He said "finite", so not necessarily. Let's say you're going to shoot for 5 bets and then quit. 1/2^5 times you make it, the rest of the time you bust out.

Their question was how to "maximize their wealth", not their EV. Nevermind a bunch of people, suppose it's one person. All-in maximizes EV and therefore average wealth. If that's what was desired then fine, but IIRC, Kelly maximizes the median wealth.

Oops that analogy doesn't hold, because the money isn't pooled together and if one person is reduced to 1/10 of BR, they won't have enough to make the wager dictated by my quote. So then the answer (for maximizing growth) might be as simple as everyone kelly-betting for themselves.

And of course the expected value of betting everything is simply your staring bankroll times (1 + your edge) to the power of the number of pre specified bets (even though you won't usually make them all ). If you were getting 3-1 on an even money shot, started with five bucks, and were willing to bet ten times, your EV is 5 x 1024 which is more than if you bet whatever Kelly says to bet.

In my finance classes I too was taught that Kelly maximizes median wealth. I believe it is true both asymptotically as well as in many finite situations.

I think a lot of responses are confusing what "horizontal' ergodicity with vertical.

If we consider time as transpiring graphically, from left to right, horizontal ergodicity would be, say, 1,000 guys go to a casino and play a given game. In vertical ergodicity, one guy play the same game for 1,000 trials.

In a game with a "classic," ( i.e. positive probability-weighted mean outcome) expectation that is positive, 1,000 guys go into a casino and the expected aggregate gain among the 1,000 guys is maximized by betting 100% of their bankroll (if they are smart, they will pool their money then split the aggregate winnings).

On the other hand, for the individual doing repeated trials (because of reinvestment) the expected growth optimal fraction will approach optimal f value, and the leverage factor (i.e. Kelly) will be the same as the expected growth optimal fraction (because we are talking about gambling, not capital markets applications/ In captital market applications the Kelly Criterion solution is often > 100%, and not equal to the expected growth-optimal fraction.)

The reason I think it is important to distinguish between horizontal and vertical ergodicity (aside from the reinvestment factor in vertical) becomes more apparent as we get into more perverted distributions of outcomes.

Consider 1000 plays where you win $1 with p=.999 and lose $1,000 and 1-p=.001. This has a "classic" negative expectation.

But let;s say you are only going to play it a few times, or one time, lets call this number of trials N. In this example, you expect to make $N. However, as N gets larger, your expectation approaches the classic expectation. The point is, what we use for expectation, the classic expectation, the difference between vertical and horizontal ergodicity is hilited, the "real" expectation of N trials approaches the classic expectation asymptotically (just as the growth optimal fraction approaches an asymptotic value -- which is NOT always the same value as one would derive using the Kelly Criterion).

Asymptotic values are useful if there are enough "vertical" trials, as a proxy for the correct value, and provided the distributions are relatively normal.

forgive my confusing horizontal and vertical on the previous post. I always do that, and have a difficult time with it evidently.

Mr. Ralph Vince himself! Welcome to the forums. I'd like to think my citation in an earlier post is what summoned you.

Yes, it did, and thank you. It;s a topic I'm both passionate about and painfully addicted to, unable to break the addiction, and appreciate the thoughts and insights of others on it, and seek them out.

I misplaced my password op here. So guys I've read the replies in this thread but most of these replies are way over my head , can someone break in down into layman's regarding the original op question and does it still stand that if you bet anything higher than 2.58% that you would lose overall?

thanks

Betting more than the growth-optimal fraction (which people refer to as Kelly here, which is not accurate as Kelly is something else, but let;s call it that for simplicity and consistency here), which is evidently a bet size of 2.58% (I'm not sure of the parameters of the bet here) does NOT garner you more profit, it garners you less. In fact, if you reduced that bet size by 1/2 %, you would make the same as if you increased it by 1/2%. So increasing it not only makes you less, you end up with wilder swings in equity.

