Kelly betting

EasyRhino

Well-Known Member
#41
Fixed full Kelly has a 13.5% risk of bankroll going to zero.
Continuous full Kelly has a 13.5% risk of bankroll going to 13.5%.

Both are bad. One is worse.
 
#42
I Agree With You, However

rukus said:
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.
I am not sharing Kasi's lunch money!:joker::whip:
 
#43
Optimal Does Not Equal Bad!

EasyRhino said:
Fixed full Kelly has a 13.5% risk of bankroll going to zero.
Continuous full Kelly has a 13.5% risk of bankroll going to 13.5%.

Both are bad. One is worse.
Both are bad?

Fixed Kelly is what it is, how comfortable is one with a 13.53% ror, 1% or 20%, it's subjective.

Continuous resizing Kelly is optimal.
Just not completely practical in the real world.

The trick is how to get as much of the positive properties as possible of resizing Kelly in the real world while trying to avoid potential real world pitfalls.
 

Kasi

Well-Known Member
#46
rukus said:
even for a fixed number of hands a pure kelly bettor should still have a theoretical ROR of 0%.
Hey I'm glad more people are getting involved in this. Especially people that don't want my milk money :) I'm afraid to go to school now lol.

OK, I get if you always bet a fraction of something, you'll always have something left. So, OK, he'll never ever go broke if you want to put it like that.

Maybe what I'm wondering about is say a "pure" Kelly bettor does play a fixed number of hands. Maybe even a fairly large number of hands. Perhaps my question would be better phrased as "After playing 1MM sessions of at most 1MM hands each session, how often would he have less than 1%, or 0.1%, whatever, of his original roll left?" Maybe even either at some point during those 1MM hands or even at the end? Would that happen in somewhat close to 135,300 sessions of the 1MM sessions?

Would it be true if he ever did get to say 1% of his roll left, wouldn't it take a heck of a long time anyway from that forward to just double it to 2% of original roll?

Would having some very small percentage of his starting roll left after 1MM hands, or even at some point during the 1MM hands, be approaching 13.53%? Like, say, how often would he stop playing if at some point he had lost at least 99%, or 99.9% if u want, of starting roll?

Isn't there anyway a 10% chance of losing 90% (not sure if that is exactly 90% or maybe "at least" 90% or not?) (or even if it's right lol) of his original roll at some point?

What is the magic of that 13.53% anyway? Why does, what does it mean that, "=exp(-2/1)" (Excel formula) equal 0.1353 and the 1 is the Kelly fraction? Does it not apply to a "pure" Kelly bettor as well as an "equivalent Kelly bettor"? Maybe at least practically speaking anyway?

I don't care how you're betting but, if you lose 99% or 99.9% of your starting roll at some point, you ain't getting rich anytime soon. Even if that 0.1% you have left is being bet in a way that will maximize its growth. If a fixed-spread Kelly bettor says to himself with a $10K starting roll, "I will play at most 1MM hands or just quit if I ever have only $100 left", after losing 99% of his roll, how often will he have to quit compared to a "pure" Kelly bettor who says the same thing with the same starting $10K roll? Would it be just as often?

BJ Avenger - never fear. I'm Kelly-betting my milk money so there will always be some left for you :grin:
 
#47
Kasi said:
Hey I'm glad more people are getting involved in this. Especially people that don't want my milk money :) I'm afraid to go to school now lol.

OK, I get if you always bet a fraction of something, you'll always have something left. So, OK, he'll never ever go broke if you want to put it like that.

Maybe what I'm wondering about is say a "pure" Kelly bettor does play a fixed number of hands. Maybe even a fairly large number of hands. Perhaps my question would be better phrased as "After playing 1MM sessions of at most 1MM hands each session, how often would he have less than 1%, or 0.1%, whatever, of his original roll left?" Maybe even either at some point during those 1MM hands or even at the end? Would that happen in somewhat close to 135,300 sessions of the 1MM sessions?

Would it be true if he ever did get to say 1% of his roll left, wouldn't it take a heck of a long time anyway from that forward to just double it to 2% of original roll?

Would having some very small percentage of his starting roll left after 1MM hands, or even at some point during the 1MM hands, be approaching 13.53%? Like, say, how often would he stop playing if at some point he had lost at least 99%, or 99.9% if u want, of starting roll?

Isn't there anyway a 10% chance of losing 90% (not sure if that is exactly 90% or maybe "at least" 90% or not?) (or even if it's right lol) of his original roll at some point?

