Kelly betting
- Thread starter squeeks
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Let's call the starting bankroll S and the current bankroll C.
Forget about blackjack, suppose you play with biased coin which has 51% for heads and 49% for tails. So, you keep betting on heads and your edge is 2%.
Kelly says that you should always bet 2% of your current bankroll. The fraction f of the current bankroll to bet, is constant, and it is f=0.02*C
Now how about all the possible formulas that suggest to bet e.g:
f=0.02*C when C=S,
f<0.02*C when C<S,
f>0.02*C when C>S.
Or more generally, where X any given number,
f=0.02*C when C=X*S,
f<0.02*C when C<X*S,
f>0.02*C when C>X*S.
Note that this set of all the possible such formulas, includes the formulas which suggest only minor increases of f as C increases, e.g. that you should raise to f=0.04 only when C=1,000,000,000,000 * S
The proof of kelly criterion, cannot proove that none of all these possible formulas is more maximising than always betting Kelly = f = 0.02*C.
Because that proof, is only the mathematical answer to the question:"SUPPOSING that you keep betting a CONSTANT fraction of your current bankroll, then what is the fraction among all possible fractions that maximises the growth of your bankroll?"
The proof of kelly criterion is quite simple: If you bet ANY constant fraction of your bankroll, then after W+L tosses, and if you have the average luck (and you WILL have approximately the average luck after too many tosses), then your current bankroll C will be:
C=((1+f)^W)((1-f)^L)S
(also see the relevant previous post of mine where I explain this equation)
So we shape the graph of the above function, where at the axis of Y we put the values for C, and at the axis of X we put the values for f. And we see that the curve which is shaped at this graph, gets its highest for the value of f which is equal to your edge.
Do you get it now? This proof says that IF you decide to bet a CONSTANT fraction of your current bankroll, then the growth of the current bankroll is maximised for f=edge. Therefore it has nothing to say mathematically for all other betting systems that do not suggest to bet a constant f. If it HAS something to say, then this needs additional mathematical proof.
Forget about blackjack, suppose you play with biased coin which has 51% for heads and 49% for tails. So, you keep betting on heads and your edge is 2%.
Kelly says that you should always bet 2% of your current bankroll. The fraction f of the current bankroll to bet, is constant, and it is f=0.02*C
Now how about all the possible formulas that suggest to bet e.g:
f=0.02*C when C=S,
f<0.02*C when C<S,
f>0.02*C when C>S.
Or more generally, where X any given number,
f=0.02*C when C=X*S,
f<0.02*C when C<X*S,
f>0.02*C when C>X*S.
Note that this set of all the possible such formulas, includes the formulas which suggest only minor increases of f as C increases, e.g. that you should raise to f=0.04 only when C=1,000,000,000,000 * S
The proof of kelly criterion, cannot proove that none of all these possible formulas is more maximising than always betting Kelly = f = 0.02*C.
Because that proof, is only the mathematical answer to the question:"SUPPOSING that you keep betting a CONSTANT fraction of your current bankroll, then what is the fraction among all possible fractions that maximises the growth of your bankroll?"
The proof of kelly criterion is quite simple: If you bet ANY constant fraction of your bankroll, then after W+L tosses, and if you have the average luck (and you WILL have approximately the average luck after too many tosses), then your current bankroll C will be:
C=((1+f)^W)((1-f)^L)S
(also see the relevant previous post of mine where I explain this equation)
So we shape the graph of the above function, where at the axis of Y we put the values for C, and at the axis of X we put the values for f. And we see that the curve which is shaped at this graph, gets its highest for the value of f which is equal to your edge.
Do you get it now? This proof says that IF you decide to bet a CONSTANT fraction of your current bankroll, then the growth of the current bankroll is maximised for f=edge. Therefore it has nothing to say mathematically for all other betting systems that do not suggest to bet a constant f. If it HAS something to say, then this needs additional mathematical proof.
