Kelly Criterion Explained
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(Editor: I saw this post about the Kelly Criterion
on another website and decided to post it here. Interesting math theory here. Warning that it's
not for the faint of heart. If you want an easier explaination, view the Kelly made simple post.)
Date: Mon, 13 Jul 1998 16:59:45 -0700
From: Tom Weideman
Newsgroups: rec.gambling.poker
Subject: Kelly criterion (long, but give it a chance!) ;-)
Joesmallie wrote:
Please correct me if I am wrong. I used to use the Kelly Criterion for
playing the horses. If I remember correctly it was "edge diveded by odds."
We then made an adjustment whereby we used a 1/2 Kelly. When you say Kelly
Criterion as applied to a poker situation are you doing the same thing?
Could you please define Kelly Criterion. Is there a web site that has a thorough
explaination. I am less than a stellar mathematician. When people start
posting formulas and such about how they derived certain calculations I
become blurry eyed. Would love to be able to follow them better but they leave me
in the dust. I have math questions all the time but hesitate to bring them up
here and bog down the poker newsgroup. Would love to have a math tutor.
Thank you.
I haven't checked the r.g.* FAQ, so this may be redundant, but since you
asked, I'll give as simplified a version as I can. I'll only show
details of the easy math, while glossing over the calculus, so non-math
types can (hopefully) follow along.
Suppose you are playing a game in which you have some sort of edge.
This edge can consist of any combination of probability and odds such
that each bet has a positive expectation. Now you ask yourself, "What
fraction of my bankroll should I wager on this game?" If your answer
is, "The amount that if I keep repeating this same strategy over a long
period of time, my bankroll increases at its maximum rate.", then your
answer is to bet "Kelly".
To see how we can find such an answer, we should look at it from the
beginning. Suppose you bet a fraction f (0 > f <>1) of your current
bankroll B, and you are getting odds of v-to-1. If you win the bet, you
get back your original bet, plus v times the amount of your bet (which
was f*B). If you lose, you lose the bet amount:
Bet size = f*B, odds = v-to-1, amount won = v*f*B, amount lost =
f*B
If you win, your new bankroll B' is going to equal your old bankroll
plus your winnings:
Win: B' = B + v*f*B = (1+v*f)*B
If you lose, your new bankroll will equal your old bankroll minus your
losses:
Lose: B' = B - f*B = (1-f)*B
Let's take a short timeout to make sure these funny looking equations
make sense by looking at an example (I'm a huge fan of concrete
examples).
Say you have $100, and you bet $20, getting 2-to-1. For the sake of
future considerations, we'll assume the game is a fair coin flip (but
the need for the probabilities does not come into play for awhile yet).
If you win this bet, you win $40, raising your bankroll to $140. The
equation above shows this correctly:
B = $100, f = 1/5 = 0.2, v = 2: B' = (1+v*f)*B = (1+2*0.2)*($100) =
$140
The "Lose" equation works in a similar manner.
Now, suppose you play the game multiple times. Assuming no "pushes"
(which I will be assuming throughout), then the "new" bankroll (which we
have called B') after a game is played becomes the "old" bankroll for
the next game to be played. We are assuming the game and the odds
offered don't change, and that we do not choose to change our strategy
(the fraction of bankroll to be wagered), so the equations given remain
the same for each game, with the previous game's B' becoming the next
game's B. Your result after winning twice in a row would be:
B after two wins = (1+v*f)*[B after one win] = (1+v*f)*[(1+v*f)*(starting B)]
B after two wins = [(1+v*f)^2]*(starting B)
A check: Suppose in the example above you won again, employing the same
strategy. Your new bankroll was $140, so you wagered 1/5 of it ($28)
and won at 2-to-1, for a total win of $56. Now your new bankroll is:
$140+$56 = $196. Plugging into the equation gives the same result:
B after two wins = [(1+v*f)^2]*(starting B) = [(1+2*0.2)^2]*($100)
B after two wins = [(1.4)^2]*($100) = [1.96]*($100) = $196
The two losses results work out the same way. What about a win and a
loss, you ask? You just multiply the starting bankroll by the "lose
factor" (1-f), and the "win factor" (1+v*f), and the result is the new
bankroll. Note that it doesn't even matter if you lost first or won
first. Back to our example:
You won the first game and lost the second: $100 + $40 = $140 after
first game, then lost $28 (you wagered 1/5 of it) in the second game for
a total of $140-$28 = $112. If you lost first, you have $100 - $20 =
$80 after the first game, and then you wager 1/5th of $80 (=$16) at
2-to-1 in the second game and win, for a total of $80 + 2*($16) = $112.
