increases.

Note that the 100x odds bettor has nearly a 1 in 4 chance of being up even after 1 million trials. That IMO is the power of the free odds bet.

If anyone has any corrections, please let me know.

A trial is defined as a completed Pass Line bet.

Expected Value is the same for all cases and is solely dependent on the number of trials.

The first Standard Deviation and Win % columns assume $1 bet on the pass line.

The second Standard Deviation and Win % columns assume $1 bet on the pass line plus 3x4x5x odds.

The third Standard Deviation and Win % columns assume $1 bet on the pass line plus 100x odds.

The Win % is based on the normal curve and is the probability that after the given number of trials, the bettor will be ahead or break even overall. Obviously the Win % should be the same after 1 trial for all betting systems, but the error is due to using the normal curve as an approximation of the possible outcomes. This approximation becomes more valid as the number of trials increases.

Pass Line | Only | 3x4x5x | Odds | 100x | Odds | ||
---|---|---|---|---|---|---|---|

Trials | EV | STDEV | Win % | STDEV | Win % | STDEV | Win % |

1 | $(0.01) | $0.71 | 49.2% | $1.01 | 49.4% | $20.68 | 50.0% |

10 | $(0.14) | $2.24 | 47.5% | $3.20 | 48.2% | $65.39 | 49.9% |

100 | $(1.41) | $7.07 | 42.1% | $10.12 | 44.4% | $206.77 | 49.7% |

1,000 | $(14.14) | $22.36 | 26.4% | $32.00 | 32.9% | $653.86 | 49.1% |

10,000 | $(141.41) | $70.71 | 2.3% | $101.20 | 8.1% | $2,067.69 | 47.3% |

100,000 | $(1,414.14) | $223.61 | 0.0% | $320.03 | 0.0% | $6,538.62 | 41.4% |

1,000,000 | $(14,141.41) | $707.11 | 0.0% | $1,012.02 | 0.0% | $20,676.93 | 24.7% |

10,000,000 | $(141,414.14) | $2,236.07 | 0.0% | $3,200.30 | 0.0% | $65,386.18 | 1.5% |

Note that the 100x odds bettor has nearly a 1 in 4 chance of being up even after 1 million trials. That IMO is the power of the free odds bet.

If anyone has any corrections, please let me know.

The ratio of people to cake is too big.

May 15th, 2011 at 9:30:33 AM
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Pass Line Only | Pass Line Only | With 3x4x5x | With 3x4x5x | With 100x | With 100X | ||
---|---|---|---|---|---|---|---|

Trials | EV | STDEV | Session Win % | STDEV | Session Win % | STDEV | Session Win % |

1 | $(0.01) | $0.71 | 49.2% | $1.01 | 49.4% | $20.68 | 50.0% |

10 | $(0.14) | $2.24 | 47.5% | $3.20 | 48.2% | $65.39 | 49.9% |

100 | $(1.41) | $7.07 | 42.1% | $10.12 | 44.4% | $206.77 | 49.7% |

1,000 | $(14.14) | $22.36 | 26.4% | $32.00 | 32.9% | $653.86 | 49.1% |

10,000 | $(141.41) | $70.71 | 2.3% | $101.20 | 8.1% | $2,067.69 | 47.3% |

100,000 | $(1,414.14) | $223.61 | 0.0% | $320.03 | 0.0% | $6,538.62 | 41.4% |

1,000,000 | $(14,141.41) | $707.11 | 0.0% | $1,012.02 | 0.0% | $20,676.93 | 24.7% |

10,000,000 | $(141,414.14) | $2,236.07 | 0.0% | $3,200.30 | 0.0% | $65,386.18 | 1.5% |

Apologies to dk, but I wanted to change some headers to understand this better. Hopefully it's right! This possibly is a very rare chart, showing changes in standard deviation using the free odds [at least I havent been able to find it elsewhere].

No one responded to the request that this be checked out for accuracy. Anyone know, then, if this is an accurate chart?

"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." Mark Twain ....................
So it's like love? I don't think craps is like love. I think craps is a fun random game, sometimes exhilarating, sometimes infuriating. If you try to control it, chances are it'll drive you nuts and you'll end up losing a lot of money.
... wait, craps *IS* like love! MathExtremist

May 16th, 2011 at 3:02:29 AM
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a little bump, not giving up on this

"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." Mark Twain ....................
So it's like love? I don't think craps is like love. I think craps is a fun random game, sometimes exhilarating, sometimes infuriating. If you try to control it, chances are it'll drive you nuts and you'll end up losing a lot of money.
... wait, craps *IS* like love! MathExtremist

May 16th, 2011 at 1:55:30 PM
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I guess maybe I way off on what you are trying to get at but.......I think were you are going wrong is using the same bank roll for both pass line bets and odds bets. If you separate the two you will see if the one person bets 1 unit on the pass line and and another places 1 unit on the pass line with 2X's odd over 3000 rolls of the dice came out statistically right they would lose the same amount of money.

If they had the same bankroll and they were both placing the same amount on every roll. Player one makes a 100 unit P/L bet and player two was playing a 20 unit P/L bet with 4 times odds(80 units) they both have 100 units in play but player two will lose less money. because only 20% of his bet is subject to negative expectations. will 100% of player ones bet is subject to negative exceptions.

But if they both made the same P/L bet. And one placed odds and one didn't in the end they would both lose the same amount.

If they had the same bankroll and they were both placing the same amount on every roll. Player one makes a 100 unit P/L bet and player two was playing a 20 unit P/L bet with 4 times odds(80 units) they both have 100 units in play but player two will lose less money. because only 20% of his bet is subject to negative expectations. will 100% of player ones bet is subject to negative exceptions.