What is the magic of that 13.53% anyway? Why does, what does it mean that, "=exp(-2/1)" (Excel formula) equal 0.1353 and the 1 is the Kelly fraction? Does it not apply to a "pure" Kelly bettor as well as an "equivalent Kelly bettor"? Maybe at least practically speaking anyway?

I don't care how you're betting but, if you lose 99% or 99.9% of your starting roll at some point, you ain't getting rich anytime soon. Even if that 0.1% you have left is being bet in a way that will maximize its growth. If a fixed-spread Kelly bettor says to himself with a $10K starting roll, "I will play at most 1MM hands or just quit if I ever have only $100 left", after losing 99% of his roll, how often will he have to quit compared to a "pure" Kelly bettor who says the same thing with the same starting $10K roll? Would it be just as often?

BJ Avenger - never fear. I'm Kelly-betting my milk money so there will always be some left for you :grin:
If you play long enough as a continuous resizing player you will reach your goal, if you are playing a positive expectation game. How can it be anything else?

If you resize your bets based on bank swings your long run is increased. The more frequently you resize the longer the time horizon, but the more optimal (log growth?)

Yes, you have a 10% chance of losing 90% of bank with Kelly resizing

The 13.53% is just the ror if you ask the question:
My total ror is 0% with Kelly resizing, but what is it at a very specific moment in time, and that is 13.53%.

For fixed bets 13.53% is just a ror number like 1% or 15% there is nothing magical about it.

fixed betting
resized betting
are two different things
resized kelly is optimal, just harder to implement

Did I answer everything?
 

Brock Windsor

Well-Known Member
#48
Kasi said:
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".
.
Those are two good points in this mess that escape a lot of folks I think. Betting double theoretical Kelly your bankroll gets infinitely close to zero. Like a frog that jumps half way to the wall with every leap, will he ever reach the wall?? Point #2 the "risk" in ROR. If you always played at %5 ROR, had a 100K BR and lived off of your winnings and never reinvested anything into your BR you will eventually lose the 100K, it is a certainty. A lot of rec counters I talk to about this don't seem to quite comprehend either point... but maybe I just socialize with too many knuckle draggers. Good discussion!
BW
 

rukus

Well-Known Member
#50
Kasi said:
Hey I'm glad more people are getting involved in this. Especially people that don't want my milk money :) I'm afraid to go to school now lol.
Ha! Ive been busy last few weeks but im back these days. this is an interesting thread and i need to go back to the beginning of it and see if there something i can contribute to the OP's thoughts.

Kasi said:
OK, I get if you always bet a fraction of something, you'll always have something left. So, OK, he'll never ever go broke if you want to put it like that.
yup, ie theoretically. isn't theory wonderful?

Kasi said:
Maybe what I'm wondering about is say a "pure" Kelly bettor does play a fixed number of hands. Maybe even a fairly large number of hands. Perhaps my question would be better phrased as "After playing 1MM sessions of at most 1MM hands each session, how often would he have less than 1%, or 0.1%, whatever, of his original roll left?" Maybe even either at some point during those 1MM hands or even at the end? Would that happen in somewhat close to 135,300 sessions of the 1MM sessions?
I have to think about this some more. need to turn the blackjack math part of my brain back on now that i am back.

Kasi said:
Would it be true if he ever did get to say 1% of his roll left, wouldn't it take a heck of a long time anyway from that forward to just double it to 2% of original roll?
shouldnt take any longer to go from 1% to 2% of original role than it would have taken to go from 100% to 200%. both would require doubling the current bankroll, which shouldnt change with the size of the bankroll if you are always betting the same fixed fractions of them at the appropriate times. so not sure how this point is significant.

Kasi said:
Would having some very small percentage of his starting roll left after 1MM hands, or even at some point during the 1MM hands, be approaching 13.53%? Like, say, how often would he stop playing if at some point he had lost at least 99%, or 99.9% if u want, of starting roll?

Isn't there anyway a 10% chance of losing 90% (not sure if that is exactly 90% or maybe "at least" 90% or not?) (or even if it's right lol) of his original roll at some point?

