Statistical Arbitrage Challenge, Prize Offered (Hard)

**tomcowley** · 12-24-07, 08:37 PM

I can't see why parlays (or if-bets) should be relevant here since bets 2 and 3 are already functionally if-win. The only meaningful attempt is to create a synthetic NE to win SB parlay, but that comes out at +110 instead of +118, if my math is right.

My solution is:

Bet 20174 on Bet 1 No (NE to win divisional).
Bet 3192 on Bet 2 Yes (NE to lose AFC).
Bet 32580 on Bet 4 Yes (NE to win Super Bowl)

EV is +$7058

Bankroll that corresponds to the maximized expected sqrt(bankroll) utility function is 103712. It's not really expected growth, since this isn't full kelly. Expected growth, I think, is 0.14%, but I could have screwed that up.

**Thremp** · 12-24-07, 08:39 PM

Die TomC.

Merry Christmas!

**tomcowley** · 12-24-07, 08:43 PM

You'll find me floating in a river if SD doesn't cover -3 tonight.

Interestingly, I did the full-kelly solution too, and it has a lower EV (+4844), but a much higher expected growth (1.45%), assuming I did it right.

**Thremp** · 12-24-07, 08:45 PM

Originally posted by tomcowley

You'll find me floating in a river if SD doesn't cover -3 tonight.

Interestingly, I did the full-kelly solution too, and it has a lower EV (+4844), but a much higher expected growth (1.45%), assuming I did it right.

Interesting. I'm having fits with my virtualization software on my lap top. Somehow it randomly decided to stop working. So I have no clue if I'm gonna be leaping off a bridge or not.

Still need to work on those math skills so I can figure WTF all this silliness contains. :/

Sad that we're still working on Christmas Eve.

**tomcowley** · 12-24-07, 08:55 PM

It's basically combinations of overbets on the most +EV lines combined with hedges on the least -EV lines.

Actually it looks like I didn't iterate enough:

Bet 20142 on Bet 1 No (NE to win divisional).
Bet 3167 on Bet 2 Yes (NE to lose AFC).
Bet 32557 on Bet 4 Yes (NE to win Super Bowl)

Half-Kelly is +$7055 / 0.14% growth

Kelly is +$3659 / 1.91% growth

**Wheell** · 12-24-07, 10:16 PM

If I am reading that correctly you lose 52699 if NE loses in the divisional round in half Kelly. WHAT!?!

For the record my 11th solution (which is wrong) is

11810 Pats to win SB
3190 Pats to win div round + Pats to lose AFC championship game.
708 NFC over Pats in SB

In other news I am stuffed, drunk, and happy with a purring cat next to me.

**tomcowley** · 12-24-07, 10:30 PM

Yup. If NE loses in divisional, you take a bath. Thankfully it's only 10% to happen

In the kelly variation, you only lose 28-29k. Sqrt(x) is more tolerant of losing half your roll than ln x.

If you just bet half of what kelly tells you, then you get 1.4% growth and a maximum 14.3k bath.

**Wheell** · 12-24-07, 11:42 PM

I cannot believe that half Kelly has you risk more than half your bankroll.

**Ganchrow** · 12-24-07, 11:59 PM

Originally posted by tomcowley

I can't see why parlays (or if-bets) should be relevant here since bets 2 and 3 are already functionally if-win.

Consider it a plot device put in place to render quasi-reasonable the assumption that odds won't vary with time.

Originally posted by tomcowley

The only meaningful attempt is to create a synthetic NE to win SB parlay, but that comes out at +110 instead of +118, if my math is right.

It is.

Originally posted by tomcowley

My solution is:

Bet 20174 on Bet 1 No (NE to win divisional).
Bet 3192 on Bet 2 Yes (NE to lose AFC).
Bet 32580 on Bet 4 Yes (NE to win Super Bowl)

EV is +$7058

While this is not the correct solution to the problem I presented, it does correctly address an ever-so-slightly modified problem. I'll be willing to accept minority blame for the confusion, but am still waiting on the correct answer.

Hint: Does it really make sense for the half-Kelly solution to be higher EV than the full-Kelly?

**tomcowley** · 12-25-07, 12:28 AM

Nope, that's why I was just looking for references on this. I'd never seen the half-kelly utility function before. And you deserve full blame since I just copied your utility formula

You used n=2 instead of n=.5, so I should be maximizing -1/x (constant irrelevant) instead of sqrt(x), which gives..

