Even Odds

Here is another attempt to present the same algorithm, with the goal of making it easier to memorize:

"Each puts in the square of their surprise, then swap."

To spell this out, I predict that some event will happen with probability 0.1, you say it is 0.25. When it happens, I am 0.9 surprised and you are only 0.75 surprised. So I put down (0.9)^2 D, you put down (0.75)^2 D, and we swap our piles of money. Since I was more surprised, I come out the loser on the deal.

"Square of the surprise" is a quantity commonly used to measure the failure rate of predicative agents in machine learning; it is also known as Brier score. So we could describe this rule as "each bettor pays the other his or her Brier score." There was some discussion of the merits of various scoring systems in an earlier post of Coscott's.

I've generalized the Even Odds Algorithm to any number of people and any number of alternatives that they want to bet on. Here is the link http://bywayofcontradiction.com/?p=118

A worked example:

Your friend says "It's totally gonna work this time". You say "uh, no it's not". She says "Okay, maybe not but like 90% yes". You say "Bet, it's only like 1 in 4," putting down $25 (.5625-.0081) = $13.86. She puts down her $25 (0.8281-.0625) = $19.14. She attempts, smug in her expected 0.91 $13.86 - 0.09 $19.14 = $10.89, while you wait arms folded for your expected 0.75 $19.14 - 0.25 $13.86 = $10.89.

Sweet! In practice I would allow any percent estimate, loser gets to keep fractional cents as a humorous consolation prize, and obviously always settle up over PayPal or successive games of zero $EV high card to even up to whatever denominations the loser has on hand, because using pennies lowers both participants $EV to negative infinity.

This doesn't punish estimates like 99.999% when the estimate should be more like 99%, unless you rescale and allow much higher than $25 bets.

A less complex way of making bets on probabilities, which I learned at a finance internship, is for one player to "make a market" on the outcome of interest--that is, offer to buy or sell contracts whose value settles to, e.g., $1 if the outcome is true and $0 if the outcome is false. If you make a market with a "spread"--e.g. buy at .40, sell at .60--then you can safely name your probability, as the counterparty in this anecdote asks you to do, without giving up a large advantage. For example, at my internship one common thing was precommitting to trade on any market of a given spread.

This has the advantage of being easier to remember and allowing multiple parties, but the disadvantage of (probably, I think) being subject to gaming and market dynamics which distract from the probabilities (but are interesting in their own right).

A pays: (A ^ 2 - (1 - B) ^ 2) * 25 = (0.36 - 0.09) * 25 = 6.57
B pays: (B ^ 2 - (1 - A) ^ 2) * 25 = (0.49 - 0.16) * 25 = 8.25

A = 0.4
B = 0.7

A pays: (A ^ 2 - (1 - B) ^ 2) * 25 = (0.16 - 0.09) * 25 = 1.75
B pays: (B ^ 2 - (1 - A) ^ 2) * 25 = (0.49 - 0.36) * 25 = 3.25

Assume p=0.5, q=0.8. Alice should bet d( (1-0.8)^2 - 0.5^2 ) = -0.21d. Am I missing something here?

This is great, thank you!

"Sure," he responds, and you both check your wallets and have 25 dollars each. "Okay," he says, "now you pick some betting odds, and I'll choose which side I want to pick."

"That's crazy," you say, "I am going to pick the odds so that I cannot be taken advantage of, which means that I will be indifferent between which of the two options you pick, which means that I will expect to gain 0 dollars from this transaction. I wont take it. It is not fair!"

At first reading I got lost here -- why would I say that this is crazy or that the bet is not fair if I expect to gain (and lose) 0 dollars? Is the reason that my opponent would expect (according to their estimation of probabilities of events) to gain from the trade, and it's not fair for all the gains to be on one side?

I claim the bet is fair if both players expect to make the same profit on average.

I like this idea. As you say, it's not the only way to define it but it does seem like a very reasonable way. The two players have come upon a situation which seems profitable to both of them and they simply agree to "split the profit".

A: 0.24
B: 0.56

E(A) = (0.5 * 0.56 - 0.5 * 24) = 0.16

E'(A) = (0.5 * 1 - 0.5 * 0.43) = 0.285

A: 0.03
B: 0.17
E(A) = 0.5 * (0.17 - 0.03) = 0.07

E'(A) = 0.5 * 1 - 0.5 * 0.18 = 0.41

What do you mean by 'average' I can think of at least four possibilities.

The standard of fairness "both players expect to make the same profit on average" implies that the average to use here is the arithmetic mean of the two probabilities, (p+q)/2. That's what gives each person the same expected value according to their own beliefs.

This is easy to verify in the example in the post. Stakes of 13.28 and 2.72 involve a betting probability of 83%, which is halfway between the 99% and 67% that the two characters had.

You can also do a little arithmetic from the equal expected values to derive that this is how it works in general: the ratio of the amounts at stake should be (p+q):(2-p-q), which is the ratio mean(p,q):[1-mean(p,q)]. Using the arithmetic mean to set the betting odds gives both parties the same subjective expected value.

If there's a simple intuitive sketch of why this has to be true, I don't have it.

The extra work involved in the method in the post is to make it incentive-compatible for each person to state their true probability by allowing the stakes to vary.

So: To make the bet fair (equal EV), the betting odds should be based on the arithmetic mean of the probabilities. If you need to make it strategy-proof, then you can use the algorithm in this post to decide how much money to bet at those odds.

he puts 2.72 on the table, and you put 13.28 on the table.

I'm confused...if the prediction does not come true (which you estimated as being 33 percent likely) you only gain $2.72? and if the most probable outcome does come true you lose 13.28?

This is indeed great. I've already set it up in a betting thread I run - I'll report back with how well it works.

Note, if p > 1-q, then switch the sense of the bet so you're working with 1-p and q instead. Or tolerate having negative wagers as being payouts instead of wagered amounts

ETA: uh, yes, I meant p < 1-q

For an explicit derivation of why this is fair:

Say that you believe the event is likely with probability p, and your betting partner tells you that it will fail with probability q. Then I am going to modify my estimate by c before I tell it to the other person. So my expected value is:

p*(q^2-(1-(p+c))^2) - (1-p)((p+c)^2 - (1-q)^2)

Naturally I want to find the local maximum for variation in c, for a fixed value of p and assuming q is out of my control. So we take the derivative with respect to c. Using Wolfram Alpha shows this is -2c. So the only local maxima possible are telling someone 0, telling them 1, or telling them the true value of p.

What if Alice is willing to bet up to d dollars, but Bob is willing to bet up to e dollars?

For what it's worth, I don't care that much about fairness. If I have to give it up to make it strategy-proof against lying about the relative amounts of money, that's fine.