What is the magic of that 13.53% anyway? Why does, what does it mean that, "=exp(-2/1)" (Excel formula) equal 0.1353 and the 1 is the Kelly fraction? Does it not apply to a "pure" Kelly bettor as well as an "equivalent Kelly bettor"? Maybe at least practically speaking anyway?
13.5% is not magical for a resizing kelly bettor other than what Rhino and Avenger already mentioned - the bettor would have 13.5% chance of reaching 13.5% of his original bankroll. alternatively, as Avenger stated, 13.5% can be the probability that you reach 13.5% of your current bankroll at any point in time when using continuous resizing kelly betting.

for a fixed kelly bettor, 13.5% is the instantaneous ROR of the bettor if he fixes his bets using a kelly fraction =1. now as soon as the fixed bettor plays a single hand and his bankroll changes up or down, his new ROR is NOT 13.5% anymore. if he properly resized his bets (using f=1) after that first hand and again planned to fix his bets, his new ROR from tha tpoint forward would be 13.5% again. BUT, obviously if the bettor continues to resize his bets after each time his BR changes, he is no longer playing fixed kelly but is actually using continuous resizing and his theoretical ROR goes to 0%, not 13.5%.

Kasi said:
I don't care how you're betting but, if you lose 99% or 99.9% of your starting roll at some point, you ain't getting rich anytime soon. Even if that 0.1% you have left is being bet in a way that will maximize its growth. If a fixed-spread Kelly bettor says to himself with a $10K starting roll, "I will play at most 1MM hands or just quit if I ever have only $100 left", after losing 99% of his roll, how often will he have to quit compared to a "pure" Kelly bettor who says the same thing with the same starting $10K roll? Would it be just as often?

BJ Avenger - never fear. I'm Kelly-betting my milk money so there will always be some left for you :grin:
i agree if you lose that much of your bank you wont be getting rich anytime soon. using proportional betting with resizing, it will take you a long time to rebuild to your original bankroll. think about it this way: on day 1, how long does it take you to double you original bankroll? let's say it was calculated as 1000 hours (i made this up for example purposes). now if you somehow unluckily draw down to 1% of your original bank, using the same proportional betting scheme, it should take you about 1000 hours to go from 1% to 2% of the original bank. then another 1000 hours to get back to 4%, another 1000 hours to get to 8%, and on and on. like you said, you wont be getting rich anytime soon - not only that but it will take you a long long time to build back up just to your original bank. if you are using fixed kelly, if you hit 1% of your original bank and do not resize, you will be going broke very shortly and wont EVER get rich.

if you define 1% as the stopping point for a fixed kelly bettor, given the stats for the game he is playing (EV/hand Std dev/hand), you can easily calculate how often he will have to stop within 1mm hands using the formulas in BJA3 (though you'll have to write some code since the formula is not closed-form). alternatively you might be able use Norm's online calculators (he might have the first version where you can calculate probability of reaching some goal above your starting bank within a given number of hands but maybe he doesnt have posted the opposite calculator - how often you will lose a certain amount within a given number of hands). give me some ev and std dev numbers and ill plug them into my own spreadsheets which do do the calculation when i get home.

i have not mathematically compared these types of results vs those of a resizing bettor (ie probability of reaching 1% of an original bank and then stopping), but you can intuitively reach the proper conclusion:

assume the two players each bet on the same box at a table and start with the same bankroll (and the table minimum bet = 1% of this starting bankroll).

Now, the resizing bettor's bankroll would grow faster than the fixed bettor if the table box experienced positive results (ie wins). if hte box experiences losses, the fixed better would deplete his bankroll faster and thus reach 1% and have to stop more often than a resizing bettor who is cutting his bets as his bankroll decreases with the losses.

as you can see, the faster increase in bet sizes (and bankroll) during wins and the lowered downside (due to lower bet levels) during losses is what makes continuous resizing the most optimal betting strategy in terms of maximizing the log of bankroll growth.

make sense?
 
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sagefr0g

Well-Known Member
#51
Kasi said:
....

What is the magic of that 13.53% anyway? Why does, what does it mean that, "=exp(-2/1)" (Excel formula) equal 0.1353 and the 1 is the Kelly fraction? Does it not apply to a "pure" Kelly bettor as well as an "equivalent Kelly bettor"? Maybe at least practically speaking anyway?

