Lifetime EV and Standard Deviations
I wouldn't just like an answer, but if you could walk me through it I'd appreciate it. For what it's worth, I have a math-based college degree but I know very little statistics jargon. Like I know what a standard deviation is, but without a cheat sheet I couldn't calculate it for you.
Let's make the following assumptions:
- You only play roulette at a HA of 5.26% twice a month for three hours each session. You bet $20 every spin at 50 spins per hour (for simplicity's sake you do even money bets, but over the course of so many outcomes the type of betting becomes unimportant, right?). You keep this betting pattern unchanged for 20 years.
I can do something as simple as calculated the EV and have access to programs like mathematica/matlab/excel, I'm inclined to believe that you might have to just do this the hard way with a million simulations. If by chance you can do it just with a pen and paper, I'm -very- interested in that.
Quote: ahiromuIn fact it was EvenBob's name that led me to wonder this question: Over a lifetime of gambling, what are the odds that I will have a positive outcome?
I wouldn't just like an answer, but if you could walk me through it I'd appreciate it. For what it's worth, I have a math-based college degree but I know very little statistics jargon. Like I know what a standard deviation is, but without a cheat sheet I couldn't calculate it for you.
Let's make the following assumptions:
- You only play roulette at a HA of 5.26% twice a month for three hours each session. You bet $20 every spin at 50 spins per hour (for simplicity's sake you do even money bets, but over the course of so many outcomes the type of betting becomes unimportant, right?). You keep this betting pattern unchanged for 20 years.
I can do something as simple as calculated the EV and have access to programs like mathematica/matlab/excel, I'm inclined to believe that you might have to just do this the hard way with a million simulations. If by chance you can do it just with a pen and paper, I'm -very- interested in that.
The formula to use would be: Std. deviation= 2b sqrt(npq ), where b = flat bet per round, n = number of rounds, p = 18/38, and q = 20/38. (p and q are the probabilities of winning and losing, respectively, an even-money roulette bet.)
Number of rounds is 300/mo, 3600/yr, total 72,000.
Plugging in, you have 2($20) ((sqrt(72,000 x 18/38 x 20/38)). This nicely rounds to (40) ((sqrt (18,000)), or 40 x 134, or $5360. This is the standard deviation from your expected loss.
Your expected loss over 72,000 trials is (-.0526) (20) (72,000), as in: (expected value x bet size x number of trials). This equals 1.052 x 72,000. Your expected loss is therefore $75,744. This reflects the fact that your expected loss from each $20 bet is slightly more than $1.
The six-sigma range is the area under the bell curve that represents three standard deviations in each direction from the mean. The mean in this case is -$75,744, and the standard deviation is $536. This means that the six-sigma range is from (-75,744+16,080) to (-75,744-16,080) ,or -91,824 to -59,664. There is a 99.8% probability that your results will fall in this range. And of course, there is a vanishingly small probability--for all intents and purposes, zero--that you will be ahead at this time.
The variance here is surprisingly low, but roulette is a very low-variance casino game. The high expected loss per bet is what kills you.
edited
Quote: odiousgambitI find this an interesting question as well, but I don't like your scenario. To me 5.26% is a foregone conclusion towards ruin. It's when the HE approaches zero, that's realisticially accomplishable. Within a standard deviation it becomes possible to have a lifetime ahead of the game, it seems. So at what point does the HE start to make that possible?
edited
That would be a function not only of the house edge, but of variance. For example, the lottery has a 40% (or worse) house edge, but SOME people are ahead for their lifetimes playing it; this is because its variance is huge.
Even a house edge of, say 0.5 percent would be impossible to overcome in the long run if the variance were relatively low. So the answer to your question is, either high variance, or a low HE, or both, would start to make a lifetime win possible. Roulette pretty much guarantees a lifetime loser over even a relatively short number of outcomes because the HE is high AND the variance is low.
Another thing: Standard Deviation % - Sorry it's late but just wanted to add this. Basically it can be used to relate how many standard deviations are needed to cover a certain amount.
Quote: ahiromu*You dropped a zero on your standard deviation - although the outcome is the same. 40*134=5360
Thank you for such a detailed explanation, exactly what I was looking for. I have an additional question, how would you augment the standard deviation equation for a bet that pays more than 1:1? In addition to odius's question about what HA leads to a reasonable chance at a positive outcome I'm also curious to see how the payout changes the standard deviation - I mean it's obvious it would increase it but the more outcomes the less it will affect it.
Another thing: Standard Deviation % - Sorry it's late but just wanted to add this. Basically it can be used to relate how many standard deviations are needed to cover a certain amount.
I edited my post to fix the error. The range of sig-sigma results is much broader now, and is more in line with what I intuitively expected.
You would modify the equation so that p was (p)(payoff), and q was (q)(payoff), for the two respective outcomes. In the roulette example, the amounts for (payoff) were 1 and 1, respectively.