7364.5 (bet 1 no)
1035.4 (bet 2 yes)
8295.8 (bet 4 yes)

EV +1891.9, 1.44% growth. Maximum bath 15.9k

**roasthawg** · 12-25-07, 12:46 AM

will ne lose in 1st rnd...bet no, wager $15,000/-600
will ne lose in afc championship...bet yes, wager $1,319/+183
will ne lose in sb...bet yes, wager $16,398.50/+437

close?

**Ganchrow** · 12-25-07, 12:55 AM

Originally posted by tomcowley

Nope, that's why I was just looking for references on this. I'd never seen the half-kelly utility function before. And you deserve full blame since I just copied your utility formula

You used n=2 instead of n=.5, so I should be maximizing -1/x (constant irrelevant) instead of sqrt(x), which gives..

7364.5 (bet 1 no)
1035.4 (bet 2 yes)
8295.8 (bet 4 yes)

EV +1891.9, 1.44% growth. Maximum bath 15.9k

As previously stated, the general Kelly utility function is:

U(X;n≠1) = ^{X^1-¹/_n} / _{(1-¹/_n)}

U(X;n=1) = ln(X)

Never take an example on faith when an exact formula is provided.

To the nearest penny I get:

1) No -600 $7,635.85
2) Yes +183 $1,034.92
4) Yes +118 $8,295.27

Anyway, a very good job. No cash prize, however, as you still didn't get the correct answer prior to midnight.

PM me your shipping info and choose from one of the following critical thinking-inducing selections:

CD set of the the Mackris v. O'Reilly Opera
I Am America (And So Can You!) by Stephen Colbert
The God Delusion by Richard Dawkins
Why Darwin Matters by Micahel Shermer
The Demon-Gaunted World: Science as a Candle in the Dark by Carl Sagan (paperback only)

**Ganchrow** · 12-25-07, 01:11 AM

Originally posted by roasthawg

will ne lose in 1st rnd...bet no, wager $15,000/-600
will ne lose in afc championship...bet yes, wager $1,319/+183
will ne lose in sb...bet yes, wager $16,398.50/+437

close?

Certainly close ... just not quite there.

**Wheell** · 12-25-07, 01:31 AM

ok, I am confused, why can you lose 15660 if the Patriots lose round 1? I had that capping at 15000.

**Ganchrow** · 12-25-07, 01:37 AM

Originally posted by Wheell

ok, I am confused, why can you lose 15660 if the Patriots lose round 1? I had that capping at 15000.

Lose $7,635.85 on bet 1 + $8,295.27 on bet 4 = $15,931.12 total loss.

**Wheell** · 12-25-07, 03:33 AM

OK, I've sobered up (I think, I drove home...) and I still don't see how a half Kelly edge can risk over 15k on a round 1 loss. I have 15000 being the max, so I wonder am I doing my kelly math wrong, am I not getting the edge correct, or what... I have the odds as being 6-1 on a 9-1 adv, and the odds being 100/218 on a 5/9 edge.

Perhaps I need to recheck the formula...

**Wheell** · 12-25-07, 03:39 AM

Never mind, I'm an idiot... but I STILL don't see why you can't lose a bit over 16k...

Sigh.

**Art Vandeleigh** · 12-25-07, 04:29 AM

Originally posted by tomcowley

(Post #40) It's basically combinations of overbets on the most +EV lines combined with hedges on the least -EV lines.

Actually it looks like I didn't iterate enough:

Bet 20142 on Bet 1 No (NE to win divisional).
Bet 3167 on Bet 2 Yes (NE to lose AFC).
Bet 32557 on Bet 4 Yes (NE to win Super Bowl)

Half-Kelly is +$7055 / 0.14% growth

Kelly is +$3659 / 1.91% growth

While it's true these weren't the final numbers, it's the first line in this quote that describes Tom's process to get the numbers - "It's basically combinations of overbets on the most +EV lines combined with hedges on the least -EV lines. "

He might as well have written "then a miracle happens here" as far as I'm concerned.

I think if the only thing Ganch does is change the numbers of this problem around (with the exact same wording), only Tom would get the answer again. I don't know if anyone learned anything about the process to get to the solution, I certainly didn't. But that's OK, that may not have been part of the intention of putting this problem up.