....
i've wondered the same thing about that number.
Schlesinger on page 155 chpt 9 of Blackjack Attack says something about it has to do with fixed-Kelly that maximizes log-growth of bankroll and leads to the 13.5% ROR number.
and some how him and some colleagues working on this SCORE stuff used the fixed-Kelly procedure i guess as a part of getting SCORE sort of standardized.
thing is Schlesinger points out you can still mess around doing fractions of Kelly and change the size of your ROR.
thing about that number that freaks me out is i guess it's just coincidental that happens to be the 'normal' number for standard error of human plus equipment error my old quantitative analysis chemistry professor used to spout about. :confused:
i guess it is what it is excepting maybe to a numeroligist. :)
 

Kasi

Well-Known Member
#52
rukus said:
i have not mathematically compared these types of results vs those of a resizing bettor (ie probability of reaching 1% of an original bank and then stopping)
Thanks for your reply and same to all.

Actually it all does make sense to me lol. I think. At least sometimes lol.

I have seen, I think anyway maybe, a "fixed-spread kelly" bettor actually being defined as the same spread, like always 1-6, but the min unit always changing compared to current roll. If so, like you say, I guess the same as a "pure" Kelly bettor in that neither could ever go broke and both would would be betting in a way that would minimize the time for doubling of original roll?

Anyway, and I don't know, I still think a "pure" continuously re-sizing better, if limited to 1MM hands or 10MM hands per "session", might have an extremely small amount left of starting roll very close to 13.53% of his total number of sessions.

And, even if so, maybe it's all mostly anyway, maybe practically speaking?, a matter of semantics, "forever" vs 10MM trials, lol.
 

MAZ

Well-Known Member
#53
Kasi said:
Thanks for your reply and same to all.

Actually it all does make sense to me lol. I think. At least sometimes lol.

I have seen, I think anyway maybe, a "fixed-spread kelly" bettor actually being defined as the same spread, like always 1-6, but the min unit always changing compared to current roll. If so, like you say, I guess the same as a "pure" Kelly bettor in that neither could ever go broke and both would would be betting in a way that would minimize the time for doubling of original roll?

Anyway, and I don't know, I still think a "pure" continuously re-sizing better, if limited to 1MM hands or 10MM hands per "session", might have an extremely small amount left of starting roll very close to 13.53% of his total number of sessions.

And, even if so, maybe it's all mostly anyway, maybe practically speaking?, a matter of semantics, "forever" vs 10MM trials, lol.
Kasi you are smarter than you think, the rest are just puppets of misinformation. Stop perpetrating the fraud kasi, you get it and you pretend not to, See ya soon.

MAZ
 

Kasi

Well-Known Member
#54
MAZ said:
Kasi you are smarter than you think, the rest are just puppets of misinformation. Stop perpetrating the fraud kasi, you get it and you pretend not to, See ya soon.MAZ
Seriously? You're kidding me right? You actually think these are rhetorical questions I'm asking and I already know the answer? It's OK if this is your way of calling me an idiot or something because, trust me, I wouldn't disagree.

A man's gotta know his limitations and I know mine. All I can do is add and subtract. Throw in multiply and divide. That's it. Know zilch about statistics, never took a course in it, probability - flunked it in 10th grade to the best of my recollection - maybe it was only a test or 2, not sure, and forget about calculus and derivatives, exponents and logarithms, let alone an "infinite series" whatever that is lol. OK, I know what it is, it's just that I usually can't solve one.

I'd never call anyone here a "puppet of misinformation" for starters just because I don't know enough to call anyone that lol.

Intentionally perpetrating a fraud just so people think I'm modest when I all along know I'm not? You think I'm that big of a jerk?

But, hey, if you are actually being sincere in some way and think I may have made a point in all this, I am glad to hear it because I respect your opinion, MAZ.

And I'd love to know what "see ya soon" means. You golfing at Ligonier CC tomorrow or something lol?
 
#55
It's Good to Agree

Kasi said:
Thanks for your reply and same to all.

Actually it all does make sense to me lol. I think. At least sometimes lol.

I have seen, I think anyway maybe, a "fixed-spread kelly" bettor actually being defined as the same spread, like always 1-6, but the min unit always changing compared to current roll. If so, like you say, I guess the same as a "pure" Kelly bettor in that neither could ever go broke and both would would be betting in a way that would minimize the time for doubling of original roll?
BLACKJACK AVENGER'S RESPONSE
:joker::whip::joker::whip::joker::whip:
Yes, a continuous resizer can have a betting spread/ramp.
A continuous resizing Kelly better can have a 1 to 4 or 1 to 5 spread etc., what makes them a continuous resizer is; as you state, as their bankroll wins or loses they adjust their bets accordingly.
I have no problem with your definition "fixed-spread Kelly" bettor

Probably no player in the real world is a continuous resizer, but if you resize at all in many ways one exhibits the characterisitcs of a constant resizer.
 
