How long does it take to get even?
December 21st, 2012 at 2:48:12 PM
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Consider my question as a math question.
My question is, if you played on a roulette wheel that has no zero's (thats a wheel containing 36 numbers with an equal amount of red and black number on it) and you played flat bets of $1 per unit per spin, and while you were playing you were 50 units down (-50 units) from 200 spins.
How long will it take for you to get back to breaking even? bare in mind you have 50/50 chance of winning each time?.
In my experience, with a similar trail but of coin tossing instead (tagging the faces of a coin one side red and one side black). I have found it can take up to 1500 coin flips to get back to breaking even from -50 units down. Is there any way to lower the amount of spins it will take on average to get back to breaking even from being -50 units down?
But what about if you were betting 24 numbers (two dozens) on the wheel instead? does betting more numbers mean breaking even will take less spins on average since you have more chance of winning therefore will cause less fluctuation?.
My question is, if you played on a roulette wheel that has no zero's (thats a wheel containing 36 numbers with an equal amount of red and black number on it) and you played flat bets of $1 per unit per spin, and while you were playing you were 50 units down (-50 units) from 200 spins.
How long will it take for you to get back to breaking even? bare in mind you have 50/50 chance of winning each time?.
In my experience, with a similar trail but of coin tossing instead (tagging the faces of a coin one side red and one side black). I have found it can take up to 1500 coin flips to get back to breaking even from -50 units down. Is there any way to lower the amount of spins it will take on average to get back to breaking even from being -50 units down?
But what about if you were betting 24 numbers (two dozens) on the wheel instead? does betting more numbers mean breaking even will take less spins on average since you have more chance of winning therefore will cause less fluctuation?.
December 21st, 2012 at 4:22:17 PM
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You never expect to be even again, that is the gambler's fallacy. However, the probability that the percentage of rolls is any small distance away from fifty percent converges to zero as the number of trials gets big. That's the law of large numbers. Most people don't understand the distinction. The rolls are serially uncorrelated. The reason you converge to 50 percent is because the sample is getting large.
We could ask two questions
Is the probability of being within 5 rolls of your expected value higher or lower as the sample size grows? The answer is lower.
Is the probability of being within 5% of your expected frequency higher or lower as the sample size grows? The answer is higher.
Anyway another question may be how many rolls eventually get to the 50/50 exact point. Not really relevant to any practical gambling question.
We could ask two questions
Is the probability of being within 5 rolls of your expected value higher or lower as the sample size grows? The answer is lower.
Is the probability of being within 5% of your expected frequency higher or lower as the sample size grows? The answer is higher.
Anyway another question may be how many rolls eventually get to the 50/50 exact point. Not really relevant to any practical gambling question.
December 21st, 2012 at 4:23:38 PM
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Last edited by: sodawater on Oct 1, 2018
December 21st, 2012 at 4:42:21 PM
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Its the old Inertial Guidance System example. A computer makes zillions of minute adjustments to an airplane's heading. Each adjustment is a correction of an existing infinitesimal error. The airplane gets to the correct destination. The system is goal oriented.
There is no goal orientation in one particularly player "getting back to even". There is no magical adjustment based on prior spins.
A more meaningful question might be who is operating that casino and how are they paying for the free drinks. If the casino is not getting its beak wet at each spin, why should it provide the wheel?
There is no goal orientation in one particularly player "getting back to even". There is no magical adjustment based on prior spins.
A more meaningful question might be who is operating that casino and how are they paying for the free drinks. If the casino is not getting its beak wet at each spin, why should it provide the wheel?
December 21st, 2012 at 4:49:38 PM
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It absolutely is not the case that you trend towards an 'even' outcome in a zero edge game. The longer you play the wider the possible range gets. However, as has been explained, the expected percentage gets into line.
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
December 21st, 2012 at 5:14:31 PM
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1. You can calculate the probability to get even within any given number of flips. If I did the math correctly, your chance to get even within 1500 flips is around 30% (only!), if you double that (3000 flips) you can get to 50%. (can somebody confirm that?)
2. The probability to get even within any given amount of flips will go down if you bet one unit total on two dozen for a higher win rate, but a lower win amount. This would reduce the variance, which would work against you when trying to get even. Betting one unit on a single number only would dramatically increase the variance and thus your probability to get even within a lower number of flips.
