Even Odds

(Cross-posted on my per­sonal blog, which has LaTeX, and is eas­ier to read.)

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, lets 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+(2r-1-r+q)(dh/​dr)=0.

Sepa­ra­tion of vari­ables gives us (1/​h)dh=1/​(-1+r+q)dr,

which in­te­grates to ln(h)=C+ln(-1+r+q) at r=p,

which sim­plifies to h=e^C(-1+r+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.