Suppose someone offers a 2% edge on a 50/50 coin toss, so that anything we bet will be paid out at 2.04x if we win. KC says we should bet up to 2% of our bankroll. So if we have a $10,000 bankrol we bet $200. Easy.

Now suppose this person will only allow us to bet in $10 increments at decreasing edges:

$10 with payout of 2.04 (2.0% edge)
$10 with payout of 2.038 (1.9% edge)
$10 with payout of 2.036 (1.8% edge)
.....down to....
$10 with payout of 2.002 (0.1% edge)

If we take all bets, we end up with $200 at an average of 1.05% which is too much given our bankroll.
If we bet down to only the 1.1% edge, we end up with $100 at an average of 1.55% edge, which is well below KC sizing.
If we take the next bet at 1.0% edge, our average is still below KC. However, our risk is currently at $100 and with a 1% edge we would normally consider ourselves maxed out and not make the next bet.

So what is the correct way to calculate KC in such a situation?

If you used 20 bets as described from 2% to 0.1% edge in steps of 0.1% you need to redo the calculation not apply it as you did because the chance to lose in 20 bets like that is tiny actually. It is no longer a 1.05% edge at 50-50.

I assume by edge x you mean you risk y and lose y with 50% chance or win (1+2x)*y with 50% chance. So i assumed the EV is what you called edge.

20 such bets result in a profile that is winning more often than 50% of the time. In fact if you simulate it it comes at 59.1%.

So you have a distribution of returns now (depending how you allocate bets over these range of edges) and you must apply the kelly process of maximization for that distribution.

Yes the avg EV is the sum of the EVs etc but it is no longer a 50-50 proposition at 1.05% edge when you take all 20 at once at 10 each.

For the methodology to maximization in general see here


https://en.wikipedia.org/wiki/Kelly_criterion

and here maybe


http://papers.ssrn.com/sol3/papers.c...act_id=2259133


In this case here just formulate the problem as how you play when you have some distribution of possible returns.

If you have various available choices i suppose one can maximize in a variety of ways depending on your constraints.

I suspect here its better than you think and betting on all 20 is not bad. Its not as bad as betting all at once in some 1.05% edge i mean.

It's all on the same flip. Just see what maximizes E[ln(bankroll)] at each step along the way and that'll tell you when to stop adding on.

Also see it this way; If you had a 0.1% edge only in the worse case then i see that as getting 1.002 profit when you win and losing 1 when you lose the bet with 50% each so EV is 0.1% per 1 unit risked. (do you define edge differently, it seems you do it the same?)

Then based on previous link Kelly is (0.5*(1+1.002)-1)/1.002=0.0000998 so 9.98$ out of 10000 would be ok.

So in principle all bets you mentioned qualify even if the total had it being at the avg edge of 1.05% would be over the proper size as you can see from;

(0.5*(1+1.0105)-1)/1.0105=0.0052 justifying only 52$.


This is because you can ignore the other bets and see the single 0.1% edge for 10 alone and it passes the test. So the others will also pass it. (it doesnt matter that all happen at once because each one is independent, making the thing equivalent to a series of bets happening one after the other, if the worse passes it the others will too and it doesnt matter all are risked at the same time interval t because it is in effect 20 independent events and time of occurrence is irrelevant - its like breaking that time interval in 20 and doing one after the other 20 times faster each lol)

You dont have to imagine all at once risked at 1.05% edge because as i said the distribution is no longer a 50-50 one for the whole 200 amount.

It's adding on bets to the same coin flip.

You mean all 20 paid based on the same single flip outcome not 20 different flips each at 10?

I didnt think that was the case from OP details. If it is like that then one has to stop at some higher point say. If all 20 have to be used its aborted as substantially over Kelly.

I think then my calculation (Solve[{(0.5*(1 + b) - 1)/b == x/1000,
b == 1.02 - 0.001*(x - 1)/2}, {b, x}]) (b=avg edge for x bets added up at 10 per 10000 each )

says you stop at 8 bets at once for all the same flip (1.020 to 1.013).


I still think though OP meant 20 different flips at once.

Yeah

I meant 1 flip. All bets are coorelated 100%.

I don't think this is right. You have to look at the increments. By my math, while doing 8 increments is not an overbet against the Kelly Criterion when viewed all together as one bet, the 8th increment itself is -EU (using logarithmic utility as Kelly Criterion is based on) and therefore you should stop at 7.

Think of it this way. At 7 increments, you have bet 7 units at an average edge of 1.7%. At this edge you would want to bet more, but if you DO bet more, the extra isn't coming at 1.7%, it's coming at 1.3%, and at 1.3% you've already bet too much, even though the final bet (8 units at 1.65%) isn't an overbet.

Why would 0.8% of your bankroll at 1.3% edge be considered an overbet? Isn't this well under KC?

(Bare with me due to long post but this is necessary to correct a few things above and discuss the entire problem from fresh even. The answer to original problem is to stop at the k=10 first bets for the bankroll initially offered - but there is some additional thoughts that follow after that calculation - see next)

I want to correct my calculation above that i had spotted for a few days back now because i was using the wrong equation for b and the problem here, effectively solving a different problem.


Let me restate the problem solution and then offer some additional ideas about possible future issues that may completely undermine the entire Kelly logic here because of the specific format of bets offered that will eventually prevent us from realizing a geometric growth of bankroll, so ultimately what we are concerned with is not maximizing the growth rate using the kelly equations in links, but re-deriving the entire thing under the real life constraints of this problem. Those are that we are only offered 20 bets of $10 max each and no more and that risk of ruin may be a concern if we are not allowed to go as low in bets as we like as it is the case with the Kelly format.