3. If your bankroll is infinte, you will eventually get even.
2. The probability to get even within any given amount of flips will go down if you bet one unit total on two dozen for a higher win rate, but a lower win amount. This would reduce the variance, which would work against you when trying to get even. Betting one unit on a single number only would dramatically increase the variance and thus your probability to get even within a lower number of flips.
3. If your bankroll is infinte, you will eventually get even.
December 21st, 2012 at 5:43:56 PM
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DUHHIIIIIIIII HEARD THAT!
December 21st, 2012 at 5:47:36 PM
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Quote: Canyonero3. If your bankroll is infinte, you will eventually get even.
This is true, but the expectation of the time that it takes you to get even diverges. That is, it's infinite.
Having said that, expectation is possibly not a good thing to use in this case. It might be better to ask about the "median" amount of time it takes to get even, rather than the average, which you've answered above (although, I haven't checked your answer)
December 21st, 2012 at 5:56:25 PM
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It's been years, but I believe the following is a theorem.
For a 1 dimensional random walk a(n) starting with a(0) = 0, with
p(a(n+1)= a(n) + 1) = 0.5
p(a(n+1) = a(n) - 1) = 0.5,
for any M > 0, there exist infinitely many x, y such that a(x) > M and a(y) < -M.
I believe this is the formalization of the "Gambler's Ruin" theorem.
For a 1 dimensional random walk a(n) starting with a(0) = 0, with
p(a(n+1)= a(n) + 1) = 0.5
p(a(n+1) = a(n) - 1) = 0.5,
for any M > 0, there exist infinitely many x, y such that a(x) > M and a(y) < -M.
I believe this is the formalization of the "Gambler's Ruin" theorem.
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December 21st, 2012 at 6:48:20 PM
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Quote: Canyonero1. You can calculate the probability to get even within any given number of flips. If I did the math correctly, your chance to get even within 1500 flips is around 30% (only!), if you double that (3000 flips) you can get to 50%. (can somebody confirm that?)
2. The probability to get even within any given amount of flips will go down if you bet one unit total on two dozen for a higher win rate, but a lower win amount. This would reduce the variance, which would work against you when trying to get even. Betting one unit on a single number only would dramatically increase the variance and thus your probability to get even within a lower number of flips.
3. If your bankroll is infinte, you will eventually get even.
I think I read somewhere, to confirm if a wheel is bias or not (certain numbers being bias on the wheel) it requires around 3- 5 thousands of spins of data (correct me if iam wrong). So Iam wondering if it would require the same amount of spins for me to break even from -50 units down on the red and black game.
Im also very curious as to the most units anyone has ever been down in an experiment like mine, I've always wondered what the world record could be, can anyone be in the red as much as -500 units down playing a fair game? of course that would be very rare (may be equivalent to seeing 20 reds in a row) if it can happen, perhaps even at -250 units in the red would be very rare iam assuming, if those are rare, what would be the average units in red (if there's is such a figure) can one be down before a recovery will start to happen?
When I mentioned betting 24 numbers instead of a red and black bet. I didn't mean to bet 24 numbers to recover the -50 units I'd lost in the game of red and black. I meant it as a completely independent trial.
My initial question was, simplified,
If I had two independent trails on a table with no zeros:-
1) Playing a red and black game 50/50
2) Playing 24 numbers (betting two dozens each spin) 2/1
If in both trials I ended up losing the same amount of games, will it take me longer on average to break even playing the red and black game because iam covering less numbers?
December 21st, 2012 at 7:59:17 PM
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You'll never get back to breaking even. Your mean expected result is to break even from here on out. It was before, as well, to be even here, but you might also have been up, or as it happened, down. But that's got no bearing on the present.
In a -EV game, you're more likely to be down the longer you play (assuming you don't know the results of any part of it), since although your likely distance from the mean increases, the mean itself decreases more quickly. When there's no mean decreasing, the distance you have a given chance of being just gets further and further out.
In a -EV game, you're more likely to be down the longer you play (assuming you don't know the results of any part of it), since although your likely distance from the mean increases, the mean itself decreases more quickly. When there's no mean decreasing, the distance you have a given chance of being just gets further and further out.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
December 21st, 2012 at 9:52:07 PM
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Quote: 24BingoYou'll never get back to breaking even.
