Evidence that it is nearly always better to do arbitrage than take a risk?

[Ganchrow]

By "close near-to-the-money" I mean betting the third outcome of the event (the only outcome for which you are NOT getting a price over and above what it should be) at a price VERY NEAR to what it should be (in the above case 2.1368 when it should be 2.17). Hence closing off the arb rather than leaving it open and just betting the 2 value sides accordingly.

The point is that unless you have some exogenous reason to expect future favorable price movement (and assuming zero time-value-of-money and zero opportunity cost) you should be closing off all your hedges.

In this situation I managed to close all three sides of the arb "above-the-money". This is obviously an ideal situation - however in two cases I was only just above the money, and in outcome iii) I was well above-the-money. So I should have bet more on outcome iii) to maximize long term bankroll growth?

As I've said, if your goal is to maximize expected bankroll growth then in the case of a market presenting a riskless multiway arb your optimal strategy (again assuming zero time-value-of-money and zero opportunity cost) is to bet a proportion of your bankroll on each outcome equal to the probability of that outcome occurring. Perhaps counter-intuitively the edges on the individual bets are irrelevant. (If there are 5 people out there who express interest I'll write up a mathematical proof.)

I don't understand either why a kelly bettor in that situation would bet so big on an outcome which is undervalued when he can only get $1000 on the "value price". Why not just hypothesise that as $1000 is your max bet on outcome iii) that your bankroll is equal to $1000/0.222222 = $4500? Seems counterintuitive that this $1000 value bet would lead you to bet almost $200k on an outcome that is undervalued (albeit slightly).

The $200K bet serves in part as a hedge for the positive expectation bet placed on bet 2.

The reason why the bettor shouldn't just "hypothesize that as $1000 is [his] max bet on outcome ... that [his] bankroll is equal to $1000/0.222222 = $4500" is because it has no economic or mathematical rationale. One can't just hypothesize bankroll size as part of an effective betting strategy.

I'm interested in best way to achieve bankroll growth.

That's exactly what Kelly does.

Assuming the total bankroll were utilized on this bet this would imply expected long-term bankroll growth of 0%.

I don't understand why this is?

Very good question. (We refer to this either as the bankroll either growing (shrinking, really) to zero, or growing at a rate of -100%. This can get a little confusing as sometimes people might refer to this as a growth rate of 0, which is ambiguous at best.)

Let's say you have a bet at even odds that occurs with probability 75%. If a bettor were to wager his entire bankroll on the outcome he'd have a 75% chance of doubling his bankroll and a 25% chance of bankrupting himself.

Assuming he were to continue to wager his entire bankroll on the same bet until he were to go bankrupt or until he were to have placed and won 3 bets, then prior tto his first wager he'd have a 42.1875% prob of ending up with 8x his money and a 57.8125% prob of bankrupting himself.

10 bets? 1,024x his initial with probability 5.6314%, bankruptcy with probability 94.3686%.

20 bets? 1,048,576x his initial bankroll with prob 0.3171%, bankruptcy with prob 99.6829%.

100 bets? 1,267,650,600,228,230,000,000,000,000,00 0x his initial bankroll with prob 0.0000000000321%, bankruptcy with prob 99.9999999999679%.

You get the idea ...

The point is that as the number of bets placed increases, the probability of bankruptcy approaches certainty. So even though the expected value of any number of these bets is positive, the expected bankroll growth from making them is zero. (The idea is that were you to repeat the bet a large number of times, and then repeat that experiment a large number of times, while the most common outcome would be bankruptcy, the extremely infrequent times the player didn't go bankrupt would greatly skew the average. Because we all only live one life, the fact that an alternate you in some parallel universe is richer than Bill Gates while you're eating dried cat food is probably small comfort.)

<hr>
The equation for a single win/lose bet given a two-outcome event:
p = win probability
o = decimal odds offered on bet
x = bet size as percentage of bankroll

E(growth) = (1+(o-1)*x)^p * (1-x)^1-p

And for two mutually exclusive win/lose bets given a three-outcome event:
p₁ = win probability of event 1
p₂ = win probability of event 2
o₁ = decimal odds offered on bet 1
o₂ = decimal odds offered on bet 2
x₁ = bet 1 size as percentage of bankroll
x₂ = bet 2 size as percentage of bankroll

E(growth) =
(1 + (o₁-1)*x₁ - x₂)^p₁ *
(1 + (o₂-1)*x₂ - x₁)^p₂ *
(1 - x₁ - x₂)^1-p₁-p₂

[wack]

Awesome post as ever Ganch. Too much to quote so I'll try to be clear about what points I am addressing:

Thanks for clearing up the points about the $200k hedge in my example. I understand your points about the theoretical bankroll - I guess I was looking at it from a point of view of when you do an arb you would get the $1000 @ price 5.9339 and then use the $5933.90 as your base figure from which you determine what to bet on all other outcomes, then applying that "logic" (ok its not that logical!) to the Kelly staking in proportion of probability.

