Bubble machine paying DP/DC on a 12, how much does it lower the house edge?
Quote: TankoQuote: ChumpChangeWell, I'm trying to win to 35 DC bets ahead before I get to 25 DC bets behind.
link to original post
You might have better luck betting the Do side. I ran 5,155 rolls from those machines through Wincraps.
DP and one DC with 2X odds lost 96 units.
Pass Line and one Come bet with 2X odds won 31 units. Wins 76 units with zero odds and the 5 and 9.
link to original post
So playing with $25 units (table or machine), there's $25 line bets with $50 odds totaling at least $150 per shooter if they roll more than twice and on the Don't side this set of rolls ended up with 96 x $25 or a $2,400 loss; on the Do side it was a 31 x $25 or a $775 win. For the zero odds and the 5 & 9 it'd be $100 bet at least after two rolls and the win was 76 x $25 or a $1,900 win.
I know WinCraps can be extremely streaky and can favor one side or the other over several thousand rolls. But if I was going to run a progression on the PB 5 & 9 this is an idea, PL & 1 Come with no odds or 1X odds.
But really: $2,400/$150 = 16; $775/$150 = 5.167; and $1,900/$100 = 19
Quote: ChumpChangeI know WinCraps can be extremely streaky and can favor one side or the other over several thousand rolls.
I disagree.
The 5,155 rolls I used were actual rolls from the single player bubble craps machines. Not RNG rolls.
An individual don't bet has a standard deviation of 1 (taken to four decimal places) per resolved bet. For comparison purposes we must look at the SD on a per-roll basis, which is (165/557)^.5 =~ 0.544. If you always play a DC bet, your total bets made will increase by a factor of 557/165 compared to playing DP only. After simulating millions of rolls, the per-roll SD of always playing DC is 1.15. So, not surprisingly, its SD higher (by 15%) than increasing the number of DP bets by a factor of 557/165. This is because a seven-out can cause a win of 2-5 units depending on how many numbers are covered. All of the above is with no odds bets since there's no reason to make them with a positive EV for the line bet. Incidentally, always making DC bets increases the SD much more when odds bets are added.
So your per-roll expectation of always playing DC is +1.41% +/- 1.15. To be 99% confident of a profit (2.33 SDs), you need to play (1.15 * 2.33 / .0141)^2 = 36,113 rolls. Assuming you play 8 hours per day 5 days per week at 100 rolls per hour, you need to play for 2 months (!!) in order to have a 99% chance of a profit. This aligns with my initial estimate. 1% and change advantage on an even-money game is nothing and it will take a very long grind to get an assured profit. In those 45 days of playing, you can expect to have 16 losing days (36%) and 2 losing weeks (22%). I doubt the casino would even notice that you are a long-term winner
Some data if anyone is interested: If you always bet DC with no odds you will lose 1 unit on 36% of rolls, net zero / no action on 42.7%, win 1 on 13.9%, win 2 on 2.5%, win 3 on 2.2%, win 4 on 1.7% and win 5 on 1%
Thanks for the detailed explanation. I should be able to try to reproduce your numbers and then figure out where we disagree.Quote: Ace2I posted the variance and total bet action effects of come/dc betting a couple years ago but I cant find the posts. Having the 1.41% advantage on the don't has essentially no effect on these calculations, other than flipping the expected loss to a gain.
An individual don't bet has a standard deviation of 1 (taken to four decimal places) per resolved bet. For comparison purposes we must look at the SD on a per-roll basis, which is (165/557)^.5 =~ 0.544. If you always play a DC bet, your total bets made will increase by a factor of 557/165 compared to playing DP only. After simulating millions of rolls, the per-roll SD of always playing DC is 1.15. So, not surprisingly, its SD higher (by 15%) than increasing the number of DP bets by a factor of 557/165. This is because a seven-out can cause a win of 2-5 units depending on how many numbers are covered. All of the above is with no odds bets since there's no reason to make them with a positive EV for the line bet. Incidentally, always making DC bets increases the SD much more when odds bets are added.
So your per-roll expectation of always playing DC is +1.41% +/- 1.15. To be 99% confident of a profit (2.33 SDs), you need to play (1.15 * 2.33 / .0141)^2 = 36,113 rolls. Assuming you play 40 hours per week at 100 rolls per hour, you need to play for 2 months (!!) in order to have a 99% chance of a profit. This aligns with my initial estimate. 1% and change advantage on an even-money game is nothing and it will take a very long grind to get an assured profit. In those 45 days of playing, you can expect to have 16 losing days (36%) and 2 losing weeks (22%). I doubt the casino would even notice that you are a long-term winner
Some data if anyone is interested: If you always bet DC with no odds you will lose 1 unit 36% of rolls, net zero or no action on 42.7%, win 1 on 13.9%, win 2 on 2.5%, win 3 on 2.2%, win 4 on 1.7% and win 5 on 1%
link to original post
You are losing me where you introduce a factor of 557/165. I am looking at a situation where you make one new Don't bet every single roll (Either DP or DC). On average, then, there is exactly one bet resolved per roll. Variance per bet and variance per roll are the same thing in this scenario.