That is incorrect. With probability 1, you will get back to even infinitely many times. For that matter, you will be up 1 billion units infinitely many times.
Quote:Your mean expected result is to break even from here on out. It was before, as well, to be even here, but you might also have been up, or as it happened, down. But that's got no bearing on the present.
That is correct.
The original question was, how long will it take to get back to even. This is a difficult question to answer.
The expectation of the amount of time it takes to get back to even diverges. It is larger than n for any finite n (this is easy to prove). In not-so-rigorous terms, the "average" amount of time it will take to get back to even is infinite. However, this expectation is an infinite sum, and infinite sums don't always mean anything intuitive.
Perhaps, a better question is, (again in not-so-rigorous terms) what is the "median" amount of time it will take to get back to even? In rigorous terms, what is the minimum n such that you have greater than a 50% chance of getting back to even? I don't particularly feel like crunching the numbers, and a previous poster has already answered this question (although I haven't checked the math)
This page: http://en.wikipedia.org/wiki/Random_walk#One-dimensional_random_walk contains more information. The point above that I said was "easy to prove" follows directly from a statement on this page:
Quote:If a and b are positive integers, then the expected number of steps until a one-dimensional simple random walk starting at 0 first hits b or −a is ab. The probability that this walk will hit b before −a is a/(a+b), which can be derived from the fact that simple random walk is a martingale.
December 22nd, 2012 at 12:05:14 AM
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Quote: CanyoneroIf your bankroll is infinte, you will eventually get even.
I believe even a stipulation of infinite may be wrong. But even if it is correct to say so using infinite bankroll and zero HE, it is the wrong thing to have in your mind. Gamblers want to think that in the long run they can restore a busted bankroll because 'things have to get back closer to even' ... this is just wrong thinking.
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
December 22nd, 2012 at 1:44:01 AM
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Quote: odiousgambitI believe even a stipulation of infinite may be wrong. But even if it is correct to say so using infinite bankroll and zero HE, it is the wrong thing to have in your mind. Gamblers want to think that in the long run they can restore a busted bankroll because 'things have to get back closer to even' ... this is just wrong thinking.
"Let's go home."
"Let me get even first."
This is what I like about this problem. Technically, the gambler has the math on their side, but with a finite bankroll and lifespan, and negative EV, this thinking will more likely than not lead to disaster.
A few more interesting points:
As was mentioned before, mathematically it doesn't matter if your goal is 50 units or 50000 units, it will happen eventually - but maybe not in your lifetime.
The above fact is independent from the EV of the game. Even with a game that is heavily -EV, you will get even (or up 50, or up 5000) eventually, the pobability within any given number of spins just gets lower and lower.
This why it is important for the casinos to have a lot of money, or a very high bankroll if you will. This reduces the likelyhood of ever going broke against the combined gambling bankrolls of the players. But it still can happen, and I believe it has. Anybody out there know how many casinos went broke on gambling losses in history?
December 22nd, 2012 at 2:02:48 AM
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Quote: GDN
My initial question was, simplified,
If I had two independent trails on a table with no zeros:-
1) Playing a red and black game 50/50
2) Playing 24 numbers (betting two dozens each spin) 2/1
If in both trials I ended up losing the same amount of games, will it take me longer on average to break even playing the red and black game because iam covering less numbers?
The variance of the 50/50 game (you either win 1 unit or lose 1 unit)is: (-1)²*0.5 + 1²*0.5 = 1
For 24 number it is (one game = 1 betting unit, thus you either lose 1 unit or win half a unit: (-1)²*1/3 + 0.5²*2/3 = 0.5
So you will get even faster (but not necessarily before you go broke, no matter how much money you have) in the 50/50 example. If you will, a higher variance increases your "swings", or the deviation from your expected return of 0.
If you want to win big, play a high variance game.
If you want to lose little, play a low house edge game.
Fortunately, this is not mutually exclusive.
Unfortunately, the casino makes it mutually exclusive. (jackpot bets = sucker bets)
December 22nd, 2012 at 7:36:01 AM
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Question: How long to get even?