**Ganchrow** · 12-25-07, 08:30 AM

This is the spreadsheet I used to come up with the solution.

The process is essentially the same regardless of the underlying problem. You enumerate all possible bankroll outcomes, convert the outcome to utility (such as with ln(·) with a Kelly multiplier of 1), associate probabilities with each, and then take the dot proudct of the two to come up with an expected utility figure.

You then use an optimization engine (such as the standard Excel Solver package -- installation instructions here) to maximize expected utility with respect to the decision variables subject to non-negativity and the budget constraint (i.e., the sum of all bets must not exceed 1 -- this will only bind in the case of a riskless arbitrage). You may optionally wish to include to include other constraints on bet sizes such as, perhaps maximum bets at certain books.

And that's really about it. If you're using Excel+Solver then once you Solver installed go to the Tools menu and select "Solver..." to pull up the Solver dialog and get to maximizing.

Attached Files

NE SB Kelly Example.xls (166.0 KB, 352 views)

**tomcowley** · 12-25-07, 01:29 PM

That's basically what I did as well. For each of the 8 bets, enumerate what happens on each of the 4 outcomes, then maximize the expected utility.

**Art Vandeleigh** · 12-25-07, 01:58 PM

Merry Christmas Ganch, thanks for taking the time to post this on Christmas day.

The optimal results were given as follows:
1) No -600 $7,635.85
2) Yes +183 $1,034.92
4) Yes +118 $8,295.27

Specifics of arriving at these numbers aside, I just want to make sure I've got this right, as I'm sort of popping my Kelly cherry on this problem.

According to the rules of the problem, all bets need to made simultaneously. The following is what I get for all possible NE results in the playoffs

NE loses divisional playoffs:

Yes: -$7,635.85 - $8,295.27 = - $15,931.12

No: $7,635.85 / 6 = + $1,272.64

NE loses AFC Championship:

Yes: ($1,034.92 x 1.83) + $1,272.64 - $8,295.27 = - $5,128.72

No: $1,272.64 - $1,034.92 = + $237.72

NE wins SB:

Yes: $237.72 + ($8,295.27 x 1.18) = + $10,026.14

No: -$8,295.27 + $237.72 = - $8,057.55

So if the above is correct, a half-Kelly bettor would optimize their bankroll growth by risking nearly 16% of bankroll on the first bet, and risking a bit more than 5% of bankroll on the second bet, for the "privilege" of being able to risk $8,057.55 to win $10,026.14 on the Pats winning the SB.

I think I am more risk-aversive than this.

**Ganchrow** · 12-26-07, 10:16 AM

The way I'd suggesting considering the data you (correctly) presented would be on a per-outcome basis:

In other words, don't think as much about the bet microstructure as about the macro-impact of the full set of bets as a whole. Better yet, try to consider the set as part of a much larger portfolio of +EV plays.

For example, if you were to repeat this bet 10 times your outcome matrix would look like this (return to the nearest 10%):

There's no question that there's a notable degree of risk involved, but the point is that by using an appropriate n-Kelly you maximize your EV for a given (preset) linear combination of variance, skew, kurtosis, et. al. (which is really what most people actually intend when they use the vernacular catch-all phrase "risk"). The lower the value of the "n" in your n-Kelly the more importance is placed on lowering risk relative to return ... and the more bets you're able to place the more this becomes apparent.

Compare the above outcome set to that of a typical advantage dog-player risking risking 2.5% of his bankroll on 20 +300 shots with a 5% edge.