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hmmmmm
Kelly outperforms betting half kelly, or anything less then Kelly
Kelly outperforms betting 1.5 kelly, or anything over Kelly
Kelly outperforms betting double kelly
Kelly outperforms betting over double kelly
Kelly outperforms flat betting
So Kelly outperforms everything?
Isn't that why it is optimal?:joker::whip:
Kelly outperforms betting half kelly, or anything less then Kelly
Kelly outperforms betting 1.5 kelly, or anything over Kelly
Kelly outperforms betting double kelly
Kelly outperforms betting over double kelly
Kelly outperforms flat betting
So Kelly outperforms everything?
Isn't that why it is optimal?:joker::whip:
Wow, I finally think I'm seeing your argument, does this paraphrase it?
All of the maths re Kelly betting is geared to betting a continuously resized edge*bankroll/variance.
You're saying that the maths don't even address the subject of betting edge*bankroll/variance*UnknownVariable?
I agree, I don't think the maths address that.
However, at an intuitive level, any value of UnknownVariable that changes the bet size from full Kelly is going to give you a sub optimal result. In extreme examples, it might lead one to bet zero when there is an advantage, which is never optimal.
And Remember that for decision making purposes, all that matters is the bankroll and edge right now, the results of all previous wagers is effectively discarded.
All of the maths re Kelly betting is geared to betting a continuously resized edge*bankroll/variance.
You're saying that the maths don't even address the subject of betting edge*bankroll/variance*UnknownVariable?
I agree, I don't think the maths address that.
However, at an intuitive level, any value of UnknownVariable that changes the bet size from full Kelly is going to give you a sub optimal result. In extreme examples, it might lead one to bet zero when there is an advantage, which is never optimal.
And Remember that for decision making purposes, all that matters is the bankroll and edge right now, the results of all previous wagers is effectively discarded.
Not necessarily. What if you play half-Kelly under certain conditions and full-Kelly all other times? What if you mix up your strategies according to certain criteria? That requires further proof than what you stated.
When the unknown variable is not constant, yes. If the unknown variable is constant then the maths have covered this. There is a wide body of research that involves constantly betting half-Kelly, quarter-Kelly, etc. These are all “pure” strategies.
However, if your Kelly fraction is not constant then your growth rate will change. For example, if you start with 100 units betting full Kelly, you might increase to 1.25 Kelly after winning 25 units or drop down to 0.75 Kelly after losing 25 units. This is called a “mixed” strategy because you are alternating between different strategies. Using a mixed strategy like this can affect your win rate dramatically. In certain games a mixed strategy is more optimal than a pure strategy.
For casino games it might make sense for a “poor” player to use a smaller Kelly fraction as his minimum bet approaches the table minimums. After all, going broke doesn’t necessarily mean that you lose all your money, only that you can no longer afford to make a bet. If the optimal bet for your bankroll is $23 but all of the tables are $25 minimums, you are effectively “broke” for that particular bet. You either have to look for a better opportunity or adjust your Kelly fraction. Both decisions will affect your bankroll’s growth rate.
Essentially, the strategy described above is similar to the ploppy theory of “bet more when you are ahead and less when you are behind.” This may have merit for certain players, but I don’t think the growth rate is better except in some extreme cases. That’s just my opinion though. I would be interested to hear some arguments in favor of it.
-Sonny-
When the unknown variable is not constant, yes. If the unknown variable is constant then the maths have covered this. There is a wide body of research that involves constantly betting half-Kelly, quarter-Kelly, etc. These are all “pure” strategies.
However, if your Kelly fraction is not constant then your growth rate will change. For example, if you start with 100 units betting full Kelly, you might increase to 1.25 Kelly after winning 25 units or drop down to 0.75 Kelly after losing 25 units. This is called a “mixed” strategy because you are alternating between different strategies. Using a mixed strategy like this can affect your win rate dramatically. In certain games a mixed strategy is more optimal than a pure strategy.
For casino games it might make sense for a “poor” player to use a smaller Kelly fraction as his minimum bet approaches the table minimums. After all, going broke doesn’t necessarily mean that you lose all your money, only that you can no longer afford to make a bet. If the optimal bet for your bankroll is $23 but all of the tables are $25 minimums, you are effectively “broke” for that particular bet. You either have to look for a better opportunity or adjust your Kelly fraction. Both decisions will affect your bankroll’s growth rate.