Getting Even = Revenge.
Revenge is a dish best served cold.
Answer: When you least expect it.
Getting Even = Revenge.
Revenge is a dish best served cold.
Answer: When you least expect it.
Someday, joor goin' to see the name of Googie Gomez in lights and joor goin' to say to joorself, "Was that her?" and then joor goin' to answer to joorself, "That was her!" But you know somethin' mister? I was always her yuss nobody knows it! - Googie Gomez
December 22nd, 2012 at 10:49:43 AM
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Since this thread is posted in the Math section, let`s get into some math.
If you only play even bets on a no-zero roulette wheel and don`t ever vary bet size, your game a so called "random walk" since you either move up +1 unit or down -1 unit after each spin (and independent of your "position", i.e. your balance). Random walk processes have been extensively studied in the past by mathematicians. So why not ustheir results to answer the original question (how to come to break even after being down -50 units).
In your problem, your "goal" is to break even, meaning you want to gain 50 units. So b=50. Now to answer your question you need to know your bankroll, the amount of money you are willing to lose in order to hit your goal. The more you are going to put at risk the more likely you are going to hit your goal.
If you are willing to lose another 100 units (then being down -150 in total), a=100.
You will then hit your goal (break even) with probability a/(a+b) = 100/150 = 66.6%. The average number of bets you need to place (to either break even or bust) is a*b = 100 * 50 = 5000.
Since your game is zero-EV, you can dramaticaly reduce the number of spins for the same risk of ruin by playing bets on the roulette table with different amounts.
If you have 100 to lose, and want to win 50, i would place 50 on the dozen bet 1-12 and 50 on the dozen bet 13-24. You wil hit with 66.6% (as before) in a single spin.
If you only play even bets on a no-zero roulette wheel and don`t ever vary bet size, your game a so called "random walk" since you either move up +1 unit or down -1 unit after each spin (and independent of your "position", i.e. your balance). Random walk processes have been extensively studied in the past by mathematicians. So why not ustheir results to answer the original question (how to come to break even after being down -50 units).
Quote: Wikipedia on Random Walks
If a and b are positive integers, then the expected number of steps until a one-dimensional simple random walk starting at 0 first hits b or −a is a'b. The probability that this walk will hit b before −a is a/(a+b)
In your problem, your "goal" is to break even, meaning you want to gain 50 units. So b=50. Now to answer your question you need to know your bankroll, the amount of money you are willing to lose in order to hit your goal. The more you are going to put at risk the more likely you are going to hit your goal.
If you are willing to lose another 100 units (then being down -150 in total), a=100.
You will then hit your goal (break even) with probability a/(a+b) = 100/150 = 66.6%. The average number of bets you need to place (to either break even or bust) is a*b = 100 * 50 = 5000.
Since your game is zero-EV, you can dramaticaly reduce the number of spins for the same risk of ruin by playing bets on the roulette table with different amounts.
If you have 100 to lose, and want to win 50, i would place 50 on the dozen bet 1-12 and 50 on the dozen bet 13-24. You wil hit with 66.6% (as before) in a single spin.
December 22nd, 2012 at 11:30:16 AM
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"The best revenge is a life well lived"Quote: s2dbakerQuestion: How long to get even?
Getting Even = Revenge.
Revenge is a dish best served cold.
Answer: When you least expect it.
Answer: You are always even.
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December 22nd, 2012 at 12:43:37 PM
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Quote: MangoJ
Quote: Wikipedia on Random Walks
If a and b are positive integers, then the expected number of steps until a one-dimensional simple random walk starting at 0 first hits b or −a is ab. The probability that this walk will hit b before −a is a/(a+b)
In your problem, your "goal" is to break even, meaning you want to gain 50 units. So b=50. Now to answer your question you need to know your bankroll, the amount of money you are willing to lose in order to hit your goal. The more you are going to put at risk the more likely you are going to hit your goal.
If you are willing to lose another 100 units (then being down -150 in total), a=100.
You will then hit your goal (break even) with probability a/(a+b) = 100/150 = 66.6%. The average number of bets you need to place (to either break even or bust) is a*b = 100 * 50 = 5000.