Think of it as a curve, here the horizontal axis is your bet size (from 0 to 1,0) and the vertical axis, what you make. Since, when reinvestment occurs, and we make more than one play, what would otherwise be a straight line for what we would make, based on percent risked, is now curved, and the peak of that curve in this case in this example is 2.58%. So betting more than this, you end up at a lower point on the cure, a lower vertical coordinate (of how much you made).

It;s NOT that you would lose overall by betting more than Kelly, you just make less. This lesser amount, at some point as you increase your bet size, become negative and you DO lose overall.. One of the fortunate by-products of the way Kelly is calculated is that if you double the amount of the peak you get to that point (as a % of your bankroll) where you start going broke (this is so because Kelly is symmetrical, and it is symmetrical because it is not bound to the right at 1.0, [and thus, not a "fraction!"]) if you continue trading at 2 x Kelly.

I was under the impression we concluded.

Kelly maximises growth rate

Kelly is the median return

Betting over 2 Kelly causes the median punter to have a negative growth rate

The more you bet the higher your ev

No one has yet shown or provided any definitive argument as to why betting over 2 Kelly makes you less.

I still haven't gone through a proof, but since Kelly maximizes growth, moving away from it makes growth smaller, so it is intuitive that moving far enough away form the max will reach 0 since we can't go negative.

<<Kelly maximises growth rate>>

This is false - "Kelly" never results in the greatest expected growth rate. Kelly equals the expected growth optimal fraction in the "special case" (which is where what you can put up is what you can lose, which is true in most gambling situations, except blackjack btw, and rarely true in capital market situations). The expected growth optimal fraction asymptotes out at a value as the number of trials gets ever greater. It is this asymptotic value that many in the gambling community refer to as "Kelly.," but being asymptotic means it is ALWAYS a more conservative bet than that which is actually expected-growth optimal.

<<Kelly is the median return>>
I've never heard this, and see no reason why it should be. As specified, "Kelly, in the special case is the same value as the asymptotic (i.e. as the number of trials approaches infinity) expected-growth optimal fraction to wager.

<<Betting over 2 Kelly causes the median punter to have a negative growth rate>>
True, but only as the number of trials approaches infinity. Remember, at 1 trial, with a positive classical expectation, the expected growth optimal fraction is actually 1.0 (100%), which could well be double what Kelly is.

<<The more you bet the higher your ev >>
True provided there is no reinvestment. The moment you introduce reinvestment this becomes false.

<<No one has yet shown or provided any definitive argument as to why betting over 2 Kelly makes you less.>>

I'd be happy to, but that's a deeper dive, and I think a lot of guys here might get pissed, it may be off the theme of the forum itself, and I'm not here to step on toes. For me, what you guys call "Kelly," is something very misunderstood. It's something the gambling community refers to meaning a certain thing but it's been entirely misunderstood -- by not only Kelly himself, but Shannon as well. When understood in it's proper context, it is a small part of something much bigger, with far broader applications than gambling and capital markets.

For example, the "phi" point, that is, that fraction where the expected growth multiple drops below 1.0 i(in other words, you keep multiplying your stake by a value less than 1.0 with each subsequent play, and therefore, as the number of plays increase, the stake itself approaches 0) is, in a larger context, of far more importance than the peak (you can refer to the peak as Kelly since we are talking the special case).

There are many growth functions where, unlike as gamblers or traders where we want the growth to be positive, you would want it to be negative. Take, for example, the growth of a country's total debt. This is something that we would want to diminish, not grow.

Instead of a fraction to bet, we can use the cosine of the average of the data points used to determine the growth function and the standard deviation of those data points in such cases, such that as we increase standard deviation, all else being the same, it is akin to a higher value for the fraction we would be wagering. Therefore, at a high enough standard deviation, we reach the phi point in a growth function like national debt.

From the beginning, governments have looked at the debt problem as occurring on a continuum, a tug-of-war between raising taxes or cutting services, but here we now see a third and even more powerful and politically-agnostic approach.

With variance the square of standard deviation, we can state (and prove) unequivocally, that any increase in the variance in period-on-period borrowing (through policy that seeks to increase this variance), all else being equal, affects the aggregate national debt in the same manner as reduction in spending or a hike in taxes ...squared.