Essentially, the strategy described above is similar to the ploppy theory of “bet more when you are ahead and less when you are behind.” This may have merit for certain players, but I don’t think the growth rate is better except in some extreme cases. That’s just my opinion though. I would be interested to hear some arguments in favor of it.
-Sonny-
blackjack avenger
The more frequently one resizes the closer one is to continuous resizing kelly. Even the old half or double bets based on bank starts to resemble kelly if you do it a few times.
Sure, bumping up against table minimums can be a problem. However, that does not change the mathematics of the optimum kelly bets. The question is how do we employ kelly as best we can in the real world.
The more frequently one resizes the closer one is to continuous resizing kelly. Even the old half or double bets based on bank starts to resemble kelly if you do it a few times.
Sure, bumping up against table minimums can be a problem. However, that does not change the mathematics of the optimum kelly bets. The question is how do we employ kelly as best we can in the real world.
Yes, this is what I meant. The proof of kelly criterion cannot proove this betting system as worse than kelly. It can only proove overbetting kelly or undebetting kelly as worse. But overbetting kelly is always betting e.g. 2 times kelly no matter what the current bankroll is, and underbetting kelly is always betting 0.5 times kelly no matter what the current bankroll is.
By intuition I would say that your example is worse than kelly betting.
But what about the formula which suggests betting 1.25 kelly when the current bankroll is 100 times the starting bankroll and betting 0.75 kelly when the current bankroll is 1/100 times the starting bankroll? Is it better or worse than kelly betting? There is no obvious answer to that. We need a mathematical proof or a simulation. You are saying some people have prooved that some of these "mixed" betting systems are better than kelly? Where is this information?
By intuition I would say that your example is worse than kelly betting.
But what about the formula which suggests betting 1.25 kelly when the current bankroll is 100 times the starting bankroll and betting 0.75 kelly when the current bankroll is 1/100 times the starting bankroll? Is it better or worse than kelly betting? There is no obvious answer to that. We need a mathematical proof or a simulation. You are saying some people have prooved that some of these "mixed" betting systems are better than kelly? Where is this information?
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Bell Curve and Blackjack
Think of a bell curve.
At the top of the curve is Kelly and that is optimum log growth. As you look at the curve the right side is betting more then Kelly. To the left as you look at the curve is betting less then Kelly.
The growth rate of betting .5 Kelly is the same as betting 1.5 Kelly. They both have 75% the growth rate of Kelly. However, betting 1.5 Kelly entails much wilder swings. So you get the same growth rate but take more variance. This is not desirable.
Kelly is optimal.:joker::whip:
Better to underbet then overbet.
In the long run Kelly greatly outperforms flat betting
optimal still means optimal:joker::whip:
Think of a bell curve.
At the top of the curve is Kelly and that is optimum log growth. As you look at the curve the right side is betting more then Kelly. To the left as you look at the curve is betting less then Kelly.
The growth rate of betting .5 Kelly is the same as betting 1.5 Kelly. They both have 75% the growth rate of Kelly. However, betting 1.5 Kelly entails much wilder swings. So you get the same growth rate but take more variance. This is not desirable.
Kelly is optimal.:joker::whip:
Better to underbet then overbet.
In the long run Kelly greatly outperforms flat betting
optimal still means optimal:joker::whip:
Kelly Betting and the Real World
As one resizes one gets closer to Kelly (optimal) then flat betting.
Easy resizing:
First to adjust for human frailties and to delay sizing down bets if negative varaince one should bet .5 to .75 Kelly. It is better to underbet then overbet.
If your bank justifies $100 .5 Kelly bets. All one needs to do is adjust bets based on chips:
If you lose 25% of bank then bet $75 (3 greens)
If you lose 50% then bet 2 greens
If you win approx. 25% or 50% of bank then you can raise bets to $125 or $150.
or
One could decide to not raise bets or partial raising and have more units to defend against a negative swing and crashing into table minimums.