If you are talking about some other betting scenario than I am considering, then we won't agree on the variance per bet. Even rereading your post, I don't know what scenario you are considering.
I agree that it is possible to lose playing a +1.14% EV even money game. I have seldom played more than a few unhedged craps bets per session. I only play craps for quick loss rebates and I expect to lose most of the time. I am usually playing high-variance slots, but I am sometimes doing 5,000 bets per hours. I am usually playing games with big house edges before FSP considerations. I am quite unconcerned if I have a 30% chance of being a loser after 100,000 bets. Even some of the rocks on this site would not require a 99% chance of being ahead for any period less than a year.
So making a total of 1000 bets via continuous DC your SD is 36.4, which is 15% higher than making 1000 DP bets
Okay, I understand what you are saying now. You are using DP-only because it enforces the independent, uncorrelated nature of the bets. The ratio isn't really germane to the discussion, though -- just the uncorrelated nature of the results, right? You could make 1000 DP bets with 999 gaps where no bet was in effect and the ratio (rolls:bets) would change but the uncorrelated variance-per bet would be the same.Quote: Ace2For instance. If you play for ten hours (1000 rolls) making DC bets continuously, you’ll have 1000 bets resolved. To have that many resolved bets playing DP only, you must play 1000 * 557/165 = 3376 rolls on average. The expected win is the same either way, but the standard deviation for continuously betting DC is 1000^.5 * 1.15 = 36.4. The SD for DP only is 1000^.5 = 31.6 using bets resolved and also 3376^.5 * (165/557)^.5 = 31.6 using rolls.
So making a total of 1000 bets via continuous DC your SD is 36.4, which is 15% higher than making 1000 DP bets
link to original post
I can see where there are some instances where someone might care about variance per roll or variance per hour, but variance per bet is the only thing that is well defined.
Are you using a Markov chain to get your steady-state statistics?
As stated in my post, I simulated millions of rolls to get the variance for “always don’t coming”. I suppose a Markov chain is possible but it would be enormous as there are many ways to build up to six points covered before a seven-out. What do you mean by “steady state statistics”
I posted the probabilities of losing 1 unit, breaking even and winning 1, 2, 3, 4 or 5 units. Let’s see your numbers. Where do they differ from mine specifically?
Quote: Ace2Some data if anyone is interested: If you always bet DC with no odds you will lose 1 unit on 36% of rolls, net zero / no action on 42.7%, win 1 on 13.9%, win 2 on 2.5%, win 3 on 2.2%, win 4 on 1.7% and win 5 on 1%
link to original post
So 36% vs (13.9% x 1 + 2.5% x 2 + 2.2% x 3 + 1.7% x 4 + 1% x 5) = 36% vs (13.9% + 5.0% + 6.6% + 6.8% + 5%) = 36% vs 34.3%
I can tell I'm just throwing away chips on the continuous DC, so I'll lower the pace to 1, 2, or 3 bets up at one time.
If I get the table to myself, I'll try to throw craps or a 4/10 for the DP, and 5/9 for the 1st DC and a 6/8 for the 2nd DC, then get on with the 7-out.
Again, you are using Don't Pass as a short cut for saying independent Don't bets. I am sometimes playing in four different online casinos at a time. I can certainly have more than one bet in play on Don't Pass at one time, but they will be independent. I think we agree on the variance for Don't, but maybe not. My variance and EV for independently resolved Don't bets rolls is ev: 1.014142 tVar: 0.999800Quote: Ace2For the “always don’t coming” betting method, there are between zero and seven bets resolved on any given roll. With such varying bets on the table at any time, is there any way to analyze this method other than on a per-roll basis? You run a simulation and see what the total-result variance is relative to 1.41% gain on total rolls. There’s no way to quantify it as : “if I bet one unit then my variance is x”. But you can do that betting Don’t Pass only because there’s always just one bet in play
I got this result from a program I wrote a few days ago. I have a 15-year-old Monte Carlo craps program that I could modify to do the calculation similar to what you did. Instead, I wrote new code that calculated the variance from a sequence of rolls analytically. This program will only handle 12 rolls in about 15 minutes. The times increase geometrically as the number of rolls is increased. However, the results are supposed to be exact unless I made a mistake, which is always possible with new code.