Quote: mkl654321The formula to use would be: Std. deviation= 2b sqrt(npq ), where b = flat bet per round, n = number of rounds, p = 18/38, and q = 20/38. (p and q are the probabilities of winning and losing, respectively, an even-money roulette bet.)
Why are you multiplying by 2b instead of just b?
Quote: matildaWhy are you multiplying by 2b instead of just b?
From:
http://en.wikipedia.org/wiki/Gaming_mathematics#Standard_deviation
the binomial distribution, SD = sqrt (npq ),The binomial distribution assumes a result of 1 unit for a win, and 0 units for a loss, rather than -1 units for a loss, which doubles the range of possible outcomes. Furthermore, if we flat bet at 10 units per round instead of 1 unit, the range of possible outcomes increases 10 fold. Therefore,
SD (Roulette, even-money bet) = 2b sqrt(npq ), where b = flat bet per round, n = number of rounds, p = 18/38, and q = 20/38
FYI, The Wizard is mentioned at the end of the paragraph!
Enjoy
Quote: ahiromuIn fact it was EvenBob's name that led me to wonder this question: Over a lifetime of gambling, what are the odds that I will have a positive outcome?
I can do something as simple as calculated the EV and have access to programs like mathematica/matlab/excel, I'm inclined to believe that you might have to just do this the hard way with a million simulations. If by chance you can do it just with a pen and paper, I'm -very- interested in that.
Since you can use Excel and want to know the odds, there is a much easier and faster way to calculate the odds than
the longggggg way using the normal distribution/standard deviation formulas which are just a close "approximation" of the binomial distribution function anyways.
As for your example, you would need 35,000 wins from 70,000 trials to break even with an even money payout.
35,001 wins gives you a profit.
The formula to use would be, for Excel,
= 1- BINOMDIST(35000,70000,18/38,TRUE)
The binomdist formula is for "or less" so we subtract the result from 1 to get "or more".
Now the resulting percentage is greater than 1 X10^(-32) in my Excel or 1 in >100,000,000,000,000,000,000,000,000,000,000
Let us use 10,000 trials for an example.
1- BINOMDIST(5000,10000,18/38,TRUE) = 0.000000065678662
or 1 in 15,225,645
Odds then would be 15,224,644 to 1 of showing a profit.
to compare another bet.
The pass line in Craps.
p = 244/495 (49.2929%)
in Excel:
1- BINOMDIST(35000,70000,244/495,TRUE) = 0.0000900365594
or 1 in 11,107
or 10 on 111,070
or 100 in 1,110,700
or 1,000 on 11,107,000
Quote: guido111
Let us use 10,000 trials for an example.
1- BINOMDIST(5000,10000,18/38,TRUE) = 0.000000065678662
or 1 in 15,225,645
Odds then would be 15,224,644 to 1 of showing a profit.
Thank you for doing this. Shocking. So you have basically NO chance of coming out ahead in Roulette over a lifetime of play.
Better, but still not great.Quote:to compare another bet.
The pass line in Craps.
p = 244/495 (49.2929%)
in Excel:
1- BINOMDIST(35000,70000,244/495,TRUE) = 0.0000900365594
or 1 in 11,107
or 10 on 111,070
or 100 in 1,110,700
or 1,000 on 11,107,000
Quote: teddysThank you for doing this. Shocking. So you have basically NO chance of coming out ahead in Roulette over a lifetime of play.
Better, but still not great.
Yes, For American or Double Zero Roulette.
Still using ahiromu example: You bet $20 every spin at 50 spins per hour (for simplicity's sake you do even money bets, but over the course of so many outcomes the type of betting becomes unimportant, right?). You keep this betting pattern unchanged for 20 years...
(this assumes that you make the same $ bet all the time, it does not matter how much you bet)
The probabilities of showing a profit from:
Single Zero Roulette...
Let us use 70,000 trials for an example.
1- BINOMDIST(35000,70000,18/37,TRUE) = 0.00000000000041122661 or 1 in 2,431,749,258,839 of showing a profit.
Let us use 30,000 trials for an example.
1- BINOMDIST(15000,30000,18/37,TRUE) = 0.0000013802 or 1 in 724,530 of showing a profit.
Let us use 10,000 trials for an example.
1- BINOMDIST(5000,10000,18/37,TRUE) = 0.003331 or 1 in 300 of showing a profit.
Quote: guido111Yes, For American or Double Zero Roulette.
or 1 in 2,431,749,258,839 of showing a profit.
And we are so fortunate to know that one- mrjjj, who has retired to his half restaurant...
And for simplicity sake, all bets are the same.
For other games and other bets with more than 2 outcomes you will fail miserably in determining actual results. Blackjack for example can not be calculated using the binomial formulas since there can be pushes, double downs, splits, blackjack payouts higher than even money, etc.
Ah, thank goodness for computers.
Quote: teddysThank you for doing this. Shocking. So you have basically NO chance of coming out ahead in Roulette over a lifetime of play.
The way most people play, they have no chance of coming out ahead after one session of play. The don't understand the nature of random outcomes and don't quit when they're ahead, but expect to keep winning indefinitely.
