Please restore my sanity (kelly criterion) - Gambling and Probability

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Please restore my sanity (kelly criterion)

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01-13-2019 , 03:20 PM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

There is an advantageous bet that you can repeat forever. If you bet x amount, one of 3 things can happen:

a) Win w*x with probability p=.59375, where w = .42174
b) Lose L*x with probability q=.395833, where L = .272
c) Lose x with probability r=.010417

I tried to answer this: what % (call it f) of your bankroll should you wager each time to maximize expected growth (or median future wealth)?

My attempt:

g = (1+.42174f)^.59375 * (1-.272f)^.395833 * (1-f)^.010417

Numeric optimization in R:

Code:

kelly = function(p,w) {
  fn = function(f,p,w) prod((1+f*w)^p)
  optimize(fn,p=p,w=w,lower=0,upper=1,maximum=TRUE,tol=.Machine$double.eps)$maximum
}
w <- c(.42174, -.272, -1)
p <- c(.59375, .395833, .010417)
paste(kelly(p,w))

Result: 0.793143075975138

My own optimization algorithm in Julia:

Code:

@fastmath function kelopti(p::Float64, r::Float64, w::Float64, L::Float64, firstep=1/16)
    q = 1-p-r
    f, g, step, upper = 0.0, 0.0, firstep, 1.0
    while(true)
        newf = f+step
        while(true)
            (0 ≤ newf ≤ upper) ? break : step/=2
            newf = f+step
        end
        newg = (1+w*f)^p * (1-L*f)^q * (1-f)^r
        if newg ≤ g
            step /= -2
            step==0.0 && return f
        end
        f = newf
        g = newg
    end
end

kelopti(.59375, .010417, .42174, .272) = 0.7931431029303357

But f can't be >p, can it? So I tried an analytical solution. I took the derivative of ln(g), set it to 0 and used Sagemath to solve for f:

f = {Lpw + Lp + Lqw + w - L - qw - √[(qw + L - Lp - Lpw - Lqw - w)² - 4Lw(p + q + pw - Lq - 1)] } / (2Lw)

which gives an answer of 0.7931430966114893

Lastly, I tried a random simulation but same result.

It appears that all roads point to f>p, but how in the hell can that be? I can't bring myself to believe it, but I also don't know where my mistake is. My guess would be the growth equation, but if so, I have a fundamental misunderstanding of growth.

Here's why I refuse to believe f>p. When there are only 2 possible outcomes, the kelly equation is f = p - (1-p)/w. Here, if you remove Outcome C, observe that the kelly stake will be smaller than p:

p' = p/(p+q) ≈ .6
w' = w/L ≈ 1.5505
f' = .6 - .4/1.5505 = .342

The effect of inserting a 3rd outcome worse than B should be to decrease f, not increase it.

Can someone explain what's going on? Tyvm

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01-15-2019 , 11:31 AM

nickthegeek

journeyman

Join Date: Sep 2011 Posts: 273

Not an expert on the topic by any mean, but the main point I guess it's that you rarely lose your entire bet. The f<p condition mostly holds if you lose f when you lose. Here, assuming that you bet f = 0.7931, you are actually risking f*L, which is about 0.21, way less than p.

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01-16-2019 , 04:04 PM

jukofyork

Carpal \'Tunnel

Join Date: Sep 2004 Posts: 11,749

Quote:

Originally Posted by heehaww

So I tried an analytical solution

It's really easy to use Wolfram Alpha for this:

Solve[D[0.59375*Log[1+0.42174*x]+0.395833*Log[1-0.272*x]+0.010417*Log[1-x],x]==0,x]

(it sometimes gets confused if you try to add too many outcomes though... if it does, then try to bunch up the expression and remove all unnecessary whitespace between terms)

Quote:

But f can't be >p, can it?

f can even be >1! This indicates that you should use leverage (this isn't common for gambling-related use of Kelly, but for the quant-related use it is, eg: http://epchan.blogspot.com/2006/10/h...d-you-use.html).

