# An Elegant and Useful Theorem about Wagering

Before stating and proving the elegant theorem, let's have some preliminaries.

### MECE Events and their Probabilities

A set (S) of events is said to be "mutually exclusive and collectively exhaustive" (MECE) if exactly one of the events will occur. For example, S = { s1, s2, s3, ... s33 } might be the set of possible answers to the question "Who will win next year's Superbowl", and (p1, p2, p3, ... p33 ) the associated probabilities, pk = Probability (sk will come true). Each probability will be non-negative; the 33 probabilities will sum to 1; and the joint probabilities Probability (si and sj are both true) are all zero whenever i and j are unequal.

Why do I show 33 possible outcomes when there are only 32 teams in the NFL? Because to avoid confusion I want to ensure that the probabilities sum to exactly 100.0%, not to 99.999999%. Extra teams might be added between now and then; the NFL might go bankrupt; whatever -- I'm leaving room for "s33 = 'None of the above'."

### The Bernoulli-Kelly Criterion

If you have a million dollars and Bill Gates offers to let you gamble on a fair coin toss where he's paying off 6-5, would you bet the whole million? Sure, 50% of the time you finish with \$2.2 million, but the other 50% -- you'll be in the poorhouse begging for bread and water! Betting the whole million would maximize your expected winnings (\$100,000) but would be very foolish.

Instead of maximizing expected bankroll it might be better to maximize its expected logarithm. Or stated differently, rather than maximizing the (weighted) arithmetic mean of your resultant bankroll, you should maximize the (weighted) geometric mean (or, equivalently, its power). In our simple example, this is the same as maximizing E(b) = (1 - b)*(1 + 1.2b) where b is your bet-size in millions. Set E'(b) = 0 to find the extremum. It's at b = 0.083333; you're best off betting just \$83,333. If you use Kelly's Criterion. I won't try to explain or justify Kelly's Criterion which is named after John Kelly (1923-1965), an associate of Claude Shannon. But I'll rename it the Bernoulli-Kelly Criterion to emphasize that the Criterion isn't really new -- Daniel Bernoulli had written on the basic principle in the 18th century.

## Wagering Puzzle

So we have a set S = {s1, s2, ... s33} of 33 MECE events and two gamblers, Bill and Alice. (sk might be the event that some particular team wins the next Superbowl; I think it would be more fun to bet on who will be the Democratic nominee for President.) Each of the gamblers assigns probabilities to the events. To simplify and avoid quibbles we'll assume that Bill publishes the 33 probabilities (qk) that he believes in and promises not to change them. These 33 probabilities sum to 1 exactly.

Alice is welcome to study Bill's {qk} and use his expertise to help refine her own estimates, but let's assume she finishes with estimates {pk} that are different from Bill's estimates {qk}. Otherwise there'd be no interesting gambling to happen.

Now Bill (who happens to be Bill Gates and is just having fun -- he knows that Alice's winnings or losses will have no real effect on his fortune) offers to sell tickets. If his q4 = 0.29 for example, he'll sell Alice tickets on s4 at a price of \$0.29. The tickets will be worth \$1 if s4 succeeds, zero otherwise. Note that, since the {qk} sum to one, these are fair bets; there is no "vigorish." If Alice spends the portion qk of her bankroll on each of the sk tickets, she will be guaranteed to break exactly even at the end.

Alice can buy as many tickets as she likes, subject to only one constraint: She must pay cash; when she's out of money she must stop buying.

Assuming that she applies the Bernoulli-Kelly Criterion and seeks to maximize the expected geometric mean of her resultant bankroll, How should Alice Bet?

We'll assume that Alice's starting bankroll is 1 in some units, e.g. \$1 million. The solution will be Alice's 33 bets (or ticket purchases) (b1, b2, ... b33). Her total bet is Σ bi. Her bankroll after the winning event is chosen (call it sw) will be 1 - Σ bi + bw/qw. Alice believes that event will occur with probability pw.

## Solution to the Wagering Puzzle

The amazing solution is that Alice's wagers should be exactly bk = pk for each k. Note that the odds (qk) that Bill is offering seem to be ignored in this solution!

There are other solutions that are just as good. A bet list of (a·qk) will break even on any result, where a is any positive number, so Alice might choose to wager bk = pk - a·qk on each of 32 events and wager zero on event st, where st is whichever event minimizes a = pt / qt

I'm not sure when and where this Theorem was first published: I'll call it Shearer's Wagering Theorem since James B. Shearer presented it as the IBM Ponder puzzle in January 2007.

## Proof of Shearer's Wagering Theorem

From the preceding observation we can arrange that Σ bi = 1. In this case observe that an optimal bet list will have bk > 0 whenever pk > 0; otherwise sk would yield a non-positive final bankroll, which would destroy our geometric mean!

Now, 1 - Σ bi + bw/qw becomes 0 + bw/qw. The weighted mean of bankroll logarithms over all possibilities is Σ pi log (bi/qi). Finding the 33 optimal (bk) requires setting the 33 partial derivatives -- one for each bk -- to zero. As we've observed, there are infinitely many solutions of which one satisfies Σ bi = 1. Let's enforce that satisfaction, which will allow an easy application of the Euler-Lagrange Theorem.

So, we find the partial derivative Fk with respect to bk, of Σ pi log (bi/qi). This is Fk = pk / bk. (When you work the details you'll get qk in both numerator and denominator and they cancel.) To this we'll add, in accordance with the Euler-Lagrange Theorem, the partial derivative of Σ bi -- just 1 -- times an "undetermined multiplier;" and find the extremum by setting this sum (Fk + λ) to zero. After rearrangement this is simply pk = λ bk.

λ is the same for all events, Σ pk = 1 because these are the probabilities of MECEs and Σ bi = 1 by the constraint we imposed above. This means λ=1 and we're left with bk = pk for all k. Q.E.D.

Copyright © 2019 by James D. Allen

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