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!