Here is it is anyway ignoring these real life concerns for now;


We play a game that is described by the equation https://en.wikipedia.org/wiki/Kelly_criterion

f=((b+1)*p-1)/b (equation 1)

where p is the chance to win say 0.5 here and b is what we win when we win and 1 what we lose when we lose. So b is 1.04 to 1.002 and of course corresponds to edges (eg EV per unit bet) 0.020 to 0.001 for 20 available bet possibilities offered.


That formula comes from saying what is the most likely outcome after n bets (central term);

If we risk fraction x of bankroll B each time then we will be going to (1+x*b) *B when we win and (1-x)*B when w lose.

So over time after n bets

B~(1+x*b)^(n*p)*(1-x)^((1-p)*n)

So log B ~ n*p*log(1+b*x)+(1-p)*n*log(1-x) so the derivative set to 0 (to maximize growth rate) gives ;

(-n (1 - p))/(1 - x) + (b n p)/(1 + b x)=0 and n cancels and we get


x -> (-1 + p + b p)/b or x=(p*(b+1)-1)/b as above.



Now look what happens if we select to use the first n of those bets where each one has b_i=1.04-0.002*(i-1) i from 1 to n (n from 1 to 20 whatever is the proper stop level).

The effective bet of n of those is essentially risking n*10 and has average payout per bet 1/n*(Sum(1.04-0.002*(i-1), {i,1,n})= 1/n*(n*1.04-0.002*n*(n-1)/2)=
=1.04-0.001*(n-1)

So we basically risk n*10 (each bet is 10) total and have an effective win of +(1.04-0.001*(n-1))*10*n when we win and lose 10*n when we lose.

So the effective b (or avg b same thing) in above (equation 1) is (1.04-0.001*(n-1)) when we take n bets.


And we need that to be f=10*n/10000 at the ideal n.

So we have the system b=1.04-0.001*(n-1) and ((b+1)*0.5-1)/b=n/1000

This has as solution x -> 13.4199, b -> 1.02758 implying n=13

Now if we have n=13 the total bet is 130 and the avg payout is 1.04-0.001*(13-1)=1.028

So if i have a bet of 130 i effectively have a result of +133.64 when i win and -130 when i lose.

Now see what Kelly says for a bet that you risk 130 to win 1.028*130=133.64

It says risk (0.5*(1+1.028)-1)/1.028=0.0136 or for 10000 it's $136. We risk 130, close enough.


Now it would appear that the discussion has concluded that n=13 is the right stop level. But it is not correct for the following reason.

What the approximate solution above found as roughly 13 is what would be true if we could realize the true Kelly level here precisely. The possible b_i here are quantized and the effective b is not changing continuously.

So we cant go to perfect Kelly fraction. And going close to it (vs less close in other cases of other number of bets) may not be the correct answer in terms of maximizing the growth rate.

To see what the correct answer is we need to go back to the original maximization here and forget about equation 1 .

It should go like this;

Since we cannot realize a precise Kelly fraction (as in 1) what we have is the growth rate for different bet combinations. And it is those that are not true maxima that we need to compare and find the best of them.


What we have therefore is the function (log of central term in bankroll after N>>1 trials)

N*p*log(1+b*x)+(1-p)*N*log(1-x)

And b=1.04-0.001*(k-1) k is how many offered bets we take. (so avg b) and x= k*10/10000=k/1000

So we have to maximize p*log(1+(1.04-0.001*(k-1)))*k/1000+(1-p)*log(1-k/1000)

p=1/2 here.

We want to find the k that maximizes ; log(1+(1.04-0.001*(k-1))*k/1000)+log(1-k/1000)

here is a list for all k from 1 to 14

Table[F[k], {k, 1, 14}]

{0.0000389592, 0.0000738413, 0.000104653, 0.000131399, 0.000154088, \
0.000172725, 0.000187316, 0.000197868, 0.000204387, 0.000206879, \
0.000205349, 0.000199804, 0.00019025, 0.000176692}

As you see at k=10 ie stopping at the 10th bet edge offered (and betting therefore just 100 ) is the point that the growth rate is maximized.

So although k=13 gives the closer value to the kelly number and the true kelly bet size for the edge we have at n=10 is much larger than 100, we are unable to realize these precisely in the structure offered, since each bet is capped at 10 and we move to lower edge after that.


So we simply need to compare growth rates and find the best for all 20 choices of bets available. This happens if we stop at 10.

Notice that difference between 10 and 13 is not huge but it is there.

13 does indeed get us closer to the true Kelly bet size for that collection of bets effective edge but this is not the maximum growth rate.



What we need to do now to finish this is take into account the fact that as we increase our bankroll from profits we will be able to play more bets that become available to us.

So the true maximization here is even more complicated than what already described.

It is however likely true that sticking with k=10 initially will take us faster to the bankroll target that will allow us to go to k=11 ,12 etc and saturate eventually all bets faster than other choices.

An issue of risk of ruin however here may be interesting to consider too. By that i mean that since real Kelly problems do not have risk of ruin (you can bet as low as you want until you recover from any abyss), the same is not true here.

So it is important to take risk of ruin to as low as possible here even at the cost of initial growth rate in order to secure eventually maximum long term profits (realized when we bet all possible 20 bets at the same time and we need to as close to 100% of the time get there eventually).

Notice that because the bets go up to 200 here and stop after that the long term growth rate goes to 0 because the bankroll eventually grows linearly with time.

So this is not a completely traditional clean Kelly problem here (with the log of bankroll) and we may need to be even more thorough to maximize the typical long term outcomes realized under these constraints here.