Scoring, Part IV

In the previous post, we found that if we disagree to the odds of an event happening then it may be difficult to offer an appropriate bet.

Can we come to a suitable arrangement? Certainly. If I believe I am offering a fair bet with $H\approx 0.080793\ldots$, I should be willing to take the opposite side of my bet (at $p=1\%$) as payment. That is, I charge nothing up front. If the event comes to pass, I pay $\lg{0.49}-\lg{0.01}$ and otherwise pay $\lg{0.51}-\lg{0.99}$. On average I expect I will pay $$ 0.01\lg{\frac{0.49}{0.01}}+0.99\lg{\frac{0.51}{0.99}}\approx -0.891214\ldots $$ (That is, that I will win about 89 cents.) Yet you expect I will pay you $$ 0.49\lg{\frac{0.49}{0.01}}+0.51\lg{\frac{0.51}{0.99}}\approx 2.263172\ldots $$ So we both think that we will win on average.

In fact, this works in general. Let us suppose I get to pick a $q_1$ so that I will pay you $-\lg{q_1}$ if the event happens and $-\lg{\!\left(1-q_1\right)}$ if it does not. Likewise, you will get to pick a $q_2$ so that you will have to pay me $-\lg{q_2}$ if the event happens and $-\lg{\!\left(1-q_2\right)}$ if not. Now, if I believe the actual probability is $p$, then I expect to win $$ w_1\!\left(q_1,q_2\right)=p\lg{\frac{q_1}{q_2}}+\left(1-p\right)\lg{\frac{1-q_1}{1-q_2}} $$ which is maximized when $$ 0=\frac{\partial w_1\!\left(q_1,q_2\right)}{\partial q_1}=\frac{p}{q_1}-\frac{1-p}{1-q_1} $$ regardless of your choice $q_2$. Thus, I always maximize my expected winnings by choosing $q_1=p$. When I do, my expected winnings are $$ w_1\!\left(p,q_2\right)=p\lg{\frac{p}{q_2}}+\left(1-p\right)\lg{\frac{1-p}{1-q_2}} $$ What is the least, then, I can win by picking $q_1=p$? My winnings are minimized when $$ 0=\frac{\partial w_1\!\left(p,q_2\right)}{\partial q_2}=-\frac{p}{q_2}+\frac{1-p}{1-q_2} $$ or when you choose $q_2=p$. This results in expected winnings $w_1\!\left(p,p\right)=0$. In other words, so long as I choose $q_1=p$ I maximize my expected winnings and I expect you to pay me no matter what $q_2$ you choose!

Of course, the structure of the bet is completely symmetric, so provided you choose $q_2$ to equal what you believe to be the correct probability, then you maximize your expected winnings and expect me to pay you no matter what $q_1$ I choose.

In the end, the winner will be the one of us least surprised by the outcome! Mathe-magical!