Puzzle III is phrased in terms of the Super Bowl, but no knowledge of American football is relevant. The player's knowledge of football is encapsulated in his probability estimates (p1, p2, p3, p4, ...) Instead of betting on the Super Bowl, the bet could be on how many felony convictions some specific politician has before the end of 2024, or whatever.
III. Betting on the Super Bowl
There are 32 teams that might win the Super Bowl, so we will say that the future is partitioned into 33 possibilities: E1, E2, E3, E4, ... , E33. I've added a 33rd case, "None of the above" to accommodate weirdnesses and ensure we have a perfect partition. Your task will be to determine the bets x1, x2, x3, x4, ... , x33 you should place to maximize your risk-adjusted reward.
Warren Buffett once offered $1 billion to anyone who guessed the NCAA men's basketball tournament perfectly, so we'll impose on him again to fashion this betting problem. Both he and you each come up with 33 probability guesses p1, p2, p3, p4, ... , p33 where pk is the guessed probability that team #k will win the Super Bowl. (To avoid confusion Buffett's estimates will be denoted q1, q2, ... rather than p1, p2...) Buffett reveals his final estimates BEFORE you finalize yours, so you're free to use his expertise to modify your own guesses. Buffett's 33 probabilities must sum to exactly 1. The same constraint applies to yours.
Buffett offers to accept wagers without vigorish or commission. If he judges Dallas to be 25% to win, he offers 3:1 odds. If you spend $10 for a Dallas "ticket" the ticket will be worth $40 if Dallas wins. More generally, Buffett offers (1-q) : (q) for any event whose likelihood he judges to be q.
If your probability estimates are exactly the same as Buffett's you may as well buy no tickets at all. Instead you'll be trying to exploit the flaws in Buffett's estimates. Assume that you have total confidence in your own probability estimates.
You have 1 unit of money to bet; assume that's all the wealth you have in the world. (Perhaps 1 unit = $10,000.) After you place your bets, and the NFL season and Super Bowl play out, all but one of your tickets will be worthless. You'll have A dollars where A is the value of your winning ticket (if any) plus the change left over if you didn't spend the entire 1 unit on tickets. Your goal is to maximize the Expected Value of the Logarithm of A. (This is the same as maximizing the "weighted geometric mean" of the various outcomes.) Note that this is different than many betting puzzles where the goal is to maximize the expected value of A itself (or the weighted arithmetic mean). However this is NOT a perverse objective. This criterion for reward-vs-risk is well-known, and is often called the "Kelly Criterion." It is associated with John L. Kelly Jr. and, centuries earlier, with Daniel Bernoulli.
Please ask questions if any of this is unclear. :-(
The description supposes N=33 possible outcomes, but N can be any finite plural integer; and a specific example below will use N=3.
So we are given the bookie's probability estimates ∑q
i = 1 and our own estimates ∑p
i = 1. If w is the winning outcome, a winning ticket for which we pay x
w will be worth (x
w / q
w).
We want to derive our optimal bets x
i with 0 ≤ ∑x
i ≤ 1. Specifically we want to maximize the weighted geometric mean
Maximize ∏ ((xi / qi + R)pi)
where R = 1 - ∑x
i is the residual bankroll. Since logarithm is monotonic, this is equivalent to maximizing the expected logarithm of bankroll:
Maximize ∑ (pi (log (xi / qi) + R))
Note that R may become troublesome. It would be nifty if R=0.
Remark 1.
If we set our bets as x
1 = q
1, x
2 = q
2, x
3 = q
3, etc. the net effect will be the same as buying no tickets at all. All but one of the tickets will become worthless, but the winning ticket will be worth (x
w / q
w), which will be 1, since we've used x
i = q
i for our bets.
Similarly, if we set x
1 = R⋅q
1, x
2 = R⋅q
2, x
3 = R⋅q
3, etc., the value of these tickets will sum to R both before and after the event.
Example:
Suppose our probability estimates are
p1 = ⅙
p2 = ⅓
p3 = ½
And the bookie's estimates are
q1 = ½
q2 = ⅓
q3 = ⅙
As we will see below, our optimal bets are
x1 = 0
x2 = 2/9
x3 = 4/9
Applying Remark 1 we see that the bet set
x1 = R/2
x2 = 2/9 + R/3
x3 = 4/9 + R/6
is also optimal (for any 0 ≤ R ≤ 3/9)
Thus there are an infinity of solutions. In this example, any 0 ≤ x
1 ≤ 1/6 will be optimal, depending on the other x
i. This will also interfere with solution.
The remedy for both problems is simply to enforce ∑x
i = 1; i.e. to insist that R = 0. Remark 1 assures us that this is possible. Now we have a unique solution and a simple solving method using a Lagrange multiplier:
∂ (∑ (pi log (xi / qi)) + λ ⋅ (∑xi - 1)) / ∂x1 = 0
∂ (∑ (pi log (xi / qi)) + λ ⋅ (∑xi - 1)) / ∂x2 = 0
∂ (∑ (pi log (xi / qi)) + λ ⋅ (∑xi - 1)) / ∂x3 = 0
...
∂ (∑ (pi log (xi / qi)) + λ ⋅ (∑xi - 1)) / ∂x33 = 0
∑xi = 1
Terms without x
j are irrelevant to ∂ / ∂x
j so these equations simplify dramatically
∂ (pj log (xj / qj) + λ ⋅ xj) / ∂xj = 0 ; for j = 1,2,3,...,33
∑xi = 1
Taking the indicated derivative, ∂ (p
j log (x
j / q
j) + λ ⋅ x
j) / ∂x
j = 0 becomes
pj (qj / xj) / qj + λ = 0 ; for j = 1,2,3,...,33
The q
j terms cancel! We are left with 33 equations of the form
pj / xj = - λ
Since ∑x
i = 1 and ∑p
i = 1 it is easily seen that λ = -1 and every x
j is equal to the corresponding probability estimate p
j.
Does this surprise? The bookie's odds qj are irrelevant to the optimal bets (when R=0 is enforced).
We can simply ignore the bookie's odds and buy tickets in exact proportion to our own probability estimates.