This real world approximation of Kelly will outperform flat betting in the long run given similar ror's.
As one resizes one gets closer to Kelly (optimal) then flat betting.
Easy resizing:
First to adjust for human frailties and to delay sizing down bets if negative varaince one should bet .5 to .75 Kelly. It is better to underbet then overbet.
If your bank justifies $100 .5 Kelly bets. All one needs to do is adjust bets based on chips:
If you lose 25% of bank then bet $75 (3 greens)
If you lose 50% then bet 2 greens
If you win approx. 25% or 50% of bank then you can raise bets to $125 or $150.
or
One could decide to not raise bets or partial raising and have more units to defend against a negative swing and crashing into table minimums.
This real world approximation of Kelly will outperform flat betting in the long run given similar ror's.
Well, here's my take on it. Your formula above is exponential growth (G), right?
Maximizing your growth for logarithmic growth for that forrmua would be ln(G) = ln(1+f)*W + ln(1-f)*L. It's growth is maximized by f=1 but the expected growth is maximized by f=.02 in your example. Logarithmic growth per hand = ln(1+f)*W/N + ln(1-f)*L/N. But a Kelly better does not maximize actual growth, he maximizes the log of his expected growth. He can't maximize actual growth becasue he doesn't know if he will win.
While W/N will converge to expected value p over time, and so your actual results will converge to your expected logarithmic growth per hand, also W/N will be normally distributed with a mean of p so the median growth per hand = the mean growth per hand. So a Kelly better is maximizing both mean and median growth per hand. That's the Kelly fraction 1. I think mean growth is higher than median growth.
Some people assume W=p*N and maximize G. That's not right.
People like to maximize different things. A square-root better would bet about twice Kelly, a risk-neutral better would bet everything. They may maximize actual growth but not expected growth.
I think Kelly takes into account risk vs reward - I think that's why an AP guy betting to a spread, sometimes even playing -EV hands, obviously not betting a constant fraction of current roll, can still achieve the max log growth of his expectation. He cannot have a risk of 0 because he could lose every hand while still betting in a way that maximizes his expected log growth. It turns out that the risk that optimizes the log of expected growth is 13.53%. (the log of -2/Kelly fraction 1).
Not sure if this answers your basic question. But this is what I think Kelly tries to do. Other betting systems actually can have higher actual growth rates than Kelly. Whether betting a constant fraction of current roll or, like an AP guy, betting to a spread at different advantages, even in -EV situations, and different variances at each count, there's only one way that will maximize the logarithmic of growth of your expectation.
I know what your're gonna say Blackajck Avenger, lmao. I said one way and you said two lmao. I'm thinking about it lol.
But I think I may have been making the mistake that Kelly maximized actual growth rather than expected growth.
Maximizing your growth for logarithmic growth for that forrmua would be ln(G) = ln(1+f)*W + ln(1-f)*L. It's growth is maximized by f=1 but the expected growth is maximized by f=.02 in your example. Logarithmic growth per hand = ln(1+f)*W/N + ln(1-f)*L/N. But a Kelly better does not maximize actual growth, he maximizes the log of his expected growth. He can't maximize actual growth becasue he doesn't know if he will win.
While W/N will converge to expected value p over time, and so your actual results will converge to your expected logarithmic growth per hand, also W/N will be normally distributed with a mean of p so the median growth per hand = the mean growth per hand. So a Kelly better is maximizing both mean and median growth per hand. That's the Kelly fraction 1. I think mean growth is higher than median growth.
Some people assume W=p*N and maximize G. That's not right.
People like to maximize different things. A square-root better would bet about twice Kelly, a risk-neutral better would bet everything. They may maximize actual growth but not expected growth.
I think Kelly takes into account risk vs reward - I think that's why an AP guy betting to a spread, sometimes even playing -EV hands, obviously not betting a constant fraction of current roll, can still achieve the max log growth of his expectation. He cannot have a risk of 0 because he could lose every hand while still betting in a way that maximizes his expected log growth. It turns out that the risk that optimizes the log of expected growth is 13.53%. (the log of -2/Kelly fraction 1).