Quote:
As stated in my post, I simulated millions of rolls to get the variance for “always don’t coming”. I suppose a Markov chain is possible but it would be enormous as there are many ways to build up to six points covered before a seven-out. What do you mean by “steady state statistics”
The Markov chain might not be as big as it would seem at first glance. The possible transitions are limited. There are 64 states for points established, but you can only transition to a few states from many of the states
By steady state, I mean when the statistics stabilize. If your MC simulation only runs for 50 rolls, your estimate of variance will not have converged. I am only doing 12 or 13 rolls analytically, so I know my estimates have not converged.
My program prints out the number of wins in the whole sequence, not the probabilities of a N wins at each roll. I can add that to the old and new programs to compare with your results.Quote:I posted the probabilities of losing 1 unit, breaking even and winning 1, 2, 3, 4 or 5 units. Let’s see your numbers. Where do they differ from mine specifically?
link to original post
I cannot look into this right now. I have a very heavy workload this weekend. I promise that I will get back to you once I have some time. If you are right and I am wrong, I will confess to my error. I appreciate having someone's detailed results to check against.
You can make 1000 DP bets concurrently at 1000 different online tables or grind out 1000 DP bets at one table. In both cases the 1000 bets are independent and the SD is 1000^,5 * 1. You must use DP bets for comparison purposes because, unlike DC bets, they are independent
My simulation is very simple and is done in Excel. One million rows and each row is a roll of the dice playing continuous DC. With a million rolls you see how many rolls had a net result of -1, 0, 1, 2, 3, 4 or 5 units. With that you can calculate the variance off the average +1.41% per roll. To my knowledge, that is the only way to calculate variance. If your program doesn’t give you the distribution of all possible results -1 to +5, how could you calculate variance?
Note: I’ve actually simulated more like 50 million rolls. Each time I press F9 excel recalculates using another 1 million random rolls. The SD is very stable at 1.15 for every recalc. The player advantage bounces between about 1.2% and 1.6%, but with 50 recalculations it’s definitely converging on 1.41% average. 1 million rows is about the most my computer can handle without freezing up
You can get the variance per spin by calculating the overall variance of a large number of spins and dividing. That is what I was attempting to do, but I could not calculate the exact variance for more than about 15 spins. Back to the drawing board. I will try a Markov approach and also do Monte Carlo. I will let you know how it works out when I get a round to it.
Thanks again for the interesting discussion.
seems as though you've become interested in this strictly as a mathematical inquiry. The chances that there are bubble craps machines out there that have this defect and not discovered yet seem pretty slim. And the excruciating study would be readily skipped I think.
This is not a criticism, please continue
For an even money bet with no pushes, if you know the EV, then you can derive the variance for a single play. We are trying to derive the variance for correlated outcomes. The above is no longer true. The correlated variance is only defined over an infinite number of rolls. That is because a bet can technically stay unresolved forever.Quote: odiousgambitInteresting how much effort you two are putting into calculating variance. There are simple formulas out there ... Yes, I guess I understand that you can form such directly from data, being an alternate method? ... my understanding is limited.
link to original post
And yes, I am doing this just because I am a nerd. I don't even like to play craps.
I think you might be calculating variance for the normal craps game and I am doing the calculation for the game where 12 is a win, not a push.Quote: Ace2As previously stated, the SD for a DP bet is 1 taken to four decimal places. If you go to five, then it’s 0.99990. It’s 1 for all practical purposes.
link to original post
For a even-money game with no pushes and an EV=+0.141414141%, the variance is 0.9998000, right? It doesn't matter what the game is.
Okay, I wrote a simple Monte Carlo craps player. It can simulate a trillion craps rounds in 34 seconds using only one processor on an old computer.Quote: Ace2I posted the variance and total bet action effects of come/dc betting a couple years ago but I cant find the posts. Having the 1.41% advantage on the don't has essentially no effect on these calculations, other than flipping the expected loss to a gain.
An individual don't bet has a standard deviation of 1 (taken to four decimal places) per resolved bet. For comparison purposes we must look at the SD on a per-roll basis, which is (165/557)^.5 =~ 0.544. If you always play a DC bet, your total bets made will increase by a factor of 557/165 compared to playing DP only. After simulating millions of rolls, the per-roll SD of always playing DC is 1.15. So, not surprisingly, its SD higher (by 15%) than increasing the number of DP bets by a factor of 557/165. This is because a seven-out can cause a win of 2-5 units depending on how many numbers are covered. All of the above is with no odds bets since there's no reason to make them with a positive EV for the line bet. Incidentally, always making DC bets increases the SD much more when odds bets are added.
