[Ganchrow]

(Note: While I've tried to keep this as simple as possible, carefully explaining each step, this isn't a topic for those who aren't already comfortable with gambling mathematics. A basic understanding of the Kelly Criterion and the calculus of differentiation would probably be helpful, too. If anything's unclear, please feel free to ask.)

Last Saturday, based on purely quantitative criteria, I expressed my opinion that AFC -6 would be a bad bet, yielding an expected return of about -12.2% at -110. This implied that NFC +6 +100 would be a good bet, yielding an expected return of 8.0%.

Surprisingly, there was at least one person out there who believed me. Let’s call him Peter. Peter put down a fairly sizeable chunk of change on NFC +6 +100. Just how sizeable? Well it turned out to be about a quarter of Peter’s bankroll. His intention had been to substantially reduce his exposure (at a decent sized-profit) after Sunday’s games.

Well guess what? Fast-forward to today, and the Super Bowl spread is trading at AFC -6½ -109 / NFC +101. Ouch. What’s more, Peter doesn’t even have a view on the game. The only circumstance working in Peter’s favor is that he doesn’t have a pressing need for the funds, and so doesn’t mind having up to 60% of his bankroll locked in until after the game on Sunday.

So what to do? About 3 weeks ago I gave the following advice to a bettor in another thread:

Originally Posted by Ganchrow

Unless you have a view on the game itself or unless you really need to safeguard the at-risk funds just stick with the … position you already have. Without either an actual view on the game or a pressing need to safeguard the money you've already risked, your hedge would be little more than a blind gamble at a house edge [equal to the vig]”

Although the question had initially been asked with regard to a positive expectation position, it might be reasonable to think that the same advice should also hold in the case of a negative expectation position. So does it?

Well no. But it doesn't always apply to positive expectation positions either. What I should have explicitly stated in the other thread is that in general this is only true in the case of a risk-averse bettor, for bets that are sufficiently small relative to the bettor's bankroll. In the case at hand, the player has risked 25% of his bankroll on this one bet. This is not a small bet relative to his bankroll. So how much should he hedge?

Peter has two candidate bets for his hedge (remember that his current position is 25% of his bankroll on NFC +6 +100)

1. AFC -6½ -109
   [NFC -6½ +101]
2. AFC -6 -120
   [NFC -6 +107]

The first hedge candidate (AFC -6½ -109), is offered at 1.87% vigorish. The disadvantage of this bet is that it does not represent a perfect hedge. This is because Peter would be leaving himself exposed (for the size of the hedge bet) if the end result of the Super Bowl were AFC winning by exactly 6. (If you don’t understand how I came up with these figures you might want to check out http://forum.sbrforum.com/players-ta...ical-hold.html.)

The second hedge candidate (AFC -6 -120), on the other hand, is a perfect hedge. The vig on that bet, however, is 2.78%, which is considerably higher than the vig on the first candidate bet.

So it turns out that the solution is really rather straightforward if we assume logarithmic preferences (à la Kelly).

So to start off we’ll first need to identify the distinct outcome scenarios and then assign probabilities to each of them. It should be apparent that 3 distinct outcome scenarios exist, to wit:

1. NFC wins straight-up or AFC wins by less than 6
2. AFC wins by exactly 6
3. AFC wins by more than 6

Looking at the market for the first hedge candidate, we see that it implies that the probability of either outcome 1 or 2 occurring is about 48.82%, and that the probability of outcome 3 occurring is about 51.18%.

Looking at the market for the second hedge candidate, we see that the probability of outcome 1 occurring, conditioned on outcome 2 NOT occurring is 46.97%, and that the probability of outcome 3 occurring, conditioned on outcome 2 NOT occurring is 53.03%. (If you don’t understand how I came up with these figures, you may want to check out http://forum.sbrforum.com/players-ta...rcentages.html.)

So, given these probabilities we get set up the following conditional probability equivalence (you'll recall that p(x | y) is the probability of x occuring given that y has occured):

p(outcome 3 | NOT outcome 2)
= p (3) / (1 - p(2)), which gives us

p(2) = 1 - p(3) / p(3 | NOT 2)
= 1 - 51.18% / 53.03%

p(2) = 3.49%

Because we already know that p(3) = 51.18%, this gives us p(1) = 1 - 3.49% - 51.18% = 45.33%.

