Am I stupid? I can't fit these two concepts into any type of harmony.

[BeerMoney]

[ QUOTE ]
this one seems easy. your bankroll is finite, $1 trillion you said. so of course probability dictates that you will eventually go bust.

[/ QUOTE ]

Do you think a casino will eventually go bust then?

If he was able to constantly play under the same conditions his bankroll would go towards infinity.

[PairTheBoard]

[ QUOTE ]
"Interesting. If you're playing with an edge and you bet more than 2 times Kelly, your win rate suddenly becomes negative. Maybe you should present that at the next gambling conference of math. They'll probably go for it. I hear you can say anything there."

PairTheBoard

Except that I believe he is right. (But don't confuse that with the statemnet that the expected value of your bankroll will go down.)

[/ QUOTE ]

Ok. I recall now Stanford Wong's treatment Here where he explains the term "Win Rate" as BillC means it. I don't know who introduced the terminology but it seems an unfortunate one to me since people commonly refer to the Arithmetic Win Rate as the "Win Rate" as well, when they are two different things. I suspect this terminology developed in the gambling world and Not in the Academic world of Mathematics. I think Mathematicians would have come up with a better name.

Basically, the Wong Win Rate for proportional betting defines a Bankroll Growth which the Bankroll will Converge to in Probabilty over the long run. In other words, there will be a high probabilty that the Bankroll will be close to that determined by the Wong Win Rate after a long period of time. Suprisingly, for Kelly betting that limiting curve is where you would be if the return on your actual action was half what your edge says it should be.

However, even though there is a high probabilty your bankroll will be close to the Wong Win Rate Curve, the reality is that your Bankroll remains a random variable whose distribution has enough probabilty spread out into extreme ranges that your Expected Return on Action remains what it should be according to your edge.

BillC was correct when he indicated that the Wong Win Rate becomes negative when your proportional betting exceeds twice Kelly. That's really just a fancy way of saying that in the long run your probablity of being close to zero gets larger and larger. But that's not to say your Arithmetic Win Rate has become negative. Even with the high probabilty of being close to zero, the Expected Value of your Bankroll remains determined by your positive edge and your Action.

I think some qualifying word should be added to the term "Win Rate" when referring to the Win Rate Curve which the Bankroll converges to in probabilty. If Wong introduced the term I'd vote for calling it the "Wong Win Rate". At least Something to distinguish it from the more commonly understood Arithmetic Win Rate, ie. the Win Rate determined by EV.

PairTheBoard

[FlFishOn]

You bust.

[diddle]

[ QUOTE ]
You bust.

[/ QUOTE ]

No you don't. You become incredibly rich.

[BillC]

[ QUOTE ]
"Interesting. If you're playing with an edge and you bet more than 2 times Kelly, your win rate suddenly becomes negative. Maybe you should present that at the next gambling conference of math. They'll probably go for it. I hear you can say anything there."

PairTheBoard

Except that I believe he is right. (But don't confuse that with the statement that the expected value of your bankroll will go down.)

[/ QUOTE ]

I was assuming betting at more than twice Kelly, so that you might have to readjust your bets to keep the condition true. Then your bankroll has an expected negative slope=rate. Is there any other "rate" in this case? If you are betting a fixed amount, then your strategy can't be betting "more than twice Kelly".

An extreme example is where you bet your entire bankroll all the time.

The fact alluded to is that (under some assumptions) betting exactly twice Kelly results in an expected rate of bankroll growth of zero.

[PairTheBoard]

[ QUOTE ]
BillC --
Then your bankroll has an expected negative slope=rate.

[/ QUOTE ]

You really don't know what you're talking about do you?


[ QUOTE ]
BillC --
The fact alluded to is that (under some assumptions) betting exactly twice Kelly results in an expected rate of bankroll growth of zero.

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Definitely NOT an "expected rate of bankroll growth of zero".

Read my explanation above for what the Wong Win Rate means. That's the so called "Win Rate" you are referring to and it is NOT an "expected rate of bankroll growth".


Do you still Agree with him David?


PairTheBoard

[BillC]

Well, OK you have a point--maybe it is more precise to say

the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.

[PairTheBoard]

[ QUOTE ]
Well, OK you have a point--maybe it is more precise to say

the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.

[/ QUOTE ]

If you have a July magazine article please provide the link here, and if you want to cite a paper cite it here.


It's not the "the rate of growth of the expected bankroll" either.

This is always assuming you are playing with an edge. Take the case where proportional betting is More than twice Kelly. Yes, in the long run, your Bankroll will converge in Probabilty to a negatively sloping curve that asymtopes to zero. But your Expected Bankroll will always be Greater than your Initial Bankroll. Your Expected Bankroll will Always be your Inititial Bankroll plus your Edge times your Action.

Again, playing with an edge. Consider the simple case of suicide Martingale Betting where you bet your bankroll until you lose. Clearly, your Bankroll converges in probabilty to zero. ie. Your Wong Win Rate is negative. But even as your Bankroll is converging in Probabilty to Zero, your Expected Bankroll climbs throughout.

Or are you now using the term "Expected Bankroll" to mean something other than the Expected Value of your Bankroll?

PairTheBoard

[PairTheBoard]

[ QUOTE ]
BillC --
the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.


[/ QUOTE ]

I understand that under proportional betting the Bankroll can be aproximated by geometric brownian motion.

Again, you are wrong in saying, "the rate of growth of the expected bankroll is zero at twice Kelly".

However, I should correct one of my statements as well. It's Not the case that the Bankroll converges in probabilty to the curve defined by the Wong Win Rate.

Wong's Win Rate

Normalizing the initial bankroll to be 1 unit the convergence in probabilty is for this expression:

ln(Bn)/n

where Bn is the bankroll after n plays.

This converges in probabilty to what Wong calls the Bankroll's Exponential Growth Rate, G. Wong goes on to define the Win Rate, r, to be

r = exp(G) -1

The idea is that the nth root of your Bankroll is behaving like exp(G). For example, if your Bankroll was increasing by a factor of .05 on each play, after n plays your Bankroll would be (1.05)^n, just like n years of 5% compound interest. With continuous compound interest of 5% you get the exponential growth of exp(.05t). One year of 5% continuous compound interest gives you exp(.05)-1 for your Annual Percentage Rate. Inutitively, r is acting like you think of your edge acting. But it's not your edge.

For double Kelly betting r is zero. For over double Kelly betting r is negative. For Kelly Betting, r is Half your Edge. And here Wong says, "when you bet the optimal proportion of your bankroll, your expected win divided by your bet size is half of your expected arithmetic win rate."

At this point I am throwing up my hands. When he refers to "bet size" he is referring to the "proportion" not the actual action. Your expected win divided by your actual action remains equal to your edge. And to make matters worse he calls r/f your "rate of return on action under proportional betting", where f is your Kelly proportion.

This mish mash of terminology including terms like "win rate" and "expected win" as he uses it above has got me shaking my head.



PairTheBoard

[PairTheBoard]

[ QUOTE ]
Well, OK you have a point--maybe it is more precise to say

the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.

[/ QUOTE ]

Ok Bill. After looking at your article Risk Formulas I'll settle for the term "Expected Growth Rate" as in the Expected Growth Rate for the Bankroll. However, I don't see what the "Expected" can refer to except the Expected Value of a specially defined "Growth Rate" random variable.

PairTheBoard