So your per-roll expectation of always playing DC is +1.41% +/- 1.15. To be 99% confident of a profit (2.33 SDs), you need to play (1.15 * 2.33 / .0141)^2 = 36,113 rolls. Assuming you play 8 hours per day 5 days per week at 100 rolls per hour, you need to play for 2 months (!!) in order to have a 99% chance of a profit. This aligns with my initial estimate. 1% and change advantage on an even-money game is nothing and it will take a very long grind to get an assured profit. In those 45 days of playing, you can expect to have 16 losing days (36%) and 2 losing weeks (22%). I doubt the casino would even notice that you are a long-term winner
Some data if anyone is interested: If you always bet DC with no odds you will lose 1 unit on 36% of rolls, net zero / no action on 42.7%, win 1 on 13.9%, win 2 on 2.5%, win 3 on 2.2%, win 4 on 1.7% and win 5 on 1%
link to original post
After 10 trillion rolls, the statistics basically agree with your results:
-1 35.959122%
0 42.696328%
1 13.945229%
2 2.525554%
3 2.154069%
4 1.683634%
5 1.036064%
EV = 0.0141428 Var = 1.3221295
I found an error in my variance calculation in my previous program. Even though the results were far from converging, they would not have converged on the correct variance.
Not again!!!Quote: MentalI think you might be calculating variance for the normal craps game and I am doing the calculation for the game where 12 is a win, not a push.Quote: Ace2As previously stated, the SD for a DP bet is 1 taken to four decimal places. If you go to five, then it’s 0.99990. It’s 1 for all practical purposes.
link to original post
For a even-money game with no pushes and an EV=+0.141414141%, the variance is 0.9998000, right? It doesn't matter what the game is.
link to original post
Do you actually think I don't know how to calculate the variance of a single DP bet that wins on 12? Read my posts and you'll see that I'm almost always listing standard deviation because that's what's used for calculations. The square root of the variance of 0.9998 is the SD of 0.9999, which is what I’ve posted
I suggest you focus on the problem at hand instead of obsessing over the inconsequential difference between 0.9998, 0.9999 and 1. As mentioned, just use 1 and move on
Take the square root of 1.32 to get the SD of 1.15. Thanks for confirming my calculationQuote: MentalOkay, I wrote a simple Monte Carlo craps player. It can simulate a trillion craps rounds in 34 seconds using only one processor on an old computer.Quote: Ace2I posted the variance and total bet action effects of come/dc betting a couple years ago but I cant find the posts. Having the 1.41% advantage on the don't has essentially no effect on these calculations, other than flipping the expected loss to a gain.
An individual don't bet has a standard deviation of 1 (taken to four decimal places) per resolved bet. For comparison purposes we must look at the SD on a per-roll basis, which is (165/557)^.5 =~ 0.544. If you always play a DC bet, your total bets made will increase by a factor of 557/165 compared to playing DP only. After simulating millions of rolls, the per-roll SD of always playing DC is 1.15. So, not surprisingly, its SD higher (by 15%) than increasing the number of DP bets by a factor of 557/165. This is because a seven-out can cause a win of 2-5 units depending on how many numbers are covered. All of the above is with no odds bets since there's no reason to make them with a positive EV for the line bet. Incidentally, always making DC bets increases the SD much more when odds bets are added.
So your per-roll expectation of always playing DC is +1.41% +/- 1.15. To be 99% confident of a profit (2.33 SDs), you need to play (1.15 * 2.33 / .0141)^2 = 36,113 rolls. Assuming you play 8 hours per day 5 days per week at 100 rolls per hour, you need to play for 2 months (!!) in order to have a 99% chance of a profit. This aligns with my initial estimate. 1% and change advantage on an even-money game is nothing and it will take a very long grind to get an assured profit. In those 45 days of playing, you can expect to have 16 losing days (36%) and 2 losing weeks (22%). I doubt the casino would even notice that you are a long-term winner
Some data if anyone is interested: If you always bet DC with no odds you will lose 1 unit on 36% of rolls, net zero / no action on 42.7%, win 1 on 13.9%, win 2 on 2.5%, win 3 on 2.2%, win 4 on 1.7% and win 5 on 1%
link to original post
After 10 trillion rolls, the statistics basically agree with your results:This is for the game where rolling a 12 on the is a win, so will be slightly different than for the normal game where 12 is a push.-1 35.959122%
0 42.696328%
1 13.945229%
2 2.525554%
3 2.154069%
4 1.683634%
5 1.036064%
EV = 0.0141428 Var = 1.3221295
I found an error in my variance calculation in my previous program. Even though the results were far from converging, they would not have converged on the correct variance.
link to original post
So continuous DC increased the SD by 15% compared to the same number of independent DP bets. The increase would be more except that there will be plenty of seven-outs with one number covered, which is a net result of zero
Quote: Ace2Do you actually think I don't know how to calculate the variance of a single DP bet that wins on 12? Read my posts and you'll see that I'm almost always listing standard deviation because that's what's used for calculations. The square root of the variance of 0.9998 is the SD of 0.9999, which is what I’ve posted
I suggest you focus on the problem at hand instead of obsessing over the inconsequential difference between 0.9998, 0.9999 and 1. As mentioned, just use 1 and move on
link to original post
I apologize. I was focused on variance and did not read your post carefully.