Glad to see that we are on the same page (literally!) :)
So, what if you have an infinite bankroll? Now a is infinite, which is not a positive integer, so we can't use the formula. But we can say that a > a' for any positive integer a'. So the expected number of steps to break even is greater than 50 * a' for any positive integer a'. In other words, it's infinite. "Expected" means average, here, in other words, it's an infinite sum, and this sum diverges.
It's somewhat interesting, because the number of steps that you need in order to break even with 50% probability is not particularly high, but the expectation of the number of steps is still infinite.
December 22nd, 2012 at 1:40:46 PM
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Quote: AxiomOfChoiceIt's somewhat interesting, because the number of steps that you need in order to break even with 50% probability is not particularly high, but the expectation of the number of steps is still infinite.
Not so sure if we're on the same page. For a success rate of 50% you need to put another 50 at risk, so a=b=50 and the average number of spins is a*b=2500. It's not infinite.
If you do have an infinite bankroll (i.e. you print your own money) your success rate is 100% (not 50%), so it *will* happen almost every try. However the distribution of the length of this random walk will be heavily long-tailed in a way that the mean value (the expected number of spins) will diverge.
There is nothing peculiar about an infinite bankroll. You can reach your goal (moreover *any* goal) almost surely (with 100% probability). It might however be a waste of time though :)
Edit: Sorry Axiom (we are on the same page). Didn't read page #2 of the thread before answering, since nothing useful came up on #1.
December 22nd, 2012 at 4:26:23 PM
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Quote: AxiomOfChoice
So, what if you have an infinite bankroll? Now a is infinite, which is not a positive integer, so we can't use the formula. But we can say that a > a' for any positive integer a'. So the expected number of steps to break even is greater than 50 * a' for any positive integer a'. In other words, it's infinite. "Expected" means average, here, in other words, it's an infinite sum, and this sum diverges.
It's somewhat interesting, because the number of steps that you need in order to break even with 50% probability is not particularly high, but the expectation of the number of steps is still infinite.
If the expectation of the number of steps to break even from -50 units down is infinite, are there studies of similar trials available to view to support such findings, I'm not disputing the facts BTW just want to determine the spread of the numbers of spins it takes to get even to acquire a more understanding of the behavior (or the fluctuations) of the random walk. I have done a number of trials myself regarding a 50/50 probability outcome of a number of spins or coin tosses, the most its ever taken me to break even from anything from 50-100 units down, is about 5 thousand spins, incidentally that's the number of spins required to determine if a certain roulette wheel contains bias numbers or not.
In a completely differently trial of betting 1 unit each on 2 dozens (covering 24 numbers each spin on a wheel with no zeros), I've always thought, since i'll be covering more numbers, that it would take me (on average) less time to break even than in a trial betting just red and black. Since the steps are infinite to break even in both trials (with an infinite bank roll), would it mean in both different trials it would take on average the same amount of time to break even from the same amount of games lost?
December 22nd, 2012 at 5:12:35 PM
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So you are looking for the probability distribution of the number of spins to reach your goal. This is quite an interesting question. I guess it could be solved with some kind of Markovian Ansatz, maybe someone here around knows more facts about the random walk process.
Is it worthwhile to pursue a loss on a zero-EV game even with infinite bankroll ? Probably not.
To your second question: Betting 1 unit each on 2 dozens won't help you a lot I would guess. The result will be a more asymmetric random walk on short scales. But since the long run applies to any sum of random variates, the only property affecting the long run properties (and your pursuit is definetly long run) is variance.
The 2 dozen bet has a variance of 2, while the even bet has a variance of 4 (if you were betting 2 units on color). You *need* variance for your goal, so the 2 dozen bet would be slower.
Is it worthwhile to pursue a loss on a zero-EV game even with infinite bankroll ? Probably not.
To your second question: Betting 1 unit each on 2 dozens won't help you a lot I would guess. The result will be a more asymmetric random walk on short scales. But since the long run applies to any sum of random variates, the only property affecting the long run properties (and your pursuit is definetly long run) is variance.
The 2 dozen bet has a variance of 2, while the even bet has a variance of 4 (if you were betting 2 units on color). You *need* variance for your goal, so the 2 dozen bet would be slower.
