[hhjjhh]

I have a question for the math-inclined: How do you apply the Kelly
staking system to asian handicap bets?

I have trouble figuring it out, especially with quarter handicaps.
For example, say I have a soccer (football) match between team A and
team B. I estimate the following:

Home win: 15%, draw: 35%, away win: 50%.

The odds for the outcome in team B's favor is "+0/+0.5" is 1.75. The
"+0/+0.5" outcome means that if team B wins, I win the whole bet. In
case of a draw, I win only half my bet, and get the other half
refunded. If A wins, my whole bet loses.

My betting bank is 100.

Now, if I simply use a regular Kelly system on this outcome, and
simply assume it is a bet on draw+win for team B, the percentage
chance would be 85%, and since the odds of 1.75 are thus favorable, I
would now bet 65 from my bank.

But how would I do it in case of asian handicap?

And what about -0/-0.5 (lose half the bet in case of a draw), if the
odds in this case were favorable?

Thanks for any help.

[tacomax]

Welcome to the forums, hhjjhh.

Ganchrow, the forums resident Kelly expert, will be notified by pager of a Kelly-related thread and should be here soon to answer your query. If he's not got something better to do - like sleeping - that is.

[Ganchrow]

I have a question for the math-inclined: How do you apply the Kelly
staking system to asian handicap bets?

I have trouble figuring it out, especially with quarter handicaps.
For example, say I have a soccer (football) match between team A and
team B. I estimate the following:

Home win: 15%, draw: 35%, away win: 50%.

The odds for the outcome in team B's favor is "+0/+0.5" is 1.75. The
"+0/+0.5" outcome means that if team B wins, I win the whole bet. In
case of a draw, I win only half my bet, and get the other half
refunded. If A wins, my whole bet loses.

My betting bank is 100.

Now, if I simply use a regular Kelly system on this outcome, and
simply assume it is a bet on draw+win for team B, the percentage
chance would be 85%, and since the odds of 1.75 are thus favorable, I
would now bet 65 from my bank.

But how would I do it in case of asian handicap?

And what about -0/-0.5 (lose half the bet in case of a draw), if the
odds in this case were favorable?

Thanks for any help.

Very good question.

After some calculus and a couple of quadratic equations I came up with the following messy (but still correct) solutions. With some effort the answers could likely be simplified.

---

Let p = win prob
Let q = draw prob
Let o = decimal odds

The Kelly stake for pk & +0.5 (assuming edge > 0) is then:
=(p*o^2+q*o^2-p*o-3*o-q+3
+SQRT((p*o^2+q*o^2-p*o-3*o-q+3)^2-4*(o-1)^2*(-2*o*p-o*q-q+2)))/(2*(o-1)^2)

The Kelly stake for pk & -0.5 (assuming edge > 0) is then:
=(p*o-q*o+2*o+q-3
-SQRT((-p*o+q*o-2*o-q+3)^2-4*(o-1)*(2*o*p+q-2)))/(2*(o-1))

So for:
o=1.75
p=20%
q=50%

The Kelly stake for pk & +0.5 would then be 7.511%.


And for:
o=4.00
p=20%
q=50%

The Kelly stake for pk & -0.5 would then be 2.365%.

If anyone were interested I could put this in a spreadsheet.

[hhjjhh]

Thanks a lot for this solution. Great work!

There is actually one more case, which I forgot in my original post:

The +0 handicap, meaning that in case of a draw the full stake is returned, regardless of the team you bet on.

So say again that the odds on team B +0 are 2.05.

My probabilities:

A win: 20%, draw: 45%, B win: 35%

I think your solution can also be used for this case, but I am not sure exactly how to modify the formula.

Can you help again?

Yes, I know, I am a bit mathematically challenged...

[hhjjhh]

If anyone were interested I could put this in a spreadsheet.

I would be very interested in that

[Ganchrow]

The Kelly solution that includes the possibility of the draw outcome is very similar to the standard binary outcome Kelly solution. (We define p, q, and o, as above.)

