Even Odds

Let’s say that you are are at your local less wrong meet up and someone makes some strong claim and seems very sure of himself, “blah blah blah resurrected blah blah alicorn princess blah blah 99 percent sure.” You think he is probably correct, you estimate a 67 percent  chance, but you think he is way over confident. “Wanna bet?” You ask.

“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!”

“Okay,” he says, annoyed with you. “We will both write down the probability we think that I am correct, average them together, and that will determine the betting odds. We’ll bet as much as we can of our 25 dollars with those odds.”

“What do you mean by ‘average’ I can think of at least four possibilities. Also, since I know your probability is high, I will just choose a high probability that is still less than it to maximize the odds in my favor regardless of my actual belief. Your proposition is not strategy proof.”

“Fine, what do you suggest?”

You take out some paper, solve some differential equations, and explain how the bet should go.

Satisfied with your math, you share your probability, he puts 13.28 on the table, and you put 2.72 on the table.

“Now what?” He asks.

A third meet up member takes quickly takes the 16 dollars from the table and answers, “You wait.”

I will now derive a general algorithm for determining a bet from two probabilities and a maximum amount of money that people are willing to bet. This algorithm is both strategy proof and fair. The solution turns out to be simple, so if you like, you can skip to the last paragraph, and use it next time you want to make a friendly bet. If you want to try to derive the solution on your own, you might want to stop reading now.

First, we have to be clear about what we mean by strategy proof and fair. “Strategy proof” is clear. Our algorithm should ensure that neither person believes that they can increase their expected profit by lying about their probabilities. “Fair” will be a little harder to define. There is more than one way to define “fair” in this context, but there is one way which I think is probably the best. When the players make the bet, they both will expect to make some profit. They will not both be correct, but they will both believe they are expected to make profit. I claim the bet is fair if both players expect to make the same profit on average.

Now, let’s formalize the problem:

Alice believes S is true with probability p. Bob believes S is false with probability q. Both players are willing to bet up to d dollars. Without loss of generality, assume p+q>1. Our betting algorithm will output a dollar amount, f(p,q), for Alice to put on the table and a dollar amount, g(p,q) for Bob to put on the table. Then if S is true, Alice gets all the money, and if S is false, Bob gets all the money.

From Alice’s point of view, her expected profit for Alice will be p(g(p,q))+(1-p)(-f(p,q)).

From Bob’s point of view, his expected profit for Bob will be q(f(p,q))+(1-q)(-g(p,q)).

Setting these two values equal, and simplifying, we get that (1+p-q)g(p,q)=(1+q-p)f(p,q), which is the condition that the betting algorithm is fair.

For convenience of notation, we will define h(p,q) by h(p,q)=g(p,q)/(1+q-p)=f(p,q)/(1+p-q).

Now, we want to look at what will happen if Alice lies about her probability. If instead of saying p, Alice were to say that her probability was r, then her expected profit would be p(g(r,q))+(1-p)(-f(r,q)), which equals

p(1+q-r)h(r,q)+(1-p)(-(1+r-q)h(r,q))=(2p-1-r+q)h(r,q)

We want this value as a function of r to be maximized when r=p, which means that

-h+(2p-1-r+q)\frac{dh}{dr}=0,

which since r=p is the same as

-h+(-1+r+q)\frac{dh}{dr}=0.

Separation of variables gives us

\int\frac{1}{h}dh=\int\frac{1}{-1+r+q}dr,

which integrates to

ln(h)=C+ln(-1+r+q),

which simplifies to

h(r,q)=e^C(-1+r+q),

or equivalently,

h(p,q)=e^C(-1+p+q).

This gives the solution

f(p,q)=e^C(-1+p+q)(1+p-q)=e^C(p^2-(1-q)^2)

and

g(p,q)=e^C(-1+p+q)(1+q-p)=e^C(q^2-(1-p)^2).

It is quick to verify that this solution is actually fair, and both players’ expected profit is maximized by honest reporting of beliefs.

The value of the constant multiplied out in front can be anything, and the most either player could ever have to put on the table is equal to this constant. Therefore, if both players are willing to bet up to d dollars, we should define

e^C=d.

Alice and Bob are willing to bet up to d dollars, Alice thinks S is true with probability p, and Bob thinks S is false with probability q. Assuming p+q>1, Alice should put in d(p^2-(1-q)^2), while Bob should put in d(q^2-(1-p)^2). I suggest you use this algorithm next time you want to have a friendly wager (with a rational person), and I suggest you set d to 25 dollars and require both players to say an odd integer percent to ensure a whole number of cents.

Read 17 comments

  1. Pingback: Generalized Even Odds | By Way of Contradiction

  2. Expected profit should be proportional to the amount you’re risking, not the same whole amount of money.

  3. Interesting analysis. Your integration breaks down when 2p+q-1-r=0, and gives the wrong answer when 2p+q-1-r<0. These are plausible values for r.

    • Thank you. I do not think there is an issue with the integration. All I care about is local information. The integration is to ensure a maximum utility at p=r, so it does not matter if it does not makes sense when r is not near p. The integral does make sense when r is near p, because p+q-1 was assumed to be greater than 0.

  4. I get a different result when I do the integration (the line after ‘Separation of variables’) – I get a negative sign, i.e.: $\ln(h) = C – \ln(2p-1-r-q)$. Have I missed something?

    • You are correct. The 2p should have been changed to 2r before you take the integral. That along with the sign error canceled out to get the correct answer, so I did not notice! I will fix it. Thank you!

        • Yes, but when you are solving the differential equation, it does not matter if it was constant or not. All that matters is that it equals r. The distinction between r and p was important when we took the derivative and set it equal to 0, but for solving the differential equation, we treat them as the same.

          • I accept it gives the correct outcome, but I’m still puzzled by the method – it feels very odd to turn a constant into a variable (the variable we’re integrating with respect to!), which in turn changes the answer of the integration. I’m still playing with it, will report back.

  5. How about this:
    * Each player j=1..k puts d into the pot and declares a vector of probabilities (p_j,1, p_j,2, … p_j,n) for outcomes 1..n.
    * Once the event is observed to be outcome i, each player receives s(1 – (1-p_j,i)^2 – (1/(k-1))( \Sigma_{j \ne j} (1-p_Ji)^2).

    I _think_ this reduces to the same solution as above for n=2, k=2, and can be shown to be fair and strategy-proof – although I haven’t quite worked through the calculus in the general case.

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