<table border=1 cellpadding=2 cellspacing=2> <tr> <td align=left class=GANCH> Probability </td> <td align=left class=GANCH> Outcome
</td> </tr> <tr> <td align=left class=GANCH> -40% to -30% </td> <td align=right class=GANCH> 1.8396%
</td> </tr> <tr> <td align=left class=GANCH> -30% to -20% </td> <td align=right class=GANCH> 5.4542%
</td> </tr> <tr> <td align=left class=GANCH> -20% to -10% </td> <td align=right class=GANCH> 29.2678%
</td> </tr> <tr> <td align=left class=GANCH> -10% to 0% </td> <td align=right class=GANCH> 20.0688%
</td> </tr> <tr> <td align=left class=GANCH> 0% to 10% </td> <td align=right class=GANCH> 17.8578%
</td> </tr> <tr> <td align=left class=GANCH> 10% to 20% </td> <td align=right class=GANCH> 12.7123%
</td> </tr> <tr> <td align=left class=GANCH> 30% to 40% </td> <td align=right class=GANCH> 7.3527%
</td> </tr> <tr> <td align=left class=GANCH> 40% to 50% </td> <td align=right class=GANCH> 3.4894%
</td> </tr> <tr> <td align=left class=GANCH> 60% to 70% </td> <td align=right class=GANCH> 1.3662%
</td> </tr> <tr> <td align=left class=GANCH> 70% to 80% </td> <td align=right class=GANCH> 0.4421%
</td> </tr> <tr> <td align=left class=GANCH> 90% to 100% </td> <td align=right class=GANCH> 0.1180%
</td> </tr> <tr> <td align=left class=GANCH> 110% to 120% </td> <td align=right class=GANCH> 0.0258%
</td> </tr> <tr> <td align=left class=GANCH> 130% to 140% </td> <td align=right class=GANCH> 0.0046%
</td> </tr> <tr> <td align=left class=GANCH> 160% to 170% </td> <td align=right class=GANCH> 0.0007%
</td> </tr> <tr> <td align=left class=GANCH> 180% to 190% </td> <td align=right class=GANCH> 0.0001%
</td> </tr> <tr> <td align=left class=GANCH> 210% to 220% </td> <td align=right class=GANCH> 0.0000%
</td> </tr> <tr> <td align=left class=GANCH> 240% to 250% </td> <td align=right class=GANCH> 0.0000%
</td> </tr> <tr> <td align=left class=GANCH> 280% to 290% </td> <td align=right class=GANCH> 0.0000%
</td> </tr> <tr> <td align=left class=GANCH> 320% to 330% </td> <td align=right class=GANCH> 0.0000%
</td> </tr> </table>

So what we see is that even after placing 20 bets the dog-bettor has a roughly 56.6% probability of losing money. Compare this to the case of the NE half-Kelly bettor with only a 31% probability of losing money (and that's after only 10 bets).

I think this illustrates the questionable manner in which many people view risk -- they tend to only look at few of the worst case scenarios and compare those to a few of the best case scenarios. This is related to the "lottery mentality" of betting against which I've spoken out in the past. This is where people place too much value on the skewness of a distribution, frequently equating negative skew with risk.

Consider this: Which is the riskier bet?

Betting 1% ($100) of a $10,000 bankroll on a +1,000 event with 5% edge, or
betting 10% ($1,000) of a $10,000 bankroll on a -1,000 event with a 5% edge?

Most people would answer without hesitation and say that the second bet is "riskier" because a player is risking $1,000 for a chance to only win $100.

But that reasoning is incorrect and speaks to a lack of understanding of the true nature of risk. In fact the dog bet has a higher standard deviation ($323.23 vs. $229.13).

Furthermore, and probably of more real-world importance to an advantage sports bettor, if you place were to place the dog bet 100 times over (risking 1% of bankroll each time) your expectation would be to make $512.58, but your likelihood of turning any profit at all would only be less than 50% (48.76% to be more exact). Sure you might have a 0.1% chance of making more than $23,452.36, but you have a roughly 28% probability of losing 15% of your bankroll or more.

Contrast that to the expectation of betting -1,000 favorite 100 times (risking 10% of bankroll each time). That your expectation would be significantly higher ($6,466.68) should come as no surprise (after all you're risking 10 times as much per bet), but what might come as a surprise would be your median outcome. Specifically, you'd have a bit better than 50% (52.11% to be more exact) shot of turning a profit of $7,053.83 or more and a 96% probability of turning any profit at all. You're probability of losing 15% or more of your bankroll would only be 1.57%.

Still think the dog bet is less risky?

The point is that by focusing on long odds bets, while you allow yourself a small probability of windfall profit, your most likely outcomes after a large number of bets will hover much closer to zero than they would by betting at shorter odds.

**Art Vandeleigh** · 12-26-07, 04:21 PM

Thanks, you've given me a lot to think about.

I wonder why the results of math, particularly probabilty theory, need to so often be counter-intuitive.

**Ganchrow** · 12-26-07, 04:43 PM

Originally posted by Art Vandeleigh

I wonder why the results of math, particularly probabilty theory, need to so often be counter-intuitive.

It's a skewed sample, right?

It's the counter-intuitive cases that are generally the most interesting and hardest to understand, and hence are overrepresented in discussions.