So to better visualize what we currently know, we can breakdown each outcome scenario as follows:

1. NFC wins straight-up or AFC wins by less than 6
   Probability: 45.33%
   Initial bet (NFC +6 +100) wins, paying off at 1/1.
   Hedge candidate 1 (AFC -6½ -109) loses.
   Hedge candidate 2 (AFC -6 -120) loses.
2. AFC wins by exactly 6
   Probability: 3.49%
   Initial bet pushes.
   Hedge 1 loses.
   Hedge 2 pushes.
3. Outcome 3: AFC wins by more than 6
   Probability: 51.18%
   Initial bet loses.
   Hedge 1 wins, paying off at 100/109.
   Hedge 2 wins, paying off at 100/120.

Because we’re assuming logarithmic preferences, the generalized form of our expected utility function would be:

E(U) = ∑_i{p(outcome_i) * ln[1 + (outcome_i profit or loss)]} for K = 1

E(U) = ( ^K/_K-1) * ∑_i{p(outcome_i) * [1 + (outcome_i profit or loss)]^1-1/_K} for K ≠ 1

where K is the constant Kelly multiplier (so for example K=1 would imply standard Kelly, K=½, would imply half Kelly, etc.), and profit or loss is given as a percentage of the total bankroll.

So now that we have our utility function, all that's left is figuring out what our variables are, and then setting up the constraints.

It should be readily apparent that we have two unknowns, the bet size on hedge candidate 1 (call it b1) and the bet size on hedge candidate 2 (call it b2). Note that all bet sizes are given as a percentage of the total bankroll. Our constants are the decimal odds on the initial bet, which is 2, the decimal odds on hedge 1 and hedge 2, which are 1+100/109 and 1+100/120, respectively, the probabilities of each of the three outcomes, which are given above, and lastly the Kelly divisor, K, which we’ll just set to 1 (meaning standard Kelly).

The constraints are:

1. b1 ≥ 0
2. b2 ≥ 0
3. b1 + b2 ≤ 35% (35% because, as you'll recall, Peter doesn’t want more than 60% of his bankroll tied up. Because 25% of his bankroll is already being used to fund the initial bet, that leaves up to 35% of his bankroll to be used on the hedge).

(Of course if the maximum bet sizes were small enough relative to the total bankroll, we’d have to include further constraints on b1 and b2, constraining each to be less that the maximum bet, however because this is the Super Bowl, the maximums won’t come in to play for Peter.

So finally, here’s our objective function:

E(U) = 45.33% * ln(1 + 25% - b1 - b2) +
3.49% * ln (1 - b1) +
51.18% ln (1 - 25% + b1 * 100/109 + b2 * 100/120)

Which we want to maximize with respect to b1 & b2, subject to b1 ≥ 0, b2 ≥ 0, and b1 + b2 ≤ 35%.

At this point we need to decide whether we want to proceed computationally or analytically. Either methodology will obviously work. If you wanted to proceed analytically, you’d take the partial derivatives with respect to b1, b2, and the Lagrangians, set each one to zero, and solve for b1 & b2 (making sure not to forget the Kuhn-Tucker conditions for the constraints).

If you wanted to proceed computationally, you could use any mathematical package, such as Mathematica, or Microsoft Excel Solver to do all the work for you. Solver is probably the easiest route as it’s included for free with Excel.

So anyway, whatever method you choose, the result is the same. There’s one unique solution that satisfies the constraints (which incidentally, don’t bind). Namely:

b1 = 22.0602%
b2 = 0.9120%

This implies the following results under each outcome scenario:

1. NFC wins straight-up or AFC wins by less than 6: bankroll increases by 2.03%
2. AFC wins by exactly 6: bankroll decreases by 22.06%
3. AFC wins by more than 6: bankroll decreases by 4.00%.

So what we see here is that under standard Kelly, it’s correct for Peter to eliminate most of his risk using the cheaper candidate (hedge 1), even though it still leaves him the chance of a disastrous reverse middle, and only bet a small portion of his bankroll on the more expensive hedge 2, even though it would provide a perfect hedge.

The reader should note that if we had used a Kelly multiplier less than 1, what we’d see would have been a somewhat smaller bet on hedge candidate 1, and a somewhat larger bet on hedge candidate 2. With a Kelly multiplier of 0.25 (quarter Kelly), for example, the solution would be:

b1 = 7.5500%
b2 = 18.5613%

implying the following results under each scenario:

1. NFC wins straight-up or AFC wins by less than 6: bankroll decreases by 1.11%
2. AFC wins by exactly 6: bankroll decreases by 7.55%
3. AFC wins by more than 6: bankroll decreases by 2.61%

So in other words Peter would be trading off a bit of expected value and would be eliminating the possibility of a profitable outcome, so as to reduce his exposure to the “worst-case scenario” (AFC by exactly 6).