(but because you added the 3rd outcome which is a "total loss" the value will obviously never go above 1 for this case)

BUT:

Quote:

Here's why I refuse to believe f>p. When there are only 2 possible outcomes, the kelly equation is f = p - (1-p)/w. Here, if you remove Outcome C, observe that the kelly stake will be smaller than p:

p' = p/(p+q) ≈ .6
w' = w/L ≈ 1.5505
f' = .6 - .4/1.5505 = .342

The effect of inserting a 3rd outcome worse than B should be to decrease f, not increase it.

I'm not quite sure what you've worked out here, but if you just remove the 3rd "total loss" outcome, re-normalise the remaining probabilities and then use the same utilities, you get an example where use of leverage is optimal:

Solve[D[0.6*Log[1+0.42174*x]+0.4*Log[1-0.272*x],x]==0,x] = ~1.25743

Juk

Quote

01-17-2019 , 07:29 AM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

I came on here to post what I think is the explanation, and it turns out Nick already said it.

It seems instead of setting the unit to the largest possible loss, I should set it to the most probable loss.

When I do that, the long formula and my julia optimization both get 21.5734 %

(which is the same as .272 * .79314)

I'm sure that's much closer to the real answer, but I suspect the real answer has to weigh both perspectives. Maybe the real answer is simply the arithmetic weighted avg of both, 23.054 %

@juk - I don't have time right now to really process your post. I skimmed it and I'm skeptical, but I'll give it a proper read later with an open mind.

Thanks to both for responding

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01-17-2019 , 07:57 AM

jukofyork

Carpal \'Tunnel

Join Date: Sep 2004 Posts: 11,749

Quote:

Originally Posted by heehaww

@juk - I don't have time right now to really process your post. I skimmed it and I'm skeptical, but I'll give it a proper read later with an open mind.

This blog post goes into more detail about the use of kelly for asset allocation and leverage:

http://epchan.blogspot.com/2014/08/k...portfolio.html

Juk

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01-18-2019 , 10:22 AM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

@juk - thanks I'll give that a read soon. Still short on time because I'm traveling.

Quote:

Originally Posted by me

It seems instead of setting the unit to the largest possible loss, I should set it to the most probable smallest loss.

FMP, or keep using the largest loss but then multiply the f by the smallest loss.

Quote:

I'm sure that's much closer to the real answer, but I suspect the real answer has to weigh both perspectives. Maybe the real answer is simply the arithmetic weighted avg of both, 23.054 %

No, I think I get it now: both f's were correct but I misinterpreted the larger one. They actually tell the same thing, as was demonstrated earlier. The largest f applies when the smallest loss is counted as one unit, and the smallest f applies when the the largest loss is counted as one unit. Either way, we want our bet size to be such that the largest possible loss is 21.5% of our bankroll.

I've extended this to 4 outcomes and it's the same thing.

Oh and my kelopti() function needs the following line:

Code:

...
(newf==upper || newf==0.0) && return newf
f = newf
g = newg

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01-21-2019 , 03:00 PM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

No no no, my previous two posts were wrong. 79% of our bankroll is right because most of the losses are less than one unit. When I removed the 3rd outcome I also increased the unit for the more-probable loss, which I shouldn't have done because that made it a different problem altogether. And if the win amount is 1 and the loss L is smaller than 1, the two-outcome formula becomes f = p/L - q, so even with 2 outcomes f can be >p (and indeed >1 as well). Nothing about this is strange, I simply confused myself.

Thanks guys

Quote

01-22-2019 , 08:10 PM

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Quote:

Originally Posted by jukofyork

It's really easy to use Wolfram Alpha for this:

It's just as easy to use Sagemath's server - https://sagecell.sagemath.org/

Unlike WA, it has no limitations (such as computation time or character limit), because it's free and open source if you want to install it.

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