As you can see, there's so much to all of this, in all sorts of fields, and it's only now just being looked at seriously. It gets particularly confusing in that what the gambling community calls "Kelly" is really not the asymptotic expected growth-optimal fraction, but only equals the asymptotic expected growth-optimal fraction in the "special case" of most gambling situations.For the sake of clarity, I tend to be pretty orthodox in these things.

Continue making posts like the one above and I'll gladly keep reading them.

+1 This forum welcomes any and all discussion on these topics.

Mr. Vince has written extensively on these topics and his thoughts and posts are always welcome.

The entire "Kelly" milieu can be a morass of confusion and misunderstanding. Even in a straightforward gambling environment in which a gambler has a starting bankroll and faces a series of bets over time, several issues can be confusing or misunderstood:

- finite vs infinite horizon

- mean vs median

- maximizing ending bankroll vs overall long-run growth rate of bankroll

- etc.

To repeat, discussion on any of these topics in this forum is very welcome and strongly encouraged.

+2, count me among those who would welcome deep-diving posts of yours, Mr. Vince. Uttering your name may turn out to be my biggest contribution to TwoPlusTwo!

Regarding the 2f rule, I might be able to prove it for a single-wager scenario. But is it even true for say, a portfolio of simultaneous wagers? After calculating the growth-optimal allocation, would doubling each asset's stake be the phi point?

Maybe I will try to write up something reasonably concise, and try to dispel the confusion (and generate further discussion) as best I can. It might take a series of posts in fact. I see this site does not allow one to put up images, so I will have to try to describe things here (it would be so much easier to show it graphically).

As for the phi point (the point where the growth multiple "beyond the peak," per trial, is <1.0) for multiple simultaneous, consider for N games you have a curve in an N+1 dimensional manifold or "space." (thus, for one game, you have a curve on a 2D plane). Assume a coin toss that pays 2:1, you have a curve with a (asymptotic) peak at .25

For, say, two games, you have a surface in N+1 space, or 3D space. with a single peak at .23,.23 representing wagering 23% on each game each trial (and thus, you are risking 46% per trial in the aggregate now). There is a phi point around this peak, "beyond the peak on each axis (thus, an arc on this 3D surface). In other words, though you are wagering on two games simultaneously, it is possible to wager so much on one of those games (or both) that you re at or beyond phi. So it takes only ONE of the N games to get you to phi! This is very interesting in terms of what it means regarding diversification in capital markets.

A better idea would be to do this as a series of youtube videos, putting the links here. I need to think about how best to convey this, to express the totality of it simply. It;s not that confusing if expressed along a certain path.....

Youtube would take too long to put together. I see what I can do, put a graphic on a server and then link to it from here.
Ok, I'm gong to start a deep dive, across a wide swatch, in 4 or 5 length posts with graphs, trying to not get into the math (because, a large part of my focus has been coming up with simple, heuristic ways to do close approximations to the math, that are quite simple).
This should clarify a lot of this stuff here - at least that;s my goal. I just need to put to get the time for it, over the next week or so I hope. Further, I hope it generates a ton of ideas by others here.

Maybe after I get this done, if it is successful here,if it explains things clearly, Ill get someone from the art schools down here to help put together some informative youtube videos on it.

if you are placing a bet where you win 50% and get paid 1.1:1

your wins follow a symmetric binomial distribution, where the median number of wins is 50.

using proportional betting after n bets your roll would be

Roll=R0*(1+r*1.1)^(0.5*n)*(1-r)^(0.5n)

the median roll is maximised by choosing r to be Kelly

<<Betting over 2 Kelly causes the median punter to have a negative growth rate>>
In the above equation choosing r to be double kelly causes Roll to equal R0. Exceeding double kelly, causes Roll < R0. n can be anything.

<<The more you bet the higher your ev >>

Your EV after n bets involves a binomial distribution, when you win 50% this is symmetric. After n bets your EV is.

