#### Kelly betting on high variance games

by , 10-28-2012 at 10:59 AM (13056 Views)

I will give this a stab - please be Gentle .I have actually asked a similar question on the old bjinfo but wanted to revisit this as I found some conflicting answers.

Generally speaking, when betting full Kelly you divide the edge% by the pay-off a winning bet to arrive at the % of your bankroll that you should wager on your bet.

Does this apply equally regardless whether your pay-off is 1:1, 5:1, 10:1, 50:1 or 100:1? Does this rule of thumb change if you are playing with a high edge (say 30% or 40% as an example).

Firstly, I hate this terminology, because it confuses punters and it confuses me.

I someone quotes a payoff of 1:1, I assume they actually mean if you bet one bet, if you lose you get zero and if you win you get 2 units:

That is ev(lose) = -1 and ev(win) = 1.

I will take this as my starting point, such that for a payoff of 1:m we have

ev(lose) = -1 , ev(win) = m.

Now here is my first problem, this tells me nothing about

Prob(win) = p, Prob(lose) = 1 - p.

This is how casinos rip off unsuspecting punters.

I can take a guess: for Roulette we have a supposed 1:1 payoff for Red vs Black.

For us 'lucky' Australian and UK players, there is only one '0' on the wheel and so

Prob(win) = 18/37 = 0.486, Prob(lose) = 0.514.

ev(win) = 1, ev(lose) = -1.

Total ev per spin is

ev = 1 x 0.486 + (-1) x 0.514 = -1/37 = -0.027 = -2.7%

var = 0.486 x (1)^2 + 0.514 x (-1)^2 = 1.0

This gives values of

ekb = var/ev = -37

N0 = var/(ev)^2 = (1)/(1/37)^2 = 1369 spins .

Since win rate is ev = ekb/N0 per spin we get -37/1369 = -0.027 = -2.7% CORRECT !!

Since this game has negative ev, clearly you give it a wide berth.

Now translating these formulas into general form, we have for a payoff of 1:m

ev(lose) = -1 , ev(win) = m

Prob(win) = p, Prob(lose) = 1 - p.

So

ev = ev(win) x Prob(win) + ev(lose) x Prob(lose)

= m * p +(-1) * (1-p) = mp - 1 + p = (m+1)p - 1

or

ev(m,P) = P(m+1) - 1 ........................ (1)

var(m,p) = p * (m)^2 + (1-p) * (-1)^2 = p.(m)^2 + (1-p) = p.m^2 - p + 1 = p(m^2 - 1) + 1

var(m,p) = p(m^2 - 1) + 1 .............................(2)

Time for sensibility checks:

ev(m,P) > 0 iff (m+1)P - 1 > 0

(m+1)P > 1

The domain of P is [0,1] => (m+1) > 0.

By the definition of 'Payoff' = 1:m we assume m >= 1.

=> ev(m,P) > 0 iff P > 1/(m+1) ............................... (3).

For our Roulette case above P = 0.486, m=1 : 0.486 < 1/2 and so ev(1,0.486) = 2 * 0.486 - 1 = - 0.027 < 0 CORRECT !!

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Now var(m,P) > 0 always by the definition of variance.

So

var(m,P) = P(m^2 - 1) + 1 > 0 iff P(m^2 - 1) > -1 ...................... (4)

Now since the domain of P is [0,1] and m is [1,inf)

var(m,P) > 0 is self evident : CORRECT !!

Roulette check:

var(1,0.486) = 0.486(1 - 1) + 1 = 1.0 : CORRECT !!

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Now

ekb(m,P) = var(m,P)/ev(m,P)

ekb(m,P) = [ P(m^2 - 1) + 1 ] / [ P(m+1) - 1 ] > 0 iff P(m^2 - 1) > -1 ...................... (5)

N0(m,P) = var(m,P)/[ev(m,P)]^2 > 0 .................................................. ....... (6).

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For a Kelly bettor betting a unit $B we have

EKB(m,P) = $B x ekb(m,P)

EKB(m,P) = $B . [ P(m^2 - 1) + 1 ] / [ P(m+1) - 1 ] > 0 iff P(m^2 - 1) > -1 ...................... (7)

EV(m,P) = $B . [ P(m+1) - 1 ] iff P > 1/(m+1) ............................... (8).

VAR(m,P) = ($B)^2 . var(m,P)

VAR(m,P) = ($B)^2 . [ P(m^2 - 1) + 1 ] > 0 ................................(9).