December 22nd, 2012 at 7:39:15 PM
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Quote: MangoJSo you are looking for the probability distribution of the number of spins to reach your goal. This is quite an interesting question. I guess it could be solved with some kind of Markovian Ansatz, maybe someone here around knows more facts about the random walk process.
Yes that's exactly what I,am looking for, just as either red or black can not repeat itself forever in consecutive spins (even though its possible, since every spin is an independent spin and the wheel has no memory), the most i've ever encountered is 20 reds in a row, so a probability of a distribution of number of spins to get back to even from 50 units down (from an infinite bankroll) in a fair game must exist. What would also be interesting is to find the probability of distribution of the number of units someone can be down. In my 50/50 probability trials (in a fair game), the most I,ve ever been down is about 100 units.
Quote: MangoJTo your second question: Betting 1 unit each on 2 dozens won't help you a lot I would guess. The result will be a more asymmetric random walk on short scales. But since the long run applies to any sum of random variates, the only property affecting the long run properties (and your pursuit is definetly long run) is variance.
The 2 dozen bet has a variance of 2, while the even bet has a variance of 4 (if you were betting 2 units on color). You *need* variance for your goal, so the 2 dozen bet would be slower.
If on a trial of betting 1 unit on 2 dozens takes longer to break even than in a separate trial of betting on red and black from the same amount of games lost (because betting 2 dozens has a variance of 2), so does that mean on a third independent trial of betting 1 unit on a single dozen each spin (which should have a higher variance than the other separate trials) it would be faster to reach my goal?. If that is true, it will mean the fewer numbers I bet in a trial, the faster I will reach my goal of breaking even from being in the red, right?
Yes, its not worth while pursuing a loss on a zero-EV game, iam just using the game of roulette to get my point across, my question is really a math question.
December 22nd, 2012 at 11:05:07 PM
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Roulette is a great game when it comes to variance, because you can tailor it.
variance of 1 unit on single color = 1/2 * 1² + 1/2 * (-1)² = 1
variance of 1 unit on 2 different color = 1/2 * 0² + 1/2 * 0² = 0
variance of 2 unit on single color = 1/2 * 2² + 1/2 * (-2)² = 4
variance of 1 unit on single dozen = 1/3 * 2² + 2/3 * (-1)² = 2
variance of 1 unit on 2 different dozens = 2/3 * 1² + 1/3 * (-2)² = 2
variance of 2 unit on single dozen = 1/3 * 4² + 2/3 * (-2)² = 8
The color bet is not generally better than the 2 dozen bet. It also depends on the total bet size you are willing to make. If you can only do 1 unit, the dozen bet (2) is better than the color bet (1). If you can do 2 units, the color bet (4) is better than the 2dozen bet (2), but betting 2 units on a single dozen (8) is still better.
So generally speaking, bet more units on more rare events. Eventually bet table max on a single number. It won't help you reach your goal with higher probability (otherwise you could play at +EV on a no-EV game), but it will save you significant time for the same result (even if the expected number of spins is infinite).
variance of 1 unit on single color = 1/2 * 1² + 1/2 * (-1)² = 1
variance of 1 unit on 2 different color = 1/2 * 0² + 1/2 * 0² = 0
variance of 2 unit on single color = 1/2 * 2² + 1/2 * (-2)² = 4
variance of 1 unit on single dozen = 1/3 * 2² + 2/3 * (-1)² = 2
variance of 1 unit on 2 different dozens = 2/3 * 1² + 1/3 * (-2)² = 2
variance of 2 unit on single dozen = 1/3 * 4² + 2/3 * (-2)² = 8
The color bet is not generally better than the 2 dozen bet. It also depends on the total bet size you are willing to make. If you can only do 1 unit, the dozen bet (2) is better than the color bet (1). If you can do 2 units, the color bet (4) is better than the 2dozen bet (2), but betting 2 units on a single dozen (8) is still better.
So generally speaking, bet more units on more rare events. Eventually bet table max on a single number. It won't help you reach your goal with higher probability (otherwise you could play at +EV on a no-EV game), but it will save you significant time for the same result (even if the expected number of spins is infinite).
