In our previous part, we noted that a fair bet concerning an event with probability $p$ could be made with me paying you $\$S\!\left(p\right)$ up front and getting back $-\$\lg{p}$ if the event came to pass and $-\$\lg{\!\left(1-p\right)}$ if the event failed to come to pass.
We may even loosen up the game somewhat. Suppose I offer $\$H$ as a bet, but you get to pick the $q$ which determines the payouts $-\$\lg{q}$ and $-\$\lg{\!\left(1-q\right)}$. What $q$ should you pick, knowing that the probability of the event is $p$? You may maximize your net payment by setting
$$
0=\frac{\partial}{\partial q}\left[H-S\!\left(q\right)\right]=\frac{\partial}{\partial q}\left[H+p\lg{q}+\left(1-p\right)\lg{\!\left(1-q\right)}\right]=\frac{p}{q}-\frac{1-p}{1-q}
$$
or, conveniently enough, $q=p$. By choosing this way, you expect that I will pay you an amount $H-S\!\left(p\right)$.
Obviously, it is hardly worth it to me to set $H>S\!\left(p\right)$. But if I set $H=S\!\left(p\right)$ then I expect to profit any time you choose $q\neq p$.
However, this all assumes we know— or at least agree upon— $p$. Yet suppose I believe that the event is unlikely, say, $p=1\%$ and so I offer what I think is a fair bet with $H\approx 0.080793\ldots$ But if you are less certain, thinking $p=49\%$, then you expect to lose almost 92 cents on a bet which I believe fair. You may be enticed by an offer of at least $H\approx 0.999711\ldots$ but I will not make such an offer, because I expect that I will wind up paying you the same 92 cents.
In the next section, we make a deal.
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