Small errors usually mean errors in someone's calculation, so I take small errors seriously. In this case, I just misread your post.
Yes, I agree with all of your results now.
There is no simple formula for this scenario because the number of bets in play and the number being resolved varies. Correct, sometimes real-life scenarios generate interesting math problemsQuote: odiousgambitInteresting how much effort you two are putting into calculating variance. There are simple formulas out there ... Yes, I guess I understand that you can form such directly from data, being an alternate method? ... my understanding is limited
seems as though you've become interested in this strictly as a mathematical inquiry. The chances that there are bubble craps machines out there that have this defect and not discovered yet seem pretty slim. And the excruciating study would be readily skipped I think.
This is not a criticism, please continue
link to original post
I already calculated this for regular continuous Come/DC a couple years ago and posted it somewhere. Changing the 12 to win is a simple modification and has no material effect, it mostly just flips expected loss to a gain.
Nope.Quote: MentalFor an even money bet with no pushes, if you know the EV, then you can derive the variance for a single play. We are trying to derive the variance for correlated outcomes. The above is no longer true. The correlated variance is only defined over an infinite number of rolls. That is because a bet can technically stay unresolved forever.Quote: odiousgambitInteresting how much effort you two are putting into calculating variance. There are simple formulas out there ... Yes, I guess I understand that you can form such directly from data, being an alternate method? ... my understanding is limited.
link to original post
link to original post
The variance of any bet with a defined set of possible outcomes and associated probabilities can be directly and easily calculated. Does not have to even money and can have pushes.
Every multi-roll bet on the craps table could potentially go unresolved forever. That does not affect the variance calculation for a single bet, which is in terms of bet resolved irrespective of how long it takes to resolve.
Actually, the opposite is true for continuous DC. If you play 1000 rolls, you know you’ll have between 994 and 1000 bets resolved, which is pretty dang close to an average of one bet per roll. You’re adding a bet every roll and there can never be more than 6 unresolved bets at any time following a roll. Playing a single DP bet, however, you could technically go 1000 rolls without a resolution. Highly unlikely but mathematically possible. And irrelevant
Duh. Who said it needed to be even bets to calculate the variance. I said that if it is an even money game and you know the EV, then you can derive the probabilities and then the variance. You need not know anything about the game. In fact, if it is any game with only two outcomes and only one fixed payoff for winners and zero payoff for losers, this is true. As soon as you have three outcomes, you cannot derive the win probabilities simply from knowing the EV. You need to know the design of the game.Quote: Ace2Nope.Quote: MentalFor an even money bet with no pushes, if you know the EV, then you can derive the variance for a single play. We are trying to derive the variance for correlated outcomes. The above is no longer true. The correlated variance is only defined over an infinite number of rolls. That is because a bet can technically stay unresolved forever.Quote: odiousgambitInteresting how much effort you two are putting into calculating variance. There are simple formulas out there ... Yes, I guess I understand that you can form such directly from data, being an alternate method? ... my understanding is limited.
link to original post
link to original post
The variance of any bet with a defined set of possible outcomes and associated probabilities can be directly and easily calculated. Does not have to even money and can have pushes.
Quote:
Every multi-roll bet on the craps table could potentially go unresolved forever. That does not affect the variance calculation, which is in terms of bet resolved irrespective of how long it takes to resolve.
It certainly does affect the outcome of a sequence of correlated results. This is what you are missing in your simplistic treatment of the craps game.
Consider an even money game where P(win) and P(lose) are 0.5. The variance is 1.0 by your calculation. However, the game is designed such that every win is followed by a loss and vice versa. What is the overall variance of 100 consecutive bets on this game? Hint, it is not 100 * 1.0. The variance and SD are zero.
Your variance calculation for craps is not getting us any closer to knowing the SD of a large number of consecutive rolls where Don't bets are made on each roll. You are calculating the variance the number of chips in your stack after a single roll. You are not calculating multi-roll variance for a correlated stream of outcomes.
Quote:
Actually, the opposite is true for continuous DC. If you play 1000 rolls, you know you’ll have between 994 and 1000 bets resolved, which is pretty dang close to an average of one bet per roll. You’re adding a bet every roll and there can never be more than 6 unresolved bets at any time following a roll. Playing a single DP bet, however, you could technically go 1000 rolls without a resolution. Highly unlikely but mathematically possible
link to original post
The opposite of what is true?