I take your point that if you go all-in every time you see a value bet your expected LR growth is effectively 0%. Thanks for spelling it out.

I am fascinated by the fact that the edges are irrelevant. This type of alternative to either a closed arb or a value bet is something I have never considered, and would like to consider.

Maths aside, for the moment, what do you think the real-world plus and minus points are of the three strategies, arb, favoured hedge, and value bet?

Plus on arbs - i) effectively no money risk (of course other risks are involved)

minus on arbs - i) they won't be around forever. As more and more people arb it becomes more important for discerning books to close them out - unless they are betting the non-soft sides of the lines at their books almost always e.g. pinny.

plus on favoured hedge - i) you are maximising each opportunity with regard to Kelly staking, widely accepted as the best way to maximise bankroll growth. As soft opportunites get harder to find/closed quicker, maximum bankroll growth is a seriously important concern.

minus - i) some risk, despite advantages bad runs can come along, uneven "earnings" - some scenarios could have a 30% advantage and result in no winnings/small loss.

Plus for value bet - i) much easier, less work involved, no need to stake anything on undervalued lines

Minus - i) Much more risky, much larger swings in bankroll.

Anything to add to the above?

And one last question: In the worked example with the $1000 constraint on outcome iii) at 5.9339 what if we also had a constraint on outcome ii) of $1760? Basically assuming we always have a limit on a soft line and no limit on an undervalued line.

[raiders72002]

Obviously you want to take any arb that you can get since it's free money. Since most don't have unlimited funds and because of betting limits you have to decide where to invest your money.

If you feel that your expected rate of return on a bet, considering the odds, is better then taking the arb then place your bet(s) first and use all remaining funds to arb.

[Ganchrow]

I am fascinated by the fact that the edges are irrelevant. This type of alternative to either a closed arb or a value bet is something I have never considered, and would like to consider.

Yeah it's an interesting and surprising result and I still lack an intuitive justification. A pair of mutually exclusive outcomes offered at -110 apiece occurring with probability 50% and another pair of mutually exclusive outcomes offered at +100,000,000 and -1,999.96 also occurring with probability 50%, would both rate the same size pair of bets -- namely 50% of bankroll on each.

Maths aside, for the moment, what do you think the real-world plus and minus points are of the three strategies, arb, favoured hedge, and value bet?

-snip-

I think you pretty much covered this.

(As a minor side note, I'd mention that when you refer to Kelly staking by noting that it is "widely accepted as the best way to maximise bankroll growth", you're damning it with faint praise. Referring to Kelly in such a manner is s a little like referring to the number 2 as "widely accepted as equaling 1+1". The fact that Kelly staking maximizes expected bankroll is an easily proven mathematically and the level to which this fact is accepted by the public is irrelevant to its veracity.)

And one last question: In the worked example with the $1000 constraint on outcome iii) at 5.9339 what if we also had a constraint on outcome ii) of $1760? Basically assuming we always have a limit on a soft line and no limit on an undervalued line.

Assuming an arbitrarily large bankroll in practical terms, a bankroll of about $205,000 or more), optimal full-Kelly strategy under these circumstances would be be to max out outcomes ii and iii ($1,760 and $1,000 respectively) and bet nothing on outcome i. The logic behind this is straightforward -- with a given bankroll one will be behave in a risk neutral manner towards bets of sufficiently small size in comparison to that bankroll.

[wack]

Assuming an arbitrarily large bankroll in practical terms, a bankroll of about $205,000 or more), optimal full-Kelly strategy under these circumstances would be be to max out outcomes ii and iii ($1,760 and $1,000 respectively) and bet nothing on outcome i. The logic behind this is straightforward -- with a given bankroll one will be behave in a risk neutral manner towards bets of sufficiently small size in comparison to that bankroll.

Just quickly, how come $205k or more? Thanks for all the replies to the other stuff as well Ganch.

[Ganchrow]

Just quickly, how come $205k or more?

There's no simple logical answer to this ... it's just how the math happens to work itself out.

But just for fun (I had a late night) here's how it goes:

A given event has 3 mutually exclusive outcomes, call them i, ii, and iii.