Not sure if this answers your basic question. But this is what I think Kelly tries to do. Other betting systems actually can have higher actual growth rates than Kelly. Whether betting a constant fraction of current roll or, like an AP guy, betting to a spread at different advantages, even in -EV situations, and different variances at each count, there's only one way that will maximize the logarithmic of growth of your expectation.
I know what your're gonna say Blackajck Avenger, lmao. I said one way and you said two lmao. I'm thinking about it lol.
But I think I may have been making the mistake that Kelly maximized actual growth rather than expected growth.
I'm not sure what you mean, I'll read your post again when I have the time, but it seems you're saying that what I am saying is wrong because kelly uses that logarithmic formula.
In that case, you are wrong. Kelly uses that logarithmic formula simply to find the the peak of the curve of the graph of the formula C=((1+f)^W)((1-f)^L)S. In other words, that logarithmic formula is simply a derivation of the formula C=((1+f)^W)((1-f)^L)S. Therefore all my conclusions in the previous posts are correct, as that logarithmic formula which finds the "expodential growth" or whatever you may call it that it finds, again refers to always betting a constant fraction of the current bankroll (as the current bankroll increases or decreases), and has nothing to do with increasing the fraction of the current bankroll (as the current bankroll increases).
In that case, you are wrong. Kelly uses that logarithmic formula simply to find the the peak of the curve of the graph of the formula C=((1+f)^W)((1-f)^L)S. In other words, that logarithmic formula is simply a derivation of the formula C=((1+f)^W)((1-f)^L)S. Therefore all my conclusions in the previous posts are correct, as that logarithmic formula which finds the "expodential growth" or whatever you may call it that it finds, again refers to always betting a constant fraction of the current bankroll (as the current bankroll increases or decreases), and has nothing to do with increasing the fraction of the current bankroll (as the current bankroll increases).
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Kelly would prove that the above suggested system is not "optimal" (by Kelly's terms of course) given an initial roll. It will not maximize the log of your expected growth per hand/coin flip/trial of your starting roll.
Pretty much, that's the definition. If you deviate from full-Kelly, you are no longer betting full-Kelly.
In your coin-flip example, start with 1MM units, play a billion flips, maybe your bankroll is now 10MM or 100MM units, whatever, if you bet 1.25 Kelly on that 1BB + 1 flip, sorry, max log growth of expected value per hand/flip/trial is no longer achieved.
Likewise, 1BB trials later your roll is 10K units and you bet .75 Kelly.
And variance is always a factor - you could have lost your first 1BB flips in a row after all, betting the Kelly fraction each and every time, and never had to worry about it lol. (I'm gonna call that broke, even if, yes, you'd still have a millionth of a cent left lol).
Betting 1MM units, and always betting entire current roll after each win, for 1BB winning trials in a row would have maximized actual growth per trial, (exponential growth) far exceeding Kelly growth. But Kelly growth only purports to maximize the log of expected growth given risk, not the log of actual growth.
Let the risk of losing all fall where it may. Turns out this risk approaches 13.5335% risk, just as risk of losing all approaches 1-that. Neither actually ever gets there with infinity. You could rebound with that millionth of a cent left, winning the next 2BB trials in a row. Or you could get even closer to 0.
Pretty much, that's the definition. If you deviate from full-Kelly, you are no longer betting full-Kelly.
In your coin-flip example, start with 1MM units, play a billion flips, maybe your bankroll is now 10MM or 100MM units, whatever, if you bet 1.25 Kelly on that 1BB + 1 flip, sorry, max log growth of expected value per hand/flip/trial is no longer achieved.
Likewise, 1BB trials later your roll is 10K units and you bet .75 Kelly.
And variance is always a factor - you could have lost your first 1BB flips in a row after all, betting the Kelly fraction each and every time, and never had to worry about it lol. (I'm gonna call that broke, even if, yes, you'd still have a millionth of a cent left lol).
