[tukkk]

Im not good at math, I need your help to improve a bit.

So this morning i thought id try my memory a bit and think of a simple game to calculate some simpler things like edge, kelly, ev, eg, and sigma .

So i thought of an unfair dice game:
- even odds are offered
- i win if i roll a 3 or 4 or 5 or 6
- i lose if i roll a 1 or 2

And i thought of getting the most money out of it obv
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I got the things i wanted quick and solid :
-edge 33.(3)%
-kelly 33.(3)%
-EV 11.(1)%
-EG 5.83%
-variance 88.(8)%
-sigma 94.28%
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But then i thought of the game for 100 samples and when Im thinking of sigma, there appears to be a small problem for me :

expected return(n=100) = 1.0583^100 = ~289%
sigma(n=100) = squareroot(100) x sigma = 942.8% yes, but when i start thinking moneywise, i get stuck :

After 100 dice rolls the bankroll should be 289 units give or take 1 stdev of ???
The problem is how do I compute the mean bet size?

[tukkk]

To be correct, the expected return 289x which is 28900% to be exact.

But the question still is : how do i calculate bet size in variance/standard deviation formulas, when im dealing with kelly betting in this dice game

[ws]

To be correct, the expected return 289x which is 28900% to be exact.

But the question still is : how do i calculate bet size in variance/standard deviation formulas, when im dealing with kelly betting in this dice game

I think it's easier to think in terms of the logarithm of your bankroll.
Let v(k) denote your bankroll after the k-th bet and v(0) your starting capital.
Since you're betting 1/3 of your bankroll, v(k+1)=v(k)(1 - 1/3 + X_k * 2/3), where X_k ~ Bern(2/3)
In terms of logarithms you get ln[v(k+1)]-ln[v(k)] = ln(1 - 1/3 + X_k*2/3)
or equivalently: ln[v(k+1)]-ln[v(k)] = ln(1-1/3) + X_k * ln(2)
v(n)/v[0] = [v(n)/v(n-1)]*[v(n-1)/v(n-2)]*...*[v(1)/v[0]] (remaining terms on rhs cancel out)
taking logarithms of both sides and plugging in the formula above, you get
ln[v(n)] = ln(v[0]) + sum_(0 < k <= n) [ ln(1 - 1/3) + X_k * ln(2) ]
or
ln[v(n)] = ln(v[0]) + n ln(1-1/3) + ln(2) * X, where X ~ Binom(n, 2/3)
hence, the logarithm of your bankroll is approximately normal -- and the bankroll itself is (approximately) lognormal.
The "(mean) give or take a standard deviation" comment indicates that you assumed (asymptotic) normality of the later.
The bet size in your scenario (the dice game played n times) is proportional to your bankroll, but for each bet in your game, the amount risked and the amount won both correspond to constants when you perceive your bankroll logarithmically.
Your (approximate) expected return after 100 bets isn't 100%*[exp(mu) - 1]/v(0) = 28700% -- that's the median return, the expected return is 100%*[exp(mu + sigma^2/2) - 1]/v(0) = 59974000%. (mu and sigma^2 are parameters of the underlying normal).
(And standard deviation of your bankroll after 100 bets is v[0]*12484811)

[tukkk]

Your (approximate) expected return after 100 bets isn't 100%*[exp(mu) - 1]/v(0) = 28700% -- that's the median return, the expected return is 100%*[exp(mu + sigma^2/2) - 1]/v(0) = 59974000%. (mu and sigma^2 are parameters of the underlying normal).
(And standard deviation of your bankroll after 100 bets is v[0]*12484811)

My bad about typing expected return, im referring to expected growth, because that is what matters since im not betting my whole bankroll on a roll.
When im playing this game for 100 throws, im expecting ~28900% growth
And shouldnt the expected return be 1,1(1)^100 = 3765000 %

When talking about n=100 , i can add up the variance, but what bet size will i apply? It should be average bet size. Im using :
variance = (bet)^2 * (decimal odds - 1 - edge) * (1 + edge)

since the order of wins/losses doesnt count, I basically have (g=1+expected growth% and k=kelly %)

the formula for n-th bet = k((g)^(n-1))

And what Im looking

average bet = k(g^0+g^1+g^2 ... +g^n)/n

Ofcourse i can put this into excel.
The problem for me is, can I make it simpler, so i can calculate that on the go?

[ws]

When talking about n=100 , i can add up the variance, but what bet size will i apply? It should be average bet size. Im using :
variance = (bet)^2 * (decimal odds - 1 - edge) * (1 + edge)

You're trying to calculate variance of your bankroll after the n-th bet as if it was a sum of i.i.d. random variables.
When you look at changes to your bankroll additively, they're neither independent, nor identically distributed (as opposed to when you look at changes to your bankroll multiplicatively).

[tukkk]

well whats the answer then?
the way i do it in excel, i get that after 100 rolls my median bankroll should be 288,1 units give or take 1 sigma of 162,28 units

[ws]

The bet size in your scenario (the dice game played n times) is proportional to your bankroll, but for each bet in your game, the amount risked and the amount won both correspond to constants when you perceive your bankroll logarithmically.

This should say "and the amount to win" instead of "and the amount won".

well whats the answer then?
the way i do it in excel, i get that after 100 rolls my median bankroll should be 288,1 units give or take 1 sigma of 162,28 units

Try to express your answer in terms of the logarithm of your bankroll.
Only then express it in terms of geometric mean (equivalently -- median) and geometric standard deviation of your bankroll -- so that you know what they mean.