These occur with probabilities:

1. 0.46080
2. 0.31698
3. 0.22222

With bets offered at decimal odds of:

1. 2.1368
2. 3.3703
3. 5.9390

There's no maximum bet on outcome i, but the maximum bets on outcomes ii and iii are $1,760 and $1,000, respectively.

Q: How large would the player's total bankroll need to be such that the expected growth maximizing allocation for bet i would be $0?

<hr>
Solution:

We have the objective function (assuming the $1,760 and $1,000 constraints bind):

E(U) = 
0.46080 * ln($B + 1.1368 * $x - $1,760 - $1,000) +
0.31698 * ln($B - $x + 2.3703 * $1,760 - $1,000) +
0.22222 * ln($B - $x - $1,760 + 4.9390 * $1,000)

where $B is total bankroll and $x is the amount bet on outcome i.

So we maximize E(U) with respect to $x, subject to $x ≥ $0 and $x + $1,760 + $1,000 ≤ $B (and of course we know that the latter constraint won't bind for moderate levels of $B -- "moderate" being around $5,319 -- but that's another post).

Setting ^dE(U)/_d$x to zero and solving for $x (the 2^nd derivative is everywhere negative) we come up with

$x =
5.62984 × 10<sup>-11</sup> *
      ( 8.76123 × 10<sup>9</sup> ± 9.00127 × 10<sup>9</sup> *
        sqrt (B<sup>2</sup> + 797.194 B + 158902)
      + 5.28259 × 10<sup>13</sup> )
for $x ≥ $0

Setting $x = 0 and solving for $B we come up with a value of $205,124.2305 which I'm slightly surprised to discover is so close to my $205,000 figure.

This means that for bankrolls of about $205,124.2305 or greater the expected growth maximizing allocation for bet i would be $0.

[wack]

Ask a simple question.....

Seriously, thanks Ganch, good job

[larryacraig]

Do they have a "for dummies" version of this? If this is it then do they have a for complete idiots version?

[Arilou]

The proof seems (intuitively) something like this: Obviously you're going to bet your entire bankroll, since any additional bankroll could be used to make a riskless arb, making leaving any money behind a dominated strategy. The payoff for each bet then becomes your bankroll if that outcome happens, which would then become (amount wagered)*(payoff). Payoff is a constant, so since your utility function that you're maximizing is geometric in terms of money it gets pulled out and becomes a constant. Since this is now a constant amount of utility that occurs a constant percentage of the time, it can be ignored for the purposes of maximizing utility. Thus, all that matters is maximizing in any case for this set of outcomes (in terms of the probability of those outcomes) and you solve for all cases, so simply take the case in which all the outcomes have an e.v. of 0%, so since you can't increase your bankroll with zero e.v. bets no play can be better than always breaking even which you do by wagering on each outcome proportional to its probability.

That's probably not as elegant a solution as Ganchrow has in mind, but it works and it provides something of an intuitive framework for why this is the result. If in one case you multiply your bankroll by a constant, all you're doing is altering your expected geometric mean by a constant, which doesn't change what you should do at all.

However, this result points out a number of problems with maximizing bankroll growth as a strategy. I doubt that when push comes to shove anyone, including Ganchrow, would actually bet their money in this way in the more absurd example situations. One note is that if even a tiny bit of your Kelly bankroll cannot be wagered, the result breaks down and in the extreme cases breaks down rather badly.

[pico]

you guys forgot to add the probablity of books going busto and phone bills and anger towards a slow paying book.

also if you make a mistake arbing, 1 week worth of profit...gone. kinda sucks.

[Arilou]

Pico makes a point that is easy to overlook. Even the best arbitrage involves risk of miscalculation or something as simple as a typo or brain freeze. I know it's happened to me. If you use multiple books, getting the money properly configured afterwards is usually a cost even if you set your risk of being stiffed to zero. But of course, no one thought we could actually consider a horse as a three meter sphere.

[xyz]

I would like to add another risk for arbs: the book with the off line may cancel the bet. This makes me hesitant to jump on arbs that are too good with too much capital. This happened to me with SIA a few weeks back. They canceled the bet due to "incorrect line" before the game started. Before the game gave me a chance to hedge at a loss. It would be more damaging if they canceled it after the game ended.

[CarloTwoGuns]

Can you explain this in english?

[Arilou]

xyz: Yes that is a risk but it's mostly a known one. I can't remember the last time I got a wager cancelled by a book that I wasn't explicitly worried about being allowed to keep, and when a line is in that range then you simply make the bet and keep it for this very reason.

Carlo: Which part? Some parts are already in English (mostly), others can be made more clear if you tell us what needs explaining, and some of it is by its nature math and most be treated as such.