December 22nd, 2012 at 11:38:17 PM
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I've often argued that, for essentially that reason, if what you want is to win big, roulette's a better bet than craps, since unless you have, say, 10x odds and a double-zero wheel, as a percentage of your potential win, the edge is far less.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
December 23rd, 2012 at 11:24:54 AM
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@MangoJ
I really appreciate your input and others too, what I really find interesting is even though the journey back to breaking even can consist of infinite steps (with an infinite bank roll), but maybe there is also a probable ceiling to the number of steps (or spins in roulette) it takes to get back to even, because just as i mentioned previously maybe there is a probable maximum number of times a single color, be it red or black, can repeat itself continuously (which can also be infinite amount of times) but yet, in my life time, the maximum i've ever encountered is only 20 in a row, which is not much considering it could repeat itself an infinite amount of times.
Can we ever take advantage of a probable ceiling to the number of steps it takes to breaking even? Probably, but not in the game of roulette. Roulette isn't a fair game.
I really appreciate your input and others too, what I really find interesting is even though the journey back to breaking even can consist of infinite steps (with an infinite bank roll), but maybe there is also a probable ceiling to the number of steps (or spins in roulette) it takes to get back to even, because just as i mentioned previously maybe there is a probable maximum number of times a single color, be it red or black, can repeat itself continuously (which can also be infinite amount of times) but yet, in my life time, the maximum i've ever encountered is only 20 in a row, which is not much considering it could repeat itself an infinite amount of times.
Can we ever take advantage of a probable ceiling to the number of steps it takes to breaking even? Probably, but not in the game of roulette. Roulette isn't a fair game.
December 23rd, 2012 at 12:01:12 PM
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If there wee some kind of "ceiling" then yes, you could exploit it and can come back to even in finite time. But then you would turn a no-EV game into a positive EV game - which is simply impossible. Since the result is impossible, the assumption "there is some kind of ceiling" must be wrong. Pure logic, simple as that.
Another view on the same point: If red came up 20 times in a row, why would it come up with different probability on the very next spin ? Neither the wheel nor the ball keep records of their last spin. Whats different on the 21th spin ?
A third view: The fair color bet (or coin flip) is a symmetrical bet. You are up or down any specific amount for the same probability. How do you expect to break even (i.e. gain a few units) consistently, when you are equally likely to lose more units ?
There is no magical ceiling or anything other. Don't waste your time chasing ghosts.
Another view on the same point: If red came up 20 times in a row, why would it come up with different probability on the very next spin ? Neither the wheel nor the ball keep records of their last spin. Whats different on the 21th spin ?
A third view: The fair color bet (or coin flip) is a symmetrical bet. You are up or down any specific amount for the same probability. How do you expect to break even (i.e. gain a few units) consistently, when you are equally likely to lose more units ?
There is no magical ceiling or anything other. Don't waste your time chasing ghosts.
December 23rd, 2012 at 12:28:43 PM
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Last edited by: sodawater on Oct 1, 2018
December 23rd, 2012 at 1:36:58 PM
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Quote: MangoJNot so sure if we're on the same page. For a success rate of 50% you need to put another 50 at risk, so a=b=50 and the average number of spins is a*b=2500. It's not infinite.
If you do have an infinite bankroll (i.e. you print your own money) your success rate is 100% (not 50%), so it *will* happen almost every try. However the distribution of the length of this random walk will be heavily long-tailed in a way that the mean value (the expected number of spins) will diverge.
Right, my point is just that the median (the 50% point) is well defined, but the mean diverges. "Expectation" (as in, the expectation of a random variable X, usually written as E(X)) is defined as a mean, so it's correct to say that the expectation is infinite, even though it will eventually happen with probability 1 (and, for any probability p < 1, there exists some N such that it will happen within N steps with probability p)
I think that this result (the fact that the expectation is infinite) despite the fact that the rest of it is all finite and well defined is a little counter-intuitive, which is why I think it's interesting. Infinite sums are like that, though, I guess.
December 23rd, 2012 at 1:55:14 PM
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Yeah you are right. I was not aware you talk about the "median" as a measure of performance. Of course the median is always finite for any probability function. The mean in this example is diverging, which is just a divergence of the measure, not the problem itself. There are tons of probabiltiy functions with diverging mean, so what. Indeed if the expectation would be finite, you would have a money printing machine by doing nothing more than playing a fair coin flip.