Recall that the binary Kelly solution (for edge > 0) is (p*o-1)/(o-1), which is simply the ratio of edge to the the amount that would be won off a unit bet.

If we define p' = p / (1-q), which is the probability of winning conditioned on not pushing (the "conditional win probability"), then the ternary outcome Kelly solution is (p'*o-1)/(o-1), which is simply standard binary Kelly with the the conditional win probability substituted in for the raw win probability.

[Ganchrow]

If anyone were interested I could put this in a spreadsheet.

I would be very interested in that

Nothing fancy:

[hhjjhh]

Thanks a lot for this, I will be using this in my future betting!

[hhjjhh]

I have been trying with some calculations using your two formulas, and I just want to make sure I understand this completely.

In your example, you have:

So for:
o=1.75
p=20%
q=50%

The Kelly stake for pk & +0.5 would then be 2.36%.


And for:
o=4.00
p=20%
q=50%

The Kelly stake for pk & -0.5 would then be 7.51%.

But I don't get those results exactly, using your formula. In the first example, just to spell it out:

I bet on team B (the underdog), handicap +0/+0.5 at odds 1.75.

I have the probabilities:
Team A win: 30%, draw: 50%, team B win: 20%.

Therefore:
o=1.75
p=20%
q=50%

Just like in your example.

But using your formula, I get a Kelly stake of 7.51%.

Similarly, for your other example, I get 2.36%.

Actually, I get the results reversed. Am I misunderstanding something?

Thanks again.

[Ganchrow]

I bet on team B (the underdog), handicap +0/+0.5 at odds 1.75.

I have the probabilities:
Team A win: 30%, draw: 50%, team B win: 20%.

Therefore:
o=1.75
p=20%
q=50%

I had transposed the pair of numbers in the earlier post.

Post has been corrected.

You should also make sure you're using the latest version of the Asian_Handicap_Kelly.xls spreadsheet.

[hhjjhh]

Hi again Ganchrow

One more question...

How do you apply the asian handicap calculation to the simultaneous bets calculator on this website?

I suppose again here some calculations are necessary to quantifying the edge or win probability, as is needed by the calculator...

Can you help?

Thanks

[brebbles]

Appreciate this thread is almost 10 years old, but I came across it when trying to verify the method I came up with for betting +0/+0.5 and -0/-0.5 handicaps (luckily enough for me the method I was using returns the same results as Ganchrow's above!) I thought I'd share an interesting finding that I've seen come out of it regarding the implied probability of a +0/+0.5 pick winning.

For the sake of consistency I'll use the figures from the post above - so:

p = 20%
q = 50%
o = 1.75

Plugging these into Ganchrow's formula above gives a Kelly bet size of 7.51%. If you reverse engineer this into the standard Kelly formula it returns a probability of the bet winning as 60.4% (reasonable enough as you would expected it to fall between the 20% chance of the team winning, and the 70% chance of the team winning or drawing).

Now say we change the odds from 1.75 to 2.00. This gives a Kelly bet of 22.62%. Reverse engineering this into the standard Kelly formula gives a probability of the bet winning as 61.3% - still withing the bounds of winning and winning+drawing, but a different figure than before.

More interestingly, if you plug in the odds that only just give an edge (and a bet), it returns a probability of 60%. If you plug in huge odds (approaching infinity) it returns near enough 70% (the probability of winning and drawing combined).

So as opposed to the case of +0.5 handicaps and +0 handicaps, it seems here that the implied probability of a +0/+0.5 result depends somewhat on the odds on offer.

Perhaps I'm violating some law of the Kelly criteria here (or perhaps my wording of the finding is just wrong) - but as far as I've known for the 15+ years I've used the Kelly Criteria: if you know any two of the probability, odds and Kelly bet size then you should be able to work out the third value.

Interested to hear some takes on this.