Here, you seem to be talking about start/end effects. A series of rolls starts in one defined state with no bets riding and ends in another state with some unresolved bets. If N is large (I can quickly do 10 trillion rolls), these start/end effects are completely unimportant. Monte Carlo randomness is much bigger factor than how you treat start/end effects.
What you are totally ignoring is the correlation effect. This can be significant for certain games where consecutive games are not independent of each other.
Like craps don't bets. The variance per roll is something like 0.953 bets^2. You variance number of 1.3222 is not the variance of the game and will not predict the SD of a long series of rolls
I mostly agree with OG. Let us say you see 1000 rolls a day making a new Don't bet on every roll. You will only expect to win on 69% of all days. However, after 4 days your distribution is only half as wide. Now, you will win on 82% of days played.Quote: odiousgambita 1.4% edge and you can, say, bet $25 max, and can follow up a DP bet with a DC bet whenever available, I'm thinking you can get in 150 bets an hour or so. $3750*1.4% is $52.50 per hour with low variance; 8 hrs over $400Quote: Ace2Play the DP line eight hours per day everyday. After three months you have a reasonable chance of a profit. But nowhere even close to a guaranteed profit. Some days/weeks you’ll lose a ton of money. That’s what 1.4% advantage gets you
link to original post
I don't think you'd have a day where you lost money if you played 8 hours. I think you are undervaluing how low the variance is on the DP
link to original post
You have a 92.6% chance of being ahead after 10,000 rolls. You have a 99% chance of being ahead after 26,000 rolls. It is fair to say you are almost assured of a profit after three weeks. After three months and 90,000 rolls, you have essentially zero chance of being behind. Heck, your are a 99.8% favorite after just 40K rolls. Saying "After three months you have a reasonable chance of a profit." is a huge understatement.
Axel is not taking anti-correlation of the DP/DC bets into account and he is overestimating the variance. The variance of the cumulative results of N rolls is only (0.953 * N). This is somewhat less than the variance of N independent bets taken to resolution before a new Don't bet is placed.
So let's get this straight. YOU are going to lecture ME on statistical concepts and calculation methods?!?!!Quote: MentalDuh. Who said it needed to be even bets to calculate the variance. I said that if it is an even money game and you know the EV, then you can derive the probabilities and then the variance. You need not know anything about the game. In fact, if it is any game with only two outcomes and only one fixed payoff for winners and zero payoff for losers, this is true. As soon as you have three outcomes, you cannot derive the win probabilities simply from knowing the EV. You need to know the design of the game.Quote: Ace2Nope.Quote: MentalFor an even money bet with no pushes, if you know the EV, then you can derive the variance for a single play. We are trying to derive the variance for correlated outcomes. The above is no longer true. The correlated variance is only defined over an infinite number of rolls. That is because a bet can technically stay unresolved forever.Quote: odiousgambitInteresting how much effort you two are putting into calculating variance. There are simple formulas out there ... Yes, I guess I understand that you can form such directly from data, being an alternate method? ... my understanding is limited.
link to original post
link to original post
The variance of any bet with a defined set of possible outcomes and associated probabilities can be directly and easily calculated. Does not have to even money and can have pushes.Quote:
Every multi-roll bet on the craps table could potentially go unresolved forever. That does not affect the variance calculation, which is in terms of bet resolved irrespective of how long it takes to resolve.
It certainly does affect the outcome of a sequence of correlated results. This is what you are missing in your simplistic treatment of the craps game.
Consider an even money game where P(win) and P(lose) are 0.5. The variance is 1.0 by your calculation. However, the game is designed such that every win is followed by a loss and vice versa. What is the overall variance of 100 consecutive bets on this game? Hint, it is not 100 * 1.0. The variance and SD are zero.
Your variance calculation for craps is not getting us any closer to knowing the SD of a large number of consecutive rolls where Don't bets are made on each roll. You are calculating the variance the number of chips in your stack after a single roll. You are not calculating multi-roll variance for a correlated stream of outcomes.Quote:
Actually, the opposite is true for continuous DC. If you play 1000 rolls, you know you’ll have between 994 and 1000 bets resolved, which is pretty dang close to an average of one bet per roll. You’re adding a bet every roll and there can never be more than 6 unresolved bets at any time following a roll. Playing a single DP bet, however, you could technically go 1000 rolls without a resolution. Highly unlikely but mathematically possible
link to original post
The opposite of what is true?
Here, you seem to be talking about start/end effects. A series of rolls starts in one defined state with no bets riding and ends in another state with some unresolved bets. If N is large (I can quickly do 10 trillion rolls), these start/end effects are completely unimportant. Monte Carlo randomness is much bigger factor than how you treat start/end effects.
What you are totally ignoring is the correlation effect. This can be significant for certain games where consecutive games are not independent of each other.