(If anyone's interested, I’ll put together a simple spreadsheet demonstrating this process using Solver. Anyone? Anyone at all?)

So hopefully, this explains a bit of the mystery behind how to determine optimal hedge quantities. While this example is fairly simple, the same techniques could be used to determine more complicated hedges involving, for example, combinations of straight bets, parlays, and teasers.

See my simple Kelly Hedge Spreadsheet (a more complicated example is linked to from this post).

[Korchnoi]

Great post as usual. I guess Peter had risk on last weekends games that he didn't hedge? Do you still stand by last week's analysis of the bet?

Perhaps Peter should look into SkyBook who offers Colts -6.5 at -112. I know as recently as a few weeks ago they gave you a free 1/2pt as long as it wasn't on/off the 3/7, so this looks like a possibility also.

[Ganchrow]

Originally Posted by Korchnoi

Do you still stand by last week's analysis of the bet?

Yes. I most definitely do.

Originally Posted by Korchnoi

Perhaps Peter should look into SkyBook who offers Colts -6.5 at -112. I know as recently as a few weeks ago they gave you a free 1/2pt as long as it wasn't on/off the 3/7, so this looks like a possibility also.

They're still offering the half-point on wagers up to $500. Actually, Colts -6 -112, based on the Pinnacle numbers I used in the example above, would represent a (slightly) positive expectation bet (0.368%).

[Art Vandeleigh]

I don't get what you're trying to say.

Let's say Mr. X has a bankroll of $4,000

He bet 25% or $1,000 at +100 odds on NFC +6

This turned out to be a bad bet. To safely hedge at AFC -6 he must lay -120.

So if Mr. X bets $1090 on AFC -6, he would get back $908

If AFC win by 7 or more, would lose $1000-$908= $92
If AFC wins by 6, both bets push
If NFC wins SU or loses by 5 or fewer, would lose $1090-$1000=$90

So MR. X is going to lose about 90/4000=2.25% of bankroll.

What are you trying to optimize here? I'm so not with the program.

[Jay Edgar]

Originally Posted by Ganchrow

(If anyone's interested, I’ll put together a simple spreadsheet demonstrating this process using Solver. Anyone? Anyone at all?)

Yes please!

(Not that I've ever actually had to sweat a reverse middle or anything . . . .)

[Ganchrow]

Originally Posted by Art Vandeleigh

I don't get what you're trying to say.

Let's say Mr. X has a bankroll of $4,000

He bet 25% or $1,000 at +100 odds on NFC +6

This turned out to be a bad bet. To safely hedge at AFC -6 he must lay -120.

So if Mr. X bets $1090 on AFC -6, he would get back $908

If AFC win by more than 7, would lose $1000-$908= $92
If AFC wins by 6, both bets push
If NFC wins SU or loses by 5 or fewer, would lose $1090-$1000=$90

So MR. X is going to lose about 90/4000=2.25% of bankroll.

What are you trying to optimize here? I'm so not with the program.

In this example we're maximizing the player's expected utility. Putting it another way, we're attempting to find the optimal trade-off between risk (which is bad) and return (which is good).

If the player were only interested in maximizing return (in other words, if he were risk neutral) the optimal decision would be not to hedge at all. This is because both hedge candidates have negative expected value.

If the player were only interested in minimizing risk (in other words, if he were infinitely risk averse), he would behave largely as you've described. (An infinitely risk averse player would want the same financial outcome regardless of the game's outcome. He would do this by betting less on AFC -6 -120 than in your example, and more on AFC -6½ -109. So assuming that the player bet $1,000 on NFC +6 +100, the proper hedge for the infinitely risk averse player would be $1,000 on AFC -6, and $86.92 on AFC -6½. This way, regardless of the game's outcome the financial result would be a loss of $86.92.)

What I demonstrated was how a player with logarithmic preferences (which are the preferences that the Kelly criterion requires of a player for it to be that player's optimal staking strategy) would go about optimally hedging the initial bet, given the two hedge candidate choices.

The generalized form of the expected utility function (which is given in the initial post) is:

E(U) = ( ^K/_K-1) * ∑_i{p(outcome_i) * [1 + (outcome_i profit or loss)]^1-1/_K} for K ≠ 1

E(U) = ∑_i{p(outcome_i) * ln[1 + (outcome_i profit or loss)]} for K = 1

where constant K is the Kelly multiplier.