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#56
Run to the Long Run

Anyway, and I don't know, I still think a "pure" continuously re-sizing better, if limited to 1MM hands or 10MM hands per "session", might have an extremely small amount left of starting roll very close to 13.53% of his total number of sessions.

And, even if so, maybe it's all mostly anyway, maybe practically speaking?, a matter of semantics, "forever" vs 10MM trials, lol.[/QUOTE]

BLACKJACK AVENGERS RESPONSE
:joker::whip::joker::whip::joker::whip:

A continuos resizer faces a standard deviation similar to a fixed better. However, the long run is greater for the resizer. If you have a positive advantage then the more you play the more your expectation will be realized.

A continuous resizer will lose 99% of their bank in the theoretical world 1% of the time.

Continuous resizing is very difficult if not impossible in the real world.:joker::whip:
 

Kasi

Well-Known Member
#57
blackjack avenger said:
A continuos resizer faces a standard deviation similar to a fixed better. However, the long run is greater for the resizer. If you have a positive advantage then the more you play the more your expectation will be realized.
See, as soon as we may possibly agree, you only, as usual, bring up more stuff that I wonder about lmao.

Like, why would SD differ no matter how one is betting? Like a "pure" Kelly bettor has the same SD in that situation as any other bettor? Not sure, as usual what "fixed" bettor means. But say how would SD differ for a "pure" Kelly bettor who gets paid 1.01 for "Heads" in a fair coin-toss game compared to a "fixed" bettor? Even in BJ, wouldn't the SD be whatever it is for that particular play at that particular time, no matter how one may choose to bet it? Surely the variance in the fair toin coss case, whatever it is, would be the same for each bettor on each bet? Would they need to have different starting rolls so that if each were down -3SD they would have same amount of money left?

That aside, why are you saying a "re-sizer" has a longer "long-run" than a "fixed" bettor? Would not each double their roll in the fewest number of trials? Or even have 1SD=EV in the same number of trials? Would they need to begin with different original rolls to make this happen? Would actual results being -3SD after the same number of trials be a different result with same starting roll? If "Kelly" is "optimal", why would a "fixed" bettor have a longer "long-run"?

Which leads to, and maybe this is what I really don't fully understand, take a "Kelly" bettor with a $10K roll, as in Don's tables, playing-all, including -EV situations, with that 13.53% ROR and a fixed $ spread that does not change based on current bankroll remaining. If he actually could bet that fixed $ spread, even in all neg counts, he is minimizing the number of rounds to double roll given risk, isn't he? He is minimizing the time to "long-run" isn't he? How would a "pure" Kelly bettor, playing only on + counts compare to that? Would he double his roll in the same number of hours while not being able to play every hand since he only would play rounds with an advantage? In order to do so, would his starting roll be different than the $10K for Don's bettor? Is that "equivalent-kelly-betting"? Would not each have the same chance, since after all each is playing at an average +EV, have the same "the more you play the more your expectation will be realized" chance?

blackjack avenger said:
A continuous resizer will lose 99% of their bank in the theoretical world 1% of the time.
I agree with that, at least it makes sense to me - should have thought of that in the first place lol. Maybe I asked you this before but would you say that means, given that I guess it assumes one plays "forever", does that mean exactly 1% of the time or "at least" 1% of the time? In other words, if he loses 90% of orig roll 10% of the time, would the fact he loses 99% of his starting roll 1% of the time already be included in the fact he loses 10% of his roll 90% of the time? Like does it mean "1% or "10%" more" kind of thing if that even makes any sense lol.

blackjack avenger said:
Continuous resizing is very difficult if not impossible in the real world.:joker::whip:
See, even that, while "agreeing" with it lol, I'm just going to go out on a limb and say, it is, absolutely, unequivocably and utterly, 100% impossible to theoretically achieve in the real world. And this from one who can't even say with 100% certainty that the sun will rise tomorrow lol. So I guess, yet another technical disagreement after all. Along the same lines I can't agree the theoretical ROR is "0%" for a Kelly bettor. Approaching it as you get closer to infinity but never reaching it. Even every casino in Vegas, with a little "bad luck", ok maybe a lot of "bad luck" could lose every dime it has ever made from this second forward. :joker: :grin: :whip:

Just throwing out alot of questions to anyone - that's all.