Like craps don't bets. The variance per roll is something like 0.953 bets^2. You variance number of 1.3222 is not the variance of the game and will not predict the SD of a long series of rolls
link to original post
You could not get the correct answer to the continuous DC variance problem until after I posted it, at which point you copied my very straightforward method to get it. You're welcome
You bumbled around with convoluted programs for days, always looking at the wrong thing, always getting further from the correct answer. At one point you posted that your results confirmed that the variance of continuous (correlated) DC betting was actually lower than the same amount of independent bets, which totally contradicts common sense and elementary statistics. I doubt you could even quantify what your "results" actually represented, but I guarantee you used powerful computing for all of your garbage in, garbage out results.
I still can't believe you used a program to analyze trillions of sequences to see that playing continuous DC increases total bets resolved per roll (compared to DP only) by a factor <approaching> 3.327. If you had just thought about this problem for a second, you'd realize it can be calculated exactly on the back of an envelope in about two minutes. 1 + 1/6 * 4 + 2/9 * 3.6 + 5/18 * 36/11 = 557/165. Apparently you thought this figure was somehow relevant to the continuous DC variance calculation, and you were wrong again. This is a perfect example of your blinders-on, sledgehammer-on-a-thumbtack approach
You like mentioning your programming ability to look at trillions of sequences in seconds, but all the computing power in the world is worthless if you don't understand what you're trying to calculate in the first place. Besides, there are very few scenarios for which you need that kind of computing power. I've simulated many complex problems on this forum, and I can't remember ever needing more than one million trials to confidently simulate/confirm an answer, all done in simple excel spreadsheets. For this problem, the SD is totally stable at 1.15 after a million rolls...I almost got another digit as every recalculation I did was between 1.149 and 1.151. So why bother to run your trillions of rolls then? If you have a grasp of statistics and materiality, you know that a SD of 1.15 (3 significant digits) is perfectly adequate for just about any calculation
Anyway, thanks again for the lesson lol
No, I am not going to lecture you or rant like you did with insults.Quote: Ace2So let's get this straight. YOU are going to lecture ME on statistical concepts and calculation methods?!?!!
link to original post
Your calculations of the variance are flat out wrong.
I did not calculate the number of wins or losses per roll because those numbers will not get you closer to calculating the variance of a dependent string of results. Your variance formula only works for independent trials. I only calculated your per-roll W/L percentage numbers to verify my calculations.
Your estimates of how long it will take to get to 99% winning sessions are far off from the actual results. You are a smart guy, but if you plug the right numbers into the wrong formula by habit, you won't get the right answer.
Once again, try reading what I actually wrote. As clearly stated, that calculation is for playing DP line only. I made that post on the day this thread opened before we were even discussing continuous DC.Quote: MentalI mostly agree with OG. Let us say you see 1000 rolls a day making a new Don't bet on every roll. You will only expect to win on 69% of all days. However, after 4 days your distribution is only half as wide. Now, you will win on 82% of days played.Quote: odiousgambita 1.4% edge and you can, say, bet $25 max, and can follow up a DP bet with a DC bet whenever available, I'm thinking you can get in 150 bets an hour or so. $3750*1.4% is $52.50 per hour with low variance; 8 hrs over $400Quote: Ace2Play the DP line eight hours per day everyday. After three months you have a reasonable chance of a profit. But nowhere even close to a guaranteed profit. Some days/weeks you’ll lose a ton of money. That’s what 1.4% advantage gets you
link to original post
I don't think you'd have a day where you lost money if you played 8 hours. I think you are undervaluing how low the variance is on the DP
link to original post
You have a 92.6% chance of being ahead after 10,000 rolls. You have a 99% chance of being ahead after 26,000 rolls. It is fair to say you are almost assured of a profit after three weeks. After three months and 90,000 rolls, you have essentially zero chance of being behind. Heck, your are a 99.8% favorite after just 40K rolls. Saying "After three months you have a reasonable chance of a profit." is a huge understatement.
So let's assume 12 weeks of play at 40 hours per week, 100 rolls per hour. 48000 rolls * 165/557 = 14,219 DP decisions. There will be some immaterial variance in actual rolls and actual decisions over that period that we don't need to consider.
So the expectation for that three-month period is 14219 * .0141 +/- 14219^.5 * 1 = +200 units +/- 119. Therefore anything left of 1.68 SDs is a loss, giving you a 95% chance of profit and 5% chance of loss. I stand by my statement that it's a reasonable chance of profit, but with approximately a 1 in 20 chance of a loss, it's by no means a guaranteed profit. If you've ever analyzed the variance of an even-money game with a smallish edge, this result coincides with general expectations...it takes a long time for actual results to conform with expectations with any decent confidence level
You need to work on this as you continue posting responses to things I didn't write, even though my posts are very straightforward, as these calculations are not rocket scienceQuote: MentalI apologize. I was focused on variance and did not read your post carefully.
