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Old 12-19-2016, 04:02 PM   #1
confusedandlost
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Kelly Criterion Simultaneous Bets Math

I have two bets that will occur at the same time. They both have a different EV. Bet A will have a higher EV than Bet B. I understand that Kelly Criterion is to continually adjust my bets according to my bankroll. I am under the impression I have to use a different number for bankroll to calculate the second bet with the assumption that the first bet is a loss even thought they will occur at the same time. For the second bet, do I adjust the number inputted into the bankroll for the equation of the second bet.

Does it matter which bet I treat as the first bet if I do adjust the bankroll number in the second bet.

I understand that this question is inane and it's probably pointless as the edges shouldn't be that different in the calculations to matter much, but in a purely academic hypothetical isn't it obvious that I need to treat the higher EV bet as the first bet so I can make bigger bets at a bigger edge. The risk or ruin should remain the same hence the "growth of my bankroll" should only depend on higher EV bets at higher portion of the bankroll.
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Old 12-19-2016, 04:52 PM   #2
confusedandlost
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Re: Kelly Criterion Simultaneous Bets Math

After googling, I now see there is a different calculator. Any insights more into this would be greatly appreciated. I'm not really a math guy.
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Old 12-21-2016, 12:54 AM   #3
Smiling
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Re: Kelly Criterion Simultaneous Bets Math

For Kelly Criterion you size bets based on ev, risk & bankroll at the time you place the bet. Concurrent bets are each based on bankroll at the time of bet so long as they are independent of each other. You couldn't adjust size for hypothetical results. You wouldn't know whether to increase or decrease.

Not clear regarding your risk of ruin comment. Theoretic Kelly has no risk of ruin. Practical Kelly has positive risk of ruin because (I) your bankroll is not infinitely divisible, (II) your bet sizing is not perfectly optimal and (III) at some point your bankroll would be so small it's not worth your time.

Hopefully, you're using fractional Kelly because estimating ev can be tricky and the real world usually doesn't accommodate optimally sized bets.

Good luck.
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Old 12-21-2016, 12:12 PM   #4
heehaww
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Re: Kelly Criterion Simultaneous Bets Math

Simultaneous wagers are a different animal altogether, as each wager leads to a new variable in the growth equation. Afaik, you can't feed hypothetical outcomes into the single-wager formula except in the special case of when each bet has a 1:1 payout (though I've been meaning to try to find a way). Even in that case, that alternative approach isn't vastly simpler than the standard one, a system of partial derivatives. For anyone curious anyway, I can explain that approach (which I doubt is published anywhere).

Most likely, the online calculators are using numerical methods.

Quote:
Originally Posted by confusedandlost View Post
make bigger bets at a bigger edge.
The kelly % is not always larger for higher edge bets. A longshot with a humongous edge can require a smaller wager than a bet with a slim edge that isn't a longshot. If a bet only has two possible outcomes, the kelly % cannot exceed the probability of winning.
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Old 03-07-2017, 09:59 PM   #5
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Re: Kelly Criterion Simultaneous Bets Math

Quote:
Originally Posted by confusedandlost
I have two bets that will occur at the same time. They both have a different EV. Bet A will have a higher EV than Bet B. I understand that Kelly Criterion is to continually adjust my bets according to my bankroll.
To use the Kelly criterion to allocate capital among several different simultaneous investments, you need to know more than just the EVs - you also need to know the distribution of the outcomes.

Here's an example: Without loss of generality, suppose your initial capital is 1 unit, and say you will invest a fraction x of your capital in bet #1 and y in bet #2. The two bets are as follows:
  1. You either gain 5x with probability 0.4 or lose x with probability 0.6. (EV=1.4)
  2. You either gain 20y with probability 0.1, gain 4y with probability 0.2, or else lose y with probability 0.7. (EV=2.1)

After both bets resolve, there are 6 possibilities for how much capital you will be left with:
  • 1+5x+20y (with probability .4 * .1 = .04)
  • 1+5x+4y (with probability .4 * .2 = .08)
  • 1+5x-y (with probability .4 * .7 = .28)
  • 1-x+20y (with probability .6 * .1 = .06)
  • 1-x+4y (with probability .6 * .2 = .12)
  • 1-x-y (with probability .6 * .7 = .42)

The Kelly criterion is to maximize the expectation of the logarithm of your capital. So the objective here would be:

maximize .04 log(1+5x+20y) + .08 log(1+5x+4y) + .28 log(1+5x-y) + .06 log(1-x+20y) + .12 log(1-x+4y) + .42 log(1-x-y)
subject to 0 ≤ x, 0 ≤ y, x + y ≤ 1

According to Wolfram Alpha, the maximum is achieved when x≈0.2563 and y≈0.1725. So in order to maximize the rate of growth of your capital, you would invest about 25% of your capital in bet #1 and 17% in bet #2.
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Old 03-20-2017, 03:20 PM   #6
whosnext
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Re: Kelly Criterion Simultaneous Bets Math

Prompted in part by heehaww's post above, I thought I would try to solve for the simplest case. Two independent bets each with exactly two outcomes (win and lose) and each with positive EV. (Based upon heehaww's post, I think that he and probably others have previously derived these formulas so I claim no primacy or originality.)

Without loss of generality, consider your original bankroll to be 1. We are interested in determining the optimal fraction of your bankroll to bet on each of two simultaneous and independent bets.