And, no, it's not like I'll be revealing the "answers" later :grin:
 

ThodorisK

Well-Known Member
#58
In the real world it is not necessary to follow any betting system. Even if you greatly overbet kelly and lose your bankroll, you will have another bankroll in the future, and if you overbet again, you will have another bankroll in the future, and so on. Evetually you are bound to make a profit from all these trials, which will be close to the EV=(money wagered)(edge), unless you are the most unlucky person on earth. And if you want to win 1,000,000$ with a starting bankroll of 1,000$ by betting kelly, then you will die before that unless you are the most lucky person on earth.

Kelly does not maximize the growth of the bankroll. Saying this is like saying it maximizes EV. It maximises EV only if you have exactly the average luck. Overbetting kelly has the same EV as betting kelly if the money wagered is the same. EV is always (money wagered)(edge). The only difference is that by overbetting you win too much if you have a rare good luck, and you will not win that much if you do not overbet. This counterbalances (and explains) the negative EV that betting more than 2 times kelly gives when you have the average luck. If this was not the case, one could beat roulette by using a betting system that equals forcing the casino to bet more than 2 times kelly.

See the kelly formula I gave you in my previous posts in this thread, to understand what I mean.

Hm, however, as the number of played hands tends to infinity, the luck you will have will tend to the average luck. And if you have the average luck then there is a negative EV if you bet more than 2 times kelly. This is a paradox. Can roulette be beaten? LOL.
 
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#59
Can't Beat a Negative Expectation Game

ThodorisK said:
In the real world it is not necessary to follow any betting system. Even if you greatly overbet kelly and lose your bankroll, you will have another bankroll in the future, and if you overbet again, you will have another bankroll in the future, and so on. Evetually you are bound to make a profit from all these trials, which will be close to the EV=(money wagered)(edge), unless you are the most unlucky person on earth. And if you want to win 1,000,000$ with a starting bankroll of 1,000$ by betting kelly, then you will die before that unless you are the most lucky person on earth.

Kelly does not maximize the growth of the bankroll. Saying this is like saying it maximizes EV. It maximises EV only if you have exactly the average luck. Overbetting kelly has the same EV as betting kelly if the money wagered is the same. EV is always (money wagered)(edge). The only difference is that by overbetting you win too much if you have a rare good luck, and you will not win that much if you do not overbet. This counterbalances (and explains) the negative EV that betting more than 2 times kelly gives when you have the average luck. If this was not the case, one could beat roulette by using a betting system that equals forcing the casino to bet more than 2 times kelly.

See the kelly formula I gave you in my previous posts in this thread, to understand what I mean.

Hm, however, as the number of played hands tends to infinity, the luck you will have will tend to the average luck. And if you have the average luck then there is a negative EV if you bet more than 2 times kelly. This is a paradox. Can roulette be beaten? LOL.
There are formulas and thoughts expanding kelly when considering taking money out of the bankroll over time or adding to the bankroll over time.

Even in the short term, optimal means optimal.:joker::whip: To say Kelly continuous resizing is the best in the long term but not the short term, or several short terms over time is incorrect. Our play is continuous, sessions are artificial breaks in continuous play.

The reason betting over double kelly is negative EV is because the bankroll drop when losing cannot be made up by wins. Go play some practice positive hands of bj where you bet resized 25% of bank on each hand, you will see how you can have a positive edge yet the bankroll spirals down.

I put it to you, what is the style of betting that is superior to Kelly continuous resizing?

Did you agree that given average luck Kelly is superior? Well over time your luck would converge to average and Kelly would prove optimal.
 

EasyRhino

Well-Known Member
#60
ThodorisK said:
Kelly does not maximize the growth of the bankroll. Saying this is like saying it maximizes EV. It maximises EV only if you have exactly the average luck.
I suspect you may be intentionally talking in circles now.

Important disclaimer: we do not, in advance, know which hands will win and which will lose. If we did, the bankroll maximizing play would be to bet all of the bankroll on the winning hands, and none on the losing ones. There are lots of things that can be calculated by looking at historical results, but they don't give us any useful info for decision making. We need info that gives us information about future results.

All we know is the side of the relative advantage in the odds of the bets... hence Kelly betting. It is meant to play out superiorly in the long run, but there is no superior method in the short run.
 
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