Betting 1MM units, and always betting entire current roll after each win, for 1BB winning trials in a row would have maximized actual growth per trial, (exponential growth) far exceeding Kelly growth. But Kelly growth only purports to maximize the log of expected growth given risk, not the log of actual growth.
Let the risk of losing all fall where it may. Turns out this risk approaches 13.5335% risk, just as risk of losing all approaches 1-that. Neither actually ever gets there with infinity. You could rebound with that millionth of a cent left, winning the next 2BB trials in a row. Or you could get even closer to 0.
Here we go again lol.
I can agree it approaches 0 as you approach an infinite number of hands. But it never actually reaches 0.
Just as a BJ AP's "lifetime" ROR assumes he plays "forever".
It's just that everybody seems to think Kelly betting is great because it's assumed its ROR is "0" - like ROR doesn't exist or like it implies one will never actually go broke or is somehow "risk free".
Since a Kelly bettor is maximizing the log of his expectation, his expected growth, there always exists the possibility of bad luck when a fixed number of hands is assumed.
So, if you put a limit on number of hands, trials, whatever, say you play 1MM trials 1MM times, I think you'll be really really close to "0" dollars left 13.53% of the time and, effectively, broke. That's the flip-side.
So, in effect, even a pure proportional-betting Kelly bettor's ROR gets really really close to that 13.53% risk of getting really really close to no money left over a large number of trials since risk is, after all, involved. He could lose, rarely, after all, all 1MM trials lol. In effect, almost 13.53% of the time he won't have much money left. The other 86.47% of the time, he'll have a really big roll lol. Basically, he too either reaches the "long run" or goes "broke" trying.
That's what I'm thinking tonite and I'm sticking with it until tomorrow lol.
As always, looking forward to your thoughts. Anyone else's too lol.
I can agree it approaches 0 as you approach an infinite number of hands. But it never actually reaches 0.
Just as a BJ AP's "lifetime" ROR assumes he plays "forever".
It's just that everybody seems to think Kelly betting is great because it's assumed its ROR is "0" - like ROR doesn't exist or like it implies one will never actually go broke or is somehow "risk free".
Since a Kelly bettor is maximizing the log of his expectation, his expected growth, there always exists the possibility of bad luck when a fixed number of hands is assumed.
So, if you put a limit on number of hands, trials, whatever, say you play 1MM trials 1MM times, I think you'll be really really close to "0" dollars left 13.53% of the time and, effectively, broke. That's the flip-side.
So, in effect, even a pure proportional-betting Kelly bettor's ROR gets really really close to that 13.53% risk of getting really really close to no money left over a large number of trials since risk is, after all, involved. He could lose, rarely, after all, all 1MM trials lol. In effect, almost 13.53% of the time he won't have much money left. The other 86.47% of the time, he'll have a really big roll lol. Basically, he too either reaches the "long run" or goes "broke" trying.
That's what I'm thinking tonite and I'm sticking with it until tomorrow lol.
As always, looking forward to your thoughts. Anyone else's too lol.
Continuous Kelly re-sizing approaches 0 $ left as number of trials approach infinity. (I assume by continuous Kelly re-sizing you mean always betting a fixed percentage of current roll?) (For simplicity's sake, can we assume a variance of 1 for now?)
Fixed Kelly wagering also approaches 0 $ left as number of trials approach infinity.
But both will basically have no money left 13.53% of the time.
If a fixed Kelly bettor bets in such a way as to have a 13.53% ROR, he has the same RoR as a pure proportional Kelly bettor.
I think a Kelly bettor would be really really close to 0 $ left 13.53% of the time or 135,300 times after trying 1MM times, betting 1MM trials each of those 1MM times, given the starting roll predicted by Kelly.
How often do you think, with a given original unit roll as predicted by Kelly, a pure proportional Kelly bettor would be really really close to no money left if he bet 1MM fixed trials 1MM times?
Fixed Kelly wagering also approaches 0 $ left as number of trials approach infinity.
But both will basically have no money left 13.53% of the time.