EV = (Win=0, loss=n)*Roll( after winning 0 and losing n)
+(Win=1, loss=n-1)*Roll( after winning 1 and losing n-1)
+.....
+(Win=n, loss=0)*roll(after winning n)

as you make n grow to infinity, you lose all term apart from the last, which becomes,

EV = lim n~inf (0.5)^n*(1+r*1.1)^n

clearly diverges. and diverges faster with bigger r.

You can also pair off all the different outcomes, chance of losing all is the same as winning all, nut you make more when you win them all. The chance of winning n-m and losing m, is the same as winning m and losing n-m, but again you get paid more when you win n-m (0<m<n/2)

As someone who doesn't even understand basic mathematics, I actually understood most of this post, so thanks for explaining in easy to read, only thing I didn't quite understand was the bold part?

Because the gambling community (and much of the capital markets community) still believes Kelly is the expected growth-optimal fraction, because it is equal to the expected growth-optimal fraction in most gambling situations. But Kelly is really a "Leverage Factor," and hence not a fraction per se.

In capital markets applications, routinely people would arrive at conclusions where, in satisfying the Kelly Criterion (according to Kelly in his 1956 paper, geometric growth is maximized by the gambler betting a fraction such that, "At every bet he maximizes the expected value of the logarithm of his capital.") yields results greater than 1. It took me a few years to understand this, so I am not faulting Kelly or Shannon, who reviewed Kelly's paper, for missing this. The Kelly Criterion solution actually doesn't give you the fraction, but rather, what multiplier you should put on your stake.

In other words, a vale for Kelly of 2.0 means you should trade in size relative to an account twice the size. A value of .02 means you should trade the account (full-out) as though it was only 2% of the size.

Given the nature of gambling, where what you can lose is what you put up, the growth-optimal fraction is the same as Kelly, and therefore it is confusing in the world of gambling. The important point to understand is that Kelly is not the growth optimal fraction, it only equals it in the special case, and this special case is ubiquitous in the world of gambling. Capital markets, on the other hand, are fraught with great danger here because most implementors believe Kelly is the same as the expected growth-optimal fraction, and usually end up with values for Kelly which are greater than the real expected growth-optimal fraction (because people don;t understand what Kelly is and is not) and thus over-extend, are way to the right of the peak.

A deep-dive on this can be found on my site :

http://ralphvince.com/

(I'm not hustling the sight, I dont have ads there, I only use it as a repository mostly) on the "related papers" tab there should be a link to a paper in a journal along the lines of "Optimal f and the Kelly Criterion," which will go into this in-depth if anyone is interested)

I see now the reference to the median. Thank you.

It sometimes is illuminating to look at very simple examples.

Suppose you are a gambler with a starting bankroll of 1. Suppose you are faced with a series of 10 independent binary bets each of which pays 1:1 and each of which wins 60% (and loses 40%). Suppose further that you must wager a fixed fraction of your then-current bankroll on each of the ten bets. Call this fixed fraction F.

As David and others in the thread have stated, and is quite easy to see in this case, if you want to maximize the expected value of your ending bankroll probability distribution, then you should choose F=100%.

Also, as has been proven analytically for this case, if you want to maximize the median value of your ending bankroll probability distribution, then you should choose F=Kelly.

Where Kelly is given by the formula (bp-q)/b. In our simple case this is (1*.6-.4)/1 = .2 = 20%. I think it can easily be demonstrated numerically that this fraction maximizes the median outcome.

I coded this up in a simple program. Here is a table of how the Expected Value and Median Value vary depending upon the Fixed Fraction chosen.

Fixed Fraction	Expected Value	Median Value
0%	1.00	1.00
10%	1.22	1.16
20%	1.48	1.22
30%	1.79	1.16
40%	2.16	0.98
50%	2.59	0.71
60%	3.11	0.43
70%	3.71	0.20
80%	4.41	0.05
90%	5.23	0.01
100%	6.19	0.00

Of course, the issues highlighted by this simple example barely scratch the surface of the entirety of this topic. But I wanted to include it in case it helps elucidate some of the very basics.

Sry for noob question but does F=100% mean bet whole bank?