[Ganchrow]

Originally Posted by Jay Edgar

Yes please!

(Not that I've ever actually had to sweat a reverse middle or anything . . . .)

This is the example spreadsheet I put together as proof of concept. It should be fairly self explanatory (scroll down to fill in the probabilities).

For the spreadsheet to work you'll need to have solver.xla added as a visual basic reference. If it produces an error when you run it, then you'll need to do this manually. For an explanation of how to do this check out section III ("Hints for Using Solver in Macros") of this link.

Kelly Multi-Hedge Example

[Dark Horse]

If the object is to hedge the NFC +6 bet, I would consider an open teaser strategy. If 25% of bankroll is at risk on the Superbowl bet, take five or ten games you would have bet anyway, but now as the first leg of a number of open teasers; the second leg for each is the Superbowl (now down to AFC -0.5). Greek allows two weeks for open teasers to be completed.

[Ganchrow]

Originally Posted by Dark Horse

If the object is to get out of the NFC +6 bet

Technically speaking, the object is to maximize expected utility -- this may or may not entail getting out of some or all of the NFC +6 bet.

Originally Posted by Dark Horse

I would consider an open teaser strategy. If 25% of bankroll is at risk on the Superbowl bet, take five or ten games you would have bet anyway, but now as the first leg of a number of open teasers; the second leg for each is the Superbowl (now down to AFC -0.5).

Unless the implicit juice in teasing from the 6½ to the ½ is less than the juice on the straight AFC -6 bet (which there may or may not be depending on what the Greek charges for the teaser), then this would necessarily be more expensive than and would result in a riskier outcome than (by creating the large middle) what could be achieved with some combination of the two candidate hedges. In other words, it would necessarily be suboptimal for a player with log preferences.

Of course of it turned out that the teased bet were cheaper (in terms of vig) than either of the unteased bets, then the tease may well be the right play.

[Dark Horse]

The NFC +100 no juice bet means I'm averaging out around -110 for the hedge, but I'm getting a real shot at winning both sides (with AFC down to -0.5).

Wouldn't be one teased bet, but a larger number for first leg, until risk on NFC bet was matched. Basically, I would get a hugely improved line for bets I would bet anyway. Losers would be tossed out, and winners go on until risk on NFC bet was matched.

Edit - In this case, it wouldn't really have to be an open teaser format (normally so for games more than a week away), but could be regular teasers, unless you wanted to include games after the Super bowl.

I would probably combine angles. For instance, by paying the juice for the AFC -6 -120 to lower my risk (25% of bankroll on one game is a bit steep to me), continuing to play my normal bets, and also playing those normal bets as teasers for a smaller bet size until amount risked on NFC bet was matched. (A teaser for AFC -0.5 and NFC +6 includes my projected outcome for Superbowl, so that would be a nice angle.)

A mathematical analysis to the right side of the decimal point is just not realistic to me, unless I would do so for all bets. The Lord knows I don't.

[MrX]

Originally Posted by Art Vandeleigh

Let's say Mr. X has a bankroll of $4,000

Why are we bringing me into this?

-Mr. X

P.S. Ganchrow, would you please try putting a little thought and substance into your posts. We already have one JJGold around here.

[Ganchrow]

Originally Posted by Dark Horse

I'm not sure I understand what you're saying.

The NFC +100 no juice bet means I'm averaging out around -110 for the hedge, but I'm getting a real shot at winning both sides (with AFC down to -0.5).

Wouldn't be one teased bet, but a larger number for first leg, until risk on NFC bet was matched. Basically, I would get a hugely improved line for bets I would bet anyway. Losers would be tossed out, and winners go on until risk on NFC bet was matched.

It's difficult to explain this qualitatively and much easier to explain quantitatively. But before we can do this we'd need to assign probabilities to the teaser. For the sake of simplicitly we can just assume we're dealing with a one-team teaser priced equivalently as a two-team teaser -- this would be decimal odds of 1.3817, which is just the square root of 1.9091. Or we can just say we're taking the AFC on the money line, which would be equivalent to AFC -½.

So I'll ask you ... assuming that AFC -6½ wins with probability 51.18%, how often would you expect AFC -½ to win? (Paying at odds of 1.3817, it would need to be lower than 70.37% in order for the juice to be higher than on the AFC -6 single. Based on the Pinnacle money line of-250/+230, we'd expect AFC -½ to win with probability 70.21%.)

[Sam Odom]

Ganchrow is the man!

btw- I have AFC -5.5 (-110) for 10units and Chi +7 (-110) for 5units