But if they both made the same P/L bet. And one placed odds and one didn't in the end they would both lose the same amount.

June 18th, 2011 at 9:39:37 AM
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sorry, lost track of this for a while

I'll repeat that it is very unusual to find anyone with figures on variance/standard deviation in Craps with free odds , at least I don't run across them

the premise of the original post is that this "in the end" business takes many more trials than the 10,000 claimed in another thread. I am the person who said 10,000 yields "durn close". If the claim is true, 10,000 not enough, then 3000 rolls or even 3000 come-out rolls is not enough. Or 10,000. I have pretty much confirmed this for myself now, using Wincraps."Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." Mark Twain ....................
So it's like love? I don't think craps is like love. I think craps is a fun random game, sometimes exhilarating, sometimes infuriating. If you try to control it, chances are it'll drive you nuts and you'll end up losing a lot of money.
... wait, craps *IS* like love! MathExtremist

I'll repeat that it is very unusual to find anyone with figures on variance/standard deviation in Craps with free odds , at least I don't run across them

Quote:vert1276And one placed odds and one didn't in the end they would both lose the same amount.

the premise of the original post is that this "in the end" business takes many more trials than the 10,000 claimed in another thread. I am the person who said 10,000 yields "durn close". If the claim is true, 10,000 not enough, then 3000 rolls or even 3000 come-out rolls is not enough. Or 10,000. I have pretty much confirmed this for myself now, using Wincraps.

June 18th, 2011 at 11:34:43 AM
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Bottom line: Our best chance of staying competitive is just betting the ole boring pass-odds bet. Mulitiple place betting or for that matter multiple come-odds betting schemes are doomed as the Ole Ugly is just too common, I have found that out the hard way. Think of it this way, much easier to recover from losing one bet compared to 3-6 bets.

Various betting schemes for the odds bet and I do like increasing the odds bet as inside box numbers are rolled after the point is established. Personally I like starting with $5 pass and $20 odds, then increasing the odds bet one unit as every inside number is rolled and stopping at $50 odds, then start anew on every pass-point sequence. Using a $600 day session bankroll and have had good net results over many sessions thusfar.

I tend to stray away as do like the action of place betting, but place betting has overall hurt my bankroll. The pass odds only is much better and it has taken awhile for it to sink in, must be a slow learner...

Various betting schemes for the odds bet and I do like increasing the odds bet as inside box numbers are rolled after the point is established. Personally I like starting with $5 pass and $20 odds, then increasing the odds bet one unit as every inside number is rolled and stopping at $50 odds, then start anew on every pass-point sequence. Using a $600 day session bankroll and have had good net results over many sessions thusfar.

I tend to stray away as do like the action of place betting, but place betting has overall hurt my bankroll. The pass odds only is much better and it has taken awhile for it to sink in, must be a slow learner...

June 18th, 2011 at 12:49:25 PM
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That is basically it. Which is the reason the skimpy costumes, free booze and stick's spiel exist. Combined, they are designed to have an effect on your betting, not just your merriment. Someone who starts out conservatively with pass line and 2x odds is often someone who winds up making the higher house-edge bets after a bit of booze and excitement has its effect.

I hate the effect of a seven-out if I've been doing come bets as well as my initial pass line bet, but I'm told that a come bet and a passline bet are virtually identical and it makes no difference if I do a PassLine Bet at each of two tables or do a PassLine bet and then a ComeBet at the same table.

I know about the dice having no memory. We all know that. However, I do always feel that a "seven-out" is "due". It is why I often try to do PassLine, ComeBet, ComeBet, DontCome, DontCome. Its a way of saying to myself: Okay, you've got some chances if the shooter is lucky and some if he ain't. And if things work out right, you can see those Come Bets rolled, then if the seven does appear, it won't hurt you too bad to lose that Passline bet.

We all know to have as much money as possible on the Odds portion and as little as required on the Flat portion. We don't always do it, but we know the theory and don't really dispute it in any way.

On a choppy table (they all seem to be choppy) we can do well if we pretty much wind up even. Sure somebody will walk up and make a high house edge center bet and walk away with his winnings, but often he just walks away. We who play a tight Basic Strategy game and resist temptations to listen to the stickman's patter tend to do the best at the tables despite the headline making stories such as that four hour roll at Borgata by a newbie.

I hate the effect of a seven-out if I've been doing come bets as well as my initial pass line bet, but I'm told that a come bet and a passline bet are virtually identical and it makes no difference if I do a PassLine Bet at each of two tables or do a PassLine bet and then a ComeBet at the same table.

I know about the dice having no memory. We all know that. However, I do always feel that a "seven-out" is "due". It is why I often try to do PassLine, ComeBet, ComeBet, DontCome, DontCome. Its a way of saying to myself: Okay, you've got some chances if the shooter is lucky and some if he ain't. And if things work out right, you can see those Come Bets rolled, then if the seven does appear, it won't hurt you too bad to lose that Passline bet.

We all know to have as much money as possible on the Odds portion and as little as required on the Flat portion. We don't always do it, but we know the theory and don't really dispute it in any way.

On a choppy table (they all seem to be choppy) we can do well if we pretty much wind up even. Sure somebody will walk up and make a high house edge center bet and walk away with his winnings, but often he just walks away. We who play a tight Basic Strategy game and resist temptations to listen to the stickman's patter tend to do the best at the tables despite the headline making stories such as that four hour roll at Borgata by a newbie.