N0(m,P) does NOT change because $B cancels out in Equations(6),(8) and (9).

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Now a Kelly bettor determines the unit bet $B by the relationship between Bankroll, ekb(m,P) and the Kelly fraction 'k':

$B= Bankroll/( k . ekb(m,P)) where k is the Kelly fraction

So

$B = [Bankroll / k] / [ P(m^2 - 1) + 1 ] / [ P(m+1) - 1 ]

$B = [Bankroll / k] * [ P(m+1) - 1 ] / [ P(m^2 - 1) + 1 ] iff P(m^2 - 1) > -1 ...................... (10).

Again as a consistency check we have:

EKB(m,P) = $B . ekb(m,P)

EKB(m,P) = Bankroll / k ................................................(1 1)

since ekb(m,P) cancels out as it should - remember this is a much simpler game than Blackjack.

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I think we are now done and your questions can be answered.

<Generally speaking, when betting full Kelly you divide the edge% by the pay-off a winning bet to arrive at the % of your bankroll that you should wager on your bet.>

Let's not assume you are playing full Kelly just yet.

[Payoff on winning bet] = m ............................................ (12)

NOT (m+1) since the +1 is your own money.

edge% = ev(m,P) = ev(m,P) = P(m+1) - 1 from Equation(1).

edge% / m = [ P(m+1) - 1 ] / m ........................................ (13).

Bugger: I am afraid I do not see any reference to variance in Equation(13) - it cannot be used to determine a Kelly bet !!!!

I think this is the correct expression:

$B= Bankroll/( k . ekb(m,P)) where k is the Kelly fraction from above ...... (14),

where from Equation(5)

ekb(m,P) = var(m,P)/ev(m,P) .................................................. ........... (15),

now we see the unit variance - nice.

%bankroll to bet = $B/Bankroll x 100% .................................................. ..... (16).

Substituting in the above terms we get

%bankroll to bet = [Bankroll / k] * [ P(m+1) - 1 ] / [ P(m^2 - 1) + 1 ] / Bankroll * 100%

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%bankroll to bet(m,P,k) = [ P(m+1) - 1 ] / [ P(m^2 - 1) + 1 ] / k * 100% ........... (17)

and we are FINALLY done !!!!!

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Just for fun, assume you are completely mad and playing full Kelly k=1.

Say the payoff is 1:5, that is m=5.

Assume your probability of winning is 51%, so P=0.51.

From equation(1) we have unit ev of

ev(m,P) = P(m+1) - 1 = (0.51)(6) - 1 = 2.06 units per round.

From equation(2) we have unit var of

var(m,P) = P(m^2 - 1) + 1 = (0.51)(25 - 1) + 1 = 13.24 (units)^2 per round.

From equation(5) we have unit ekb of

ekb(m,P) = var(m,P)/ev(m,P) = 13.24 / 2.06 = 6.43 units

Then from equation (17) the percentage of your bankroll to bet is

%bankroll to bet(m,P,k=1) = [2.06] / [ 13.24 ] = 0.156 = 15.6% of your bankroll.

N0(m,P) = var(m,P)/[ev(m,P)]^2

N0(m,P) = [13.24 ] /[ 2.06]^2 = 3.12 rounds.

Normally this gives a Kelly doubling time of (k=1) from Patrick Sileo:

T = 0.693 * N0(m,P)/(k - k*k/2) .................................................. .. (17),

unfortunarely this is under the assumption that N0 > 1000 rounds, so it won't work.

So just assume for now your Kelly doubling time is N0 plus a bit = 4 rounds.

With a bankroll of $10000, your unit bet will be $B = $10000 x 15.6 = $1560, call it an even $1500.

So even if you wait to double your bankroll before resize, you should expect to win about $10000 in the first 4 rounds,

$20000 in the next four rounds, $40000 in the next four rounds... get the idea :-)

I hope you have underpants of steel and your name is Walter White - or else whatever casino you try this in,

your life expectancy is about 10 rounds - good luck with that.

But it is only my juicy example - but for this example, I suggest you play with a Kelly factor of k=1/100 if you want to live.

Brett.

Oops - forgot about ROR - in the example above with k=1 you will continually have an ROR of 13.6%, way over betting.

One of two things will happen, you may be ruined after 4 rounds, or you may die - whichever is preferable I guess.

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