Let there be the following two bets offered to you.

Bet 1: prob win = P; if you win you receive Ax in profit if you bet x units

Bet 2: prob win = Q, if you win you receive By in profit if you bet y units

Based upon P, Q, A, and B, we seek the optimal x and y.

Mathematically, the Kelly criterion prescribes that a bettor bet an amount that maximizes the expected logarithm of the bettor's ending bankroll. In the single bet environment, this leads to a fairly simple (and well-known) formula. We will come back to this below.

Setting partial derivatives to zero and solving yields two very messy equations that do not appear to present a closed-formed solution. However, since you can plug one equation into the other equation, you can essentially wind up with a complicated non-linear equation in one variable. So numerical techniques can be used to solve for the optimal pair in a very short amount of time.

We are going to have a bunch of really long terms. I attempted to "simplify" all of the expressions, but looking back on it maybe I should have kept the terms in their original forms. For simplicity in what follows I have omitted all of the multiplication symbols. So, for example, ABPQ stands for A*B*P*Q and AAQY stands for A*A*Q*Y.

X3TERM = AA

X2TERM = AAQY-AAPQY-AP-AAP+AAPY-ABPQY-APQY-AABPQY+APY+AABQY+2A-AY+ABY-AA-AABY

X1TERM = 2AQY-AQYY+2ABQY+AAP+AABPY-AAPY-AABPYY-P-BPY+PY+BPYY+ABBQYY+1+BY-Y-BYY-2A-3ABY+AY+ABYY-ABBYY

X0TERM = (1+BY)*(1-Y)*(-APQY+QY-PQY+AP+ABPY-ABPQY-1+P-BY+BPY+BQY-BPQY)

Let E = (3*X3TERM*X1TERM-X2TERM*X2TERM)/(3*X3TERM*X3TERM)

Let F = (2*X2TERM*X2TERM*X2TERM-9*X3TERM*X2TERM*X1TERM+27*X3TERM*X3TERM*X0TERM)/(27*X3TERM*X3TERM*X3TERM)

Then the optimal X can be expressed in terms of Y and the other variables:

X = 2*SQRT(-E/3)*cos((1/3)*arccos((3*F/2E)*SQRT(-3/E))-2*PI/3)-(X2TERM/(3*X3TERM))

Similarly, by partial derivatives we can solve for Y in terms of X.

Y3TERM = BB

Y2TERM = -BBQ+BBQX+BBPX-ABPQX-BBPQX+2B+ABX-BX-BB-ABBX+ABBPX-BQ+BQX-BPQX-ABBPQX

Y1TERM = BBQ+ABBQX-BBQX-ABBQXX+2BPX-BPXX+1+AX-X-AXX-2B-3ABX-AABXX+BX+ABXX-Q-AQX+QX+AQXX+2ABPX+AABPXX

Y0TERM = (1+AX)*(1-X)*(BQ-BPQX+ABQX-ABPQX-PQX+PX+Q-1+APX+AQX-APQX-AX)

Let G = (3*Y3TERM*Y1TERM-Y2TERM*Y2TERM)/(3*Y3TERM*Y3TERM)

Let H = (2*Y2TERM*Y2TERM*Y2TERM-9*Y3TERM*Y2TERM*Y1TERM+27*Y3TERM*Y3TERM*Y0TERM)/(27*Y3TERM*Y3TERM*Y3TERM)

Then the optimal Y can be expressed in terms of X and the other variables:

Y = 2*SQRT(-G/3)*cos((1/3)*arccos((3*H/2G)*SQRT(-3/G))-2*PI/3)-(Y2TERM/(3*Y3TERM))

As I said above, we need to solve these two equations for X and Y. One approach is to plug the X equation (X as a function of Y) into the Y equation. This yields a single highly non-linear complicated equation in one unknown (Y). Numerical techniques are available to solve this equation in a matter of seconds. Of course, once we have the optimal Y, the optimal X is immediate.

Here are three cases for which I have calculated the optimal simultaneous bets and compared them to the original optimal single-bet Kelly condition (as if each were the only bet available).

Variable Case 1 Case 2 Case 3
P
.333
.250
.400
Q
.250
.200
.200
A
4
6
3
B
5
8
6
SingleKelly1
.166
.125
.200
SingleKelly2
.100
.100
.066
SimulKelly1
.162457
.120781
.197088
SimulKelly2
.094190
.096020
.062041

where SingleKelly refers to the optimal bets under the Single-Bet Kelly Criterion and SimulKelly refers to the optimal bets under the Simultaneous-Bet Kelly Criterion.

Anyway, not sure that this is overly interesting, especially since, even after the partial derivatives are solved, numerical techniques are still required to derive the optimal bet sizes. That is, the partial derivatives are nice and everything (lol), but numerical techniques could have simply been applied to the original expectation equation without knowing anything about the partial derivatives.

Any and all comments are of course welcome.
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Old Today, 03:16 PM   #7
DarkMagus
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Re: Kelly Criterion Simultaneous Bets Math

I think khaaan has it right. Basically you just treat all of the simultaneous bets together as a single bet with a large number of outcomes (equal to the product of the number of outcomes of each individual bet). The EV equation is just

EV = profit(outcome 1) * probability(outcome 1) + profit(outcome 2) * probability(outcome 2) + ...

For two binary bets, e.g. coinflips, there'd just be four outcomes: win both, win a/lose b, win b/lose a, lose both.
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