IS ARB A MAGIC PICTURE?

[cosmicsports]

Arb is apparently very popular but at the same time it is considered an illegal play by the bookies.
I've known it for a long time, but I've never applied it. An Italian had taught me this, a long time ago, around 2000 when we were exchanging messages.
My objection to the Italian was that in order to get the arb he had to put his money in strange companies and I told him that I see it as dangerous.
The Italian did not agree, saying that there was no danger.

The bookies have now at one point banned it and anyone caught playing arb eats a penalty.

The bookies are actually losing money from the arb, theoretically at least.
Why ;
It is an apparent paradox but it has its explanation.

Suppose there are three bookies, "A", "B" and "C", as required to get arb every now and then.
The only player in the universe is you.
Case one: You play without arb, with the predictions given by the tweezer, the locker room guy, and the dude went out for a sergiani. Go for the duration and everyone is happy.
Case two: You play arb and make an annual profit of 6%. What does this mean? It means that A, B, C each got in by about 2%.
So?

But it seems strange. Because every single one of the arb's component bets is nothing more than a tweezer-dresser-dude bet out for a spin.
So how did the bookies lose?
The explanation is that each of the three wins is necessarily their fault. They paid a little more than they should have.
This is how the bookie's loss and your own profit comes out.

We don't know in advance what the mistake is.
If we knew, we would treat it as a value bet and there would be no need to arb.
Of course, in such a case, we would have to eat a penalty again, this time because of a win and not because of the arb.

But I assert the following:
If we take the "arb zone" games and play just one point at random, without arb, in the end the 6% profit will come out again.

 

[Larry]

But I assert the following:
If we take the "arb zone" games and play just one point at random, without arb, in the end the 6% profit will come out again.

You are right that you will be positive, but I think your profit will be the average of the bookies error ie 2% as you mentioned.

 

[cosmicsports]

You are right that you will be positive, but I think your profit will be the average of the bookies error ie 2% as you mentioned.

Let's look at it another way:
Let α1,α2,α3 be the returns of A, β1,β2,β3 of B, γ1,γ2,γ3 of C and let the triad of arb be α1,β2,γ3.

Probabilities matter.
The result is equivalent to a random toss of three possibilities.
But what exactly are they?
Are they the ones corresponding to the triad [α1,α2,α3], [β1,β2,β3], [γ1,γ2,γ3] or [α1,β2,γ3]?
We do not know.
I'm going to make the assumption that the correct one is either

l. [a1,a2,a3] + m . [β1μβ2,β3] + (1 - λ - μ) . [c1,c2,c3]

Where 0<=λ,μ<=1 parameters.

If for λ,μ I examine the values ​​0, 0.1, 0.2, 0.3 .... 1, i.e. change by 0.1, there are 55 cases in total.
With a simple random number simulation, it comes out what is true in each case and where the player's long-term profit will ultimately fluctuate.

If the trio arb [α1,β2,γ3] yields 6%, it does not follow that playing x-arb yields 2% and not 6%.
What exactly comes out needs to be investigated.

 

[Larry]

To put it more simply. When you play with arb, you choose the most favorable odds.
If you choose randomly, after 1 thousand times it will have such a percentage that it is the average of the bookies' rake. If the rake averages 2% positive, that's how positive you'll be.

 

[cosmicsports]

To put it more simply. When you play with arb, you choose the most favorable odds.
If you choose randomly, after 1 thousand times it will have such a percentage that it is the average of the bookies' rake. If the rake averages 2% positive, that's how positive you'll be.

But ... I said that the mixed rake should have 6%.
If it has 2%, then 2%.
But it needs some investigation.

Because it's also a matter of who has the right chances.
Let me give you an example not related to football, but similar.
In 2016, the election contest Trump-Hillary.
Initially, Hillary was the favorite, as the election approached, Trump dropped.
At some point you also found ... Hillary odds even new with Trump approaching, so
arb came out.
Here now it was obvious that the most correct possibilities were those with Trump going down, which were also the freshest.

In random arb zone games now, does this show up somewhere? Who is more valid?

 

[Larry]

But ... I said that the mixed rake should have 6%.
If it has 2%, then 2%.
But it needs some investigation.

Because it's also a matter of who has the right chances.
Let me give you an example not related to football, but similar.
In 2016, the election contest Trump-Hillary.
Initially, Hillary was the favorite, as the election approached, Trump dropped.
At some point you also found ... Hillary odds even new with Trump approaching, so
arb came out.
Here now it was obvious that the most correct possibilities were those with Trump going down, which were also the freshest.

In random arb zone games now, does this show up somewhere? Who is more valid?

The combination you say is nothing but the selection of the best returns.
Eg if 3 bookies give odds for under-over and give 1,90-2,05 the first, 1,80-2,10 the second and 1,85-2,12 the third, then you get the 1,90 first and 2,12 of the third and you become positive.
If you randomly start pairing under-over, then you will fall into the net of big numbers and become negative over time. Unless the bookies have a huge difference between them that even by chance you can become positive. Which is impossible.
In the Trump example, it's like you've played a 0-0 score and you're at half-time and it's still 0-0. In other words, there is progress, time has passed. I don't think it's arb.