Same amount as if you win first. [BTW, this less than obvious fact may
affect retirement planning for those of you thinking about Roth IRA's
vs. traditional ones. I won't digress any further on this topic.]
Using the equation gives the same result:
B after 1 win and 1 loss = (1+v*f)*(1-f)*(starting B)
B after 1 win and 1 loss = (1+2*0.2)*(1-0.2)*($100) = (1.4)*(0.8)*$100 = $112
Okay, so lets say we have won "w" times out of n total games. This
means we have lost n-w times (since we assumed no pushes). To find the
new bankroll, we need to multiply the starting bankroll by (1+v*f) a
total of w times, and multiply it by (1-f) a total of n-w times. In
other words, our "new bankroll equation" has gotten much more
complicated:
B' = [(1+v*f)^w]*[(1-f)^(n-w)]*B
The first factor in brackets is just the "win factor" multiplied by
itself w times, and the second is the "lose factor" multiplied by itself
n-w times. The two "B's" in this equation are not important, so we will
instead look at just the factor multiplying B, as this is what
determines the bankroll's growth (or lack thereof):
B'/B = [(1+v*f)^w]*[(1-f)^(n-w)]
This gives us the factor that the bankroll has changed from the
beginning (after n games). We want to look at the bankroll's growth
rate PER GAME, so if we call the average-per-game-factor "y", then after
n games, the bankroll has grown by a factor of y^n. The
average-per-game-factor is then found to be:
B'/B = y^n
y = (B'/B)^(1/n) = {[(1+v*f)^w]*[(1-f)^(n-w)]}^(1/n)
Again we return to our example. After 2 games where we won 1 and lost
1, we have gone from $100 to $112. This is an increase of a factor of
1.12 for 2 games. On average, this is an increase PER GAME of a factor
of sqrt(1.12) = 1.058 . In other words, if in the game described, we
win the same number that we lose (i.e. we assumed way back when that the
probability of winning is 1/2), on average we will increase our bankroll
each game by a factor of 1.058. Now here's the important part:
*** If we employ a different strategy (risk a different fraction "f" of
the bankroll), then this factor will also change. We seek to find the f
for which this factor is a MAXIMUM. ***
Now finding the value of f for which this function peaks is no small
matter. It involves calculus. If this intimidates you, I invite you to
jump down to below the second set of "*'s" to see the answer. I include
the calculus for the math.weenies that may find it interesting...
********
The value of f for which y(f) is a maximum is the same value for which
ln[y(f)] is a maximum, so we can equivalently seek to maximize:
z(f) = ln[y(f)] = (1/n)*[w*ln(1+v*f) + (n-w)*ln(1-f)]
The derivative is:
dz/df = (w/n)*v/(1+v*f) - [1-w/n]/(1-f)
Setting equal to zero and finding f gives:
********
f = [p*(v+1)-1]/v, where p = w/n.
Note that in the long run, the fractional number of times you will win a
game (w/n) equals the probability of winning a single game, so p =
probability of winning.
Okay, this is our answer. Given that your chance of winning is p, and
that you are receiving v-to-1 odds on your bet, then the fraction of
your bankroll that you should wager to maximize your rate of bankroll
growth is the f given above. This value can also be plugged back in
above to find out what the maximum growth rate actually comes out to be.
Let's try it for the game we've been using as an example. We have a
probability of p=0.5 of winning, and odds of v=2, so the fraction of our
bankroll we should risk each time is:
f = [p*(v+1)-1]/v = [0.5*(2+1)-1]/2 = 1/4
y = [(1+v*f)^p]*[(1-f)^(1-p)] = 1.061 (recall p=w/n)
The bankroll increases an average of 6.1% over its preceding amount
every game. [Using the "rule of 72" familiar to bankers and investors,
this means the bankroll will double roughly every 12 games.]
Most people like to remember the Kelly criterion using a mnemonic that
goes something like: "Bet the fraction of your bankroll that equals your
percentage advantage." It should be understood that this ONLY applies
to bets with even money odds (v=1). Note that with v=1, f comes out to
be equal to 2p-1, which is exactly your percentage advantage. The
origin of this mnemonic is probably from blackjack, which would explain
why the even money assumption is made. A better, more general mnemonic
would be:
"Bet the fraction of your bankroll equal your percentage advantage
divided by the 'to-1' odds."
For example, if you have a 10% advantage and you are getting 5-to-2
odds, then the fraction of your bankroll to bet is (0.10)/2.5 = 0.04 =
4%.
If you plowed through this whole post, my compliments. It's possible
that the only people willing (able?) to follow it all the way through
are the people who already understand all this, which means I was
drawing dead as I wrote it. I hate when that happens.
Tom Weideman
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