If a fixed Kelly bettor bets in such a way as to have a 13.53% ROR, he has the same RoR as a pure proportional Kelly bettor.
I think a Kelly bettor would be really really close to 0 $ left 13.53% of the time or 135,300 times after trying 1MM times, betting 1MM trials each of those 1MM times, given the starting roll predicted by Kelly.
How often do you think, with a given original unit roll as predicted by Kelly, a pure proportional Kelly bettor would be really really close to no money left if he bet 1MM fixed trials 1MM times?
I Respectfully Disagree and Take Your Lunch Money!
Kelly continuous resizing is theoretical 0% ror.
After many hands one could be losing:joker::whip:. However, if you can play forever you will reach your goal.
If you continuously resize your bank downward then how can you ever go broke in the theoretical world? If you never go broke then if you have an advantage and play long enough then you would reach your goal in time. How could it be anything else?
With fixed Kelly wagering the 13.53% ror is your starting ror. If you were to double your money your ror would drop to 1.83% moving forward provided you kept the same size bets.
If you win enough then your ror cannot be 13.53%. If you quadruple your bank then your ror is .03% moving forward. If you are playing with an edge and don't go broke but win over time then your ror would drop. How could it be anything else?
Kelly continuous resizing is theoretical 0% ror.
After many hands one could be losing:joker::whip:. However, if you can play forever you will reach your goal.
If you continuously resize your bank downward then how can you ever go broke in the theoretical world? If you never go broke then if you have an advantage and play long enough then you would reach your goal in time. How could it be anything else?
With fixed Kelly wagering the 13.53% ror is your starting ror. If you were to double your money your ror would drop to 1.83% moving forward provided you kept the same size bets.
If you win enough then your ror cannot be 13.53%. If you quadruple your bank then your ror is .03% moving forward. If you are playing with an edge and don't go broke but win over time then your ror would drop. How could it be anything else?
let me ask you this: if pure kelly betting (resizing and all) doesn't work in the "short run", or a fixed number of hands, ie you can go FULLY broke, how will that player ever make it to the long run with no money left, ie infinite # of hands? if it works in the long run it must work in the short run otherwise you would never reach the long run, right?
even for a fixed number of hands a pure kelly bettor should still have a theoretical ROR of 0%.
another way to think about this: you have a bankroll of $100. you bet a fraction of it every time, and for simplicity sake, let's say that fraction is 50%. Now, you bet once and lose. bankroll is now $50 remaining. bet 50% or $25 and lose. bankroll remaining is $25. bet 50% or 12.50, and lose. bankroll remaining is 12.50. bet 50% again and lose, bankroll remaining is 6.25. bet 50% again and lose, bankroll remaining is 3.12. bet 50% again and lose, bankroll remaining is 1.56. bet 50% again and lose, bankroll remaining is 0.78. bet 50% again and lose, bankroll remaining is 0.39, etc etc etc. you are always halving something and will never reach zero THEORETICALLY. now take into account minimum denominations of currency and table minimums and THEORETICAL goes out the window. BUT still, THEORETICALLY, a pure kelly better won't ever go broke.
even for a fixed number of hands a pure kelly bettor should still have a theoretical ROR of 0%.
another way to think about this: you have a bankroll of $100. you bet a fraction of it every time, and for simplicity sake, let's say that fraction is 50%. Now, you bet once and lose. bankroll is now $50 remaining. bet 50% or $25 and lose. bankroll remaining is $25. bet 50% or 12.50, and lose. bankroll remaining is 12.50. bet 50% again and lose, bankroll remaining is 6.25. bet 50% again and lose, bankroll remaining is 3.12. bet 50% again and lose, bankroll remaining is 1.56. bet 50% again and lose, bankroll remaining is 0.78. bet 50% again and lose, bankroll remaining is 0.39, etc etc etc. you are always halving something and will never reach zero THEORETICALLY. now take into account minimum denominations of currency and table minimums and THEORETICAL goes out the window. BUT still, THEORETICALLY, a pure kelly better won't ever go broke.