 

[cosmicsports]

The combination you say is nothing but the selection of the best returns.
Eg if 3 bookies give odds for under-over and give 1,90-2,05 the first, 1,80-2,10 the second and 1,85-2,12 the third, then you get the 1,90 first and 2,12 of the third and you become positive.
If you randomly start pairing under-over, then you will fall into the net of big numbers and become negative over time. Unless the bookies have a huge difference between them that even by chance you can become positive. Which is impossible.
In the Trump example, it's like you've played a 0-0 score and you're at half-time and it's still 0-0. In other words, there is progress, time has passed. I don't think it's arb.

What do you think ;
I played Trump on the one that had the oldest odds - from August, let's say - and Hillary on the one that had the new ones and came out arb (from what I had seen on the oddschecker in those days, I was ahead).

Let me tell you now in more detail:

Suppose the three bookies give odds

A: a1, a2, a3
B b1, b2, b3
C: c1, c2, c3

The arb returns will be the three largest:

u = sup(α1, β1, γ1)
v = sup(a2, b2, c2)
w = sup(a2, b2, c2)

and the yield of arb will be in unit:

F1 = 1 / ( 1/u + 1/v + 1/w )

The odds now.
Based on A:

p1 = (1/α1) / (1/α1 + 1/α2 + 1/α3)
p2 = (1/α2) / (1/α1 + 1/α2 + 1/α3)
p3 = (1/α3) / (1/α1 + 1/α2 + 1/α3)

Based on B q1, q2, q3, where similar formulas apply
Based on G r1 ,r2 ,r2 where again the same formulas apply.

If I play just one point out of u, v, w then depending on who is right probably out of the three it will also be my average profit.
I would like the highest average profit:

sup { p1.u, q1.u, r1.u, p2.v, q2.v, r2.v, p3.w, q3.w, r3.w }

But how can I know in advance?
So I'm probably interested in the smallest, inf, to see what happens.

It is not obvious, because in these 9 numbers the highest odds correspond to the lowest probabilities.
However, it is clear that the search for the inf is between three numbers and not one.
Because whatever s1 the lowest estimate of its probability u, s2 for its probability v and s3 for its probability w.
So I'm looking for:

F2 = inf { s1.u, s2.v, s3.w }

So this number will tell me where my pseudo-arb ends up statistically.
What is the minimum expected profit?

I've accepted that one of the three is (probably) correct, which is true because the odds derived from the bookies' estimates are typically correct over the course of the game.

Now give me some examples to tell me what F2 comes out in relation to the F1.
The pseudo-arb means what average profit it yields at worst compared to the normal arb.

 

[cosmicsports]

In the example with under-over we have and we say:

A) 1.90 - 2.05
B) 1.80 - 2.10
C) 1.85 - 2.12

There is no w here, only u and v bets.

The arb is F1 = 1.002

Chances:

A) p1 = 0.51899, p2 = 0.48101
B) q1 = 0.53846, q2 = 0.46154
C) r1 = 0.53401 , r2 = 0.45599

That's it u is 1.90
The v is 2.12

The s1 is p1 = 0.51899
The s2 is r2 = 0.45599

s1. u = 0.51899. 1.90 = 0.98608
s2. v = 0.45599. 2.12 = 0.96670

So the minimum F2 = 0.96670 < 1 (the over).
But we can also play the under which comes out slightly better 0.98608 but still <1.

It seems when the initial arb is very small (like 2 per thousand) the method throws us into the negative.
Another example, with a better rate?

 

[cosmicsports]

I'm getting off the arb a bit.
To come out positive in any bet, the average profit must be positive.
If the probability is p and the payoff is c, then pc > 1.
This is as certain as it is that if you go to the casino you will lose or your hair will fall out after 90.

Since p is certain, the criterion is also absolutely certain.

At the bookies if we use the same odds they give to make them odds then it doesn't work.
It always comes out pc < 1, in green it comes out 0.95, in OPAP it comes out 0.90.
It also verifies because the bookie odds are pretty well calculated.

Of course, all of us, since ancient times, try to get probabilities such that pc > 1, but this is not easy at all. This is why the most conservative resort to arb.

Now with two books, what happens?
It is possible to find cases where p(B) holds. c(A) > 1.
In my example with Trump p(B) is from the bet that gives Trump the lowest return (when he started to approach). But c(A) is the old best performance (when he wasn't expected yet).
Since the "B" odds are out and we see them but the "A"s are still available at some bookies, we can take advantage.
He wants us to follow and to be fast of course.

But it is enough that p(B) holds - it is a necessary and sufficient condition.
But is this differentiation between the bookies valid or was it something random?

In general, it tends to be true that the freshest is the most correct.
But there is a case that it also emerges from the facts that p(B) is a more correct probability. In Trump's case, this was true. I think we all heard back then that "Trump is getting better" before we saw this in the bookies.