Bankroll Growth Calculation

Division

Hey folks,

I'm looking for a bit of help on how to do a calculation.

If I place 100 bets at odds of 1.5 (or 1/2 of you prefer), each for 40% of my bankroll amount at the time of bet placement (previous bet always settled before next bet placed), and 80 of them win my bankroll will grow from €100 to €7,898.

What is the growth rate per bet?

Thanks.

MathsManiac

I take it you mean average growth rate per bet. (Some bets are losing, so cause negative growth while others win, causing positive growth and you're interested in how it averages out in the long run.)

If I understand what you're describing correctly, it could be described as follows:
- there is an process which iterates repeatedly (you betting)
- it's basically an exponential/geometric process (growth is proportional to the size of the quantity growing)
- 100 iterations of the process cause €100 to grow into €7898.

The appropriate formula here is €100*[1+r]^100 = 7898.
If you solve that for r you get 0.04466049406, which is 4.47%.

That is, in the long-run average sense, your bank roll is growing by 4.47% per bet.

Does that sound like what you need?

Division

Thats exactly what I am looking for, thanks.

How is that calculation done to find the 4.47%?

MathsManiac

The "1+r" in the formula is "1 plus the rate expressed as a decimal". It's the multiplier from each iteration of the process.

So, starting with 100*[1+r]^100 = 7898, I divided both sides by 100 to get
[1+r]^100 = 78.98

Next, I got the hundredth root of 78.98 (using a calculator or computer), which is 1.04466049406. That's the number that your bankroll is, on average, getting multiplied by each time you play.

In other words, itself plus 0.04466049406 of itself. Expressing this decimal 0.0447 as a percentage gives 4.47%.

weary1

I found this an interesting question and answer.
But can I ask the OP how you arrived at a percentage of 40 percent for each bet.
I notice that if you bet less or more than 40 percent of your bank on each bet then your return will be less than 7894.
Some optimisation formula exists ?

Yakuza

weary1 wrote: »

I found this an interesting question and answer.
But can I ask the OP how you arrived at a percentage of 40 percent for each bet.
I notice that if you bet less or more than 40 percent of your bank on each bet then your return will be less than 7894.
Some optimisation formula exists ?

I think it's 40% because a win after a loss will put you back in the same position as a loss after a win. The ratio 40:60 and the odds of 3:2 are complimentary.

You have X, you bet .4X and lose, you have .6X. You now bet .24X (40% of .6X) and win .12X in profit, putting you at .72X
or
You have X, you bet .4X and win (.2X profit) and now have 1.2X. You then bet .48X (40% of 1.2X) and lose and you're back at .72X

What MM has done is fine in terms of what growth rate will grow €100 to €7898 in 100 steps, but the actual progression of wealth build up is different.

Initial condition : wealth = €100. Bet €40, this will return either €60 or 0, so at step 2 there are two possibilities, you possess €120 or €60 (allowing you to bet €48 or €24, returning either €72 or €0 or €36 or €0, leaving you with wealth €144, €72 or €36), so we're at 3 possible outcomes after two steps.
Mathematically, this is a recombining binary tree and after 100 steps there are 101 possible outcomes (ignoring the fact that you won't be able to bet less than 1c if you happen to start with a bad losing streak ).

Each up-step (win) increases your wealth by a factor of 1.2 and each down-step (loss) decreases your wealth by a factor of .6.

Starting off with €100 and with 80 wins and 20 losses, you'll end up with €100 * (1.2^80) *(.6^20) = €7898.14 to the nearest cent.

MathsManiac

weary1 wrote: »

I found this an interesting question and answer.
But can I ask the OP how you arrived at a percentage of 40 percent for each bet.
I notice that if you bet less or more than 40 percent of your bank on each bet then your return will be less than 7894.
Some optimisation formula exists ?

Fascinating question indeed.

So, this begs the more general question:
How to calculate the proportion of the bankroll that should be repeatedly bet in this scenario in order to maximise the final bankroll.

Solution
Let the odds offered be expressed in the form w:1. (That is, for each €1 staked, you win back your €1 and additionally €w.) In this example, w=0.5.
Let k be the proportion of the bankroll bet. (The OP gave this as 40%, but instead we're going to try to calculate it.)
We are going to be n times and it is known that we will win r times and lose (n-r) times. (This is not quite the same as saying we have a probability of r/n of winning each time but it's what the OP described.)

Assuming w, n and r are fixed, the final bankroll can be considered a function of k and is given by:
f(k) = (1+wk)^r * (1-k)^(n-r)

We want to find the value of k that maximises f.

Let F(k) = ln(f(k)). Then f(k) is maximised when F(k) is maximised.

F(k) = r ln(1+wk) + (n-r) ln(1-k)
F'(k) = 0 implies that rw/(1+wk) - (n-r)/(1-k) = 0.

Solving this for k yields:

k = (wr+r-n)/(wn)

So, we have our general formula. We can pop our original numbers into it:
r=80, n=100, w=0.5 gives k = (40+80100)/50 = 40%.

Division

weary1 wrote: »

I found this an interesting question and answer.
But can I ask the OP how you arrived at a percentage of 40 percent for each bet.
I notice that if you bet less or more than 40 percent of your bank on each bet then your return will be less than 7894.
Some optimisation formula exists ?

Its the Kelly Criterion.

The Kelly Criterion formula is: (BP - Q) / B

B = the Decimal odds -1
P = the probability of success
Q = the probability of failure (i.e. 1-p)

weary1

Yakuza wrote: »

Mathematically, this is a recombining binary tree and after 100 steps there are 101 possible outcomes (ignoring the fact that you won't be able to bet less than 1c if you happen to start with a bad losing streak ).

I would'nt mind upping my stake to 1c on a few bets if it guaranteed I would collect 7898 after 100 bets

MathsManiac

Division wrote: »

Its the Kelly Criterion.

The Kelly Criterion formula is: (BP - Q) / B

B = the Decimal odds -1
P = the probability of success
Q = the probability of failure (i.e. 1-p)

Comfortingly, the formula I ended up with turns into that if one replaces r/n with P and (n-r)/n with Q.