(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)
- AFC -6½ -109
[NFC -6½ +101] - 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://www.sportsbookreview.com/forum/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:- NFC wins straight-up or AFC wins by less than 6
- AFC wins by exactly 6
- 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://www.sportsbookreview.com/forum/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:
- 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. - AFC wins by exactly 6
Probability: 3.49%
Initial bet pushes.
Hedge 1 loses.
Hedge 2 pushes. - 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(outcomei) * ln[1 + (outcomei profit or loss)]} for K = 1
E(U) = ( K/K-1) * ∑i{p(outcomei) * [1 + (outcomei 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:
- b1 ≥ 0
- b2 ≥ 0
- 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:
- NFC wins straight-up or AFC wins by less than 6: bankroll increases by 2.03%
- AFC wins by exactly 6: bankroll decreases by 22.06%
- 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:
- NFC wins straight-up or AFC wins by less than 6: bankroll decreases by 1.11%
- AFC wins by exactly 6: bankroll decreases by 7.55%
- 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).