What does that variance figure represent? Please enlighten us and please don't overcomplicate your response. This ain’t “rocket surgery”Quote: MentalYou variance number of 1.3222 is not the variance of the game and will not predict the SD of a long series of rolls
link to original post
Carry on.
What’s the current money line on Bovada?!Quote: SOOPOOI am organizing a cagefight between Mental and Ace. I don’t know either of them. But I’m putting Ace as the favorite.
Carry on.
link to original post
You got that number by counting the average number of wins and losses on a roll of a stream of continuous DC/DP bets. I have no idea why you think this can be used to estimate variance for correlated outcomes. I know which variance formula you plugged these numbers into because I can reproduce your calculated variance, 1.32, using sum(Pi * delta^2) where delta is (the number -1 through 5 representing units resolved on a given roll minus the EV).Quote: Ace2What does that variance figure represent? Please enlighten us and please don't overcomplicate your response. This ain’t “rocket surgery”Quote: MentalYou variance number of 1.3222 is not the variance of the game and will not predict the SD of a long series of rolls
link to original post
link to original post
Clearly, the win/loss numbers are correlated. There is a 1.036% chance that you will win 5 units on a given roll. Given that you won 5 units on a certain roll, there is a zero percent chance that you won 5 units on the previous 5 rolls or that you will win 5 units on the next five rolls. That is negative correlation. It reduces the variance. If you don't account for correlation you will never be able to use your 'variance' to calculate the SD for a large number of rolls.
Positive correlation increases variance. Negative correlation decreases variance.
If a fair even money game is 100% negatively correlated between game N and game N+1, then the results will alternate between a win and a loss. Any even number of games will net the player exactly zero. The variance of a constant is zero.
If you play one of these games picked at random, your variance is 1.0 because correlation does not matter. If you play any even number of consecutive games, then correlation reduces your variance to zero.
I meant to quote your response to odiousgambit from Day #1 of this thread. In your response, you are already talking about making new Don't bets on every roll. This shortens the time frame to be virtually assured of a win. You acknowledge DC/DP correlation, but you just flatly state that per-bet variance increases without proof. A player will generally only win six Don't Come points in one roll after a very long drought with no sevens. This is negative correlation. My intuition told me the variance would decrease. My analytic calculations told me variance decreases. My long Monte Carlo simulations confirmed it. You are claiming that SD jumps to 1.15. I believe SD decreases to around 0.975. One of us is wrong.Quote: Ace2No idea what you’re talking about.Quote: odiousgambita 1.4% edge and you can, say, bet $25 max, and can follow up a DP bet with a DC bet whenever available, I'm thinking you can get in 150 bets an hour or so. $3750*1.4% is $52.50 per hour with low variance; 8 hrs over $400Quote: Ace2Play the DP line eight hours per day everyday. After three months you have a reasonable chance of a profit. But nowhere even close to a guaranteed profit. Some days/weeks you’ll lose a ton of money. That’s what 1.4% advantage gets you
link to original post
I don't think you'd have a day where you lost money if you played 8 hours. I think you are undervaluing how low the variance is on the DP
link to original post
The variance for DP is very close to 1 just like the pass line. Both sides are essentially a coin flip
If you play always coming then you’ll get about 100 bets resolved per hour assuming 30 line decisions per hour. You will have lots of losing days and even losing weeks. The math is easy
That’s before considering that your per-bet variance increases with DC bets because their results are correlated. They all win when a seven is rolled
link to original post
Just do the simulations to estimate the SD of a long series of rolls. Until you actually do this, there is not much to discuss. You think your variance/SD equation is correct. I think it is not applicable for reasons I have stated in two previous replies.
You state that the math is easy, but the math of handling correlations properly when you don't even have the correlation function is anything but easy.
Quote: odiousgambitit goes from approx 1.4% HE to approx 1.4% Player's edge. You'd go from using free odds bets to not [you don't want extra variance], actually eliminating all other bets than DP/DC . You'd play like there's no tomorrow , because probably there isn't one
somebody started an earlier thread on the same subject and realized, I think, he was being fooled by the return of the bet. I'd only guess that in fact they are wrong about this one because players would figure it out so quickly and you'd have " all the seats were taken" as part of this story
link to original post
I would also expect that the casino would figure it out pretty quickly and fix or shut down the machines. So, in the unlikely event that this is real, play it while you can. It won't last.
But I would bet a significant amount of money that it's not real. I once saw a player complain because he made the opposite mistake. Bubble Craps machines (at least the Interblock ones that I play), take down a DC bet when a 12 rolls (they also take down winning DC bets on 2 or 3 and wining come bets on 11). This player thought he was losing on the 12.