Quote: guido111What is being done here in this thread, started by member "ahiromu", is a binomial distribution and binomial standard deviation. Both assume that there are only 2 outcomes. Win and lose.
And for simplicity sake, all bets are the same.
For other games and other bets with more than 2 outcomes you will fail miserably in determining actual results. Blackjack for example can not be calculated using the binomial formulas since there can be pushes, double downs, splits, blackjack payouts higher than even money, etc.
Ah, thank goodness for computers.
Sure, but the binomial distribution can serve as a good illustration for those more complex sets of outcomes. For instance, playing JOB 9/6 poker for 1,000,000 hands will leave you with virtually no chance of being ahead at the end, even though the HE is only -0.46%, and the outcomes vary widely (from a push to an 800-1 payout). You could construct a hypothetical game with the same variance as JOB (which I think is 19.6), but only had even money payouts, and after that million trials, only a very small number of outliers would be on the positive side.
The whole lesson to take away from this is that variance is king in the short run, but over a large number of trials, HE/EV crushes variance. The disease that infects gamblers' minds is that the short run is significant. Casinos take advantage of this fallacy.
How would this be possible? Wouldn't any even money game automatically have a variance of 1?Quote: mkl654321You could construct a hypothetical game with the same variance as JOB (which I think is 19.6), but only had even money payouts, and after that million trials, only a very small number of outliers would be on the positive side..
Quote: mkl654321The disease that infects gamblers' minds is that the short run is significant.
But it is significant, its where the bet is placed. You just have to realize where you are in the scheme of things and not get carried away.
Quote: teddysHow would this be possible? Wouldn't any even money game automatically have a variance of 1?
Not unless the outcomes were equally likely.
Quote: teddysHow would this be possible? Wouldn't any even money game automatically have a variance of 1?
I know what you're saying and I think there's a variance of 1 (let's say an even money bet on a roulette wheel without a 0) after only a single trial. You have to take into account length of play as well as spins per hour.
Quote: teddysHow would this be possible? Wouldn't any even money game automatically have a variance of 1?
Imagine a game where I pay you 35 dollars if you roll box cars on two dice, and you pay me a dollar if you roll any other number. EV = 1. Variance != 1.
Quote: ahiromuSo I cut a lot of corners, but for a lifetime of 20x craps & 25 outcomes/hour for 20 years I got an EV of -$378 with a SD of ~$20000. Again, this is for a recreational gambler that plays for six hours every month. This is very promising in that just less than half of people out there will end up positive. Also, a good 15% of people will end up very high over a lifetime... on the other hand the same amount will be 20k in the hole. Thanks guys something like this is exactly what I was looking for... also it'll keep me away from the roulette tables (it's my guilty pleasure).
I'm trying to derive your figures, since you didn't "show your work". >:-)
25 * 6 * 12 * 20 = 36,000 outcomes
If this is based on $5 pass, 20X odds, the ev would -.0707/outcome, which would be over $2500 expected loss. Something is very wrong here! You appear to have cut a lot more than corners. An ev of -$378 represents more like 5300-5400 bets.
I ran my C program to figure this stuff for 36,000 $5 pass bets, taking 20X odds. The results:
ev: -$2545
SD: $19,733
The ev is 12.9% of the SD, so there is still roughly a 44% probability of breaking even or better. At +2 SD, a player would be ahead almost $37,000, with a probability of .023. Of course, the same number of players would be expected to be behind by about $42,000!
Each bet has an ev of -$.0707, with a standard deviation of $104. At this level of variance, it continues to swamp ev for a long, long time. Since the ev increases with the number of bets and the SD with the square root thereof, we can figure when the ev is equal in magnitude to one SD:
.0707 * x = 104 * x^.5
Divide both sides by square root of x
.0707 * x^.5 = 104
Divide both sides by .0707
x^.5 = 1471
Square both sides
x = 2,163,841
So, even after over two million bets, it would only take one standard deviation's worth of "good luck" to match the negative ev and break even.
BTW, here's how I derived (the program did) the SD for this bet:
outcome - mean diff squared ways gross variance
5 - -.0707 = 5.0707 25.712 440 11313.28
-5 - -.0707 = 4.9293 24.298 220 5345.56
-105 - -.0707 = 105.0707 11039.851 784 8655243.1
125 - -.0707 = 125.0707 15642.679 250 3910669.7
155 - -.0707 = 155.0707 24046.921 176 4232258.0
205 - -.0707 = 205.0707 42053.991 110 4625939.0
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21440768 / 1980 = 10828.67
Variance is 10828.67, standard deviation is square root = 104.0609.
Variance is weighted sum of squared differences from mean. I weighted the outcomes using the "perfect 1980".
Of course, this is an extreme example of variance. For just two hours' play at 25 resolutions/hour, the SD is over $700 creating a 2+% risk of losing over $1400, and not many players are willing to accept that risk.
Cheers,
Alan Shank
Woodland, CA
