You select your horse for a race. The horse has a parameter $q \in [-0.3,+0.3]$. Its probability of winning depends on the weather, which can be either Rainy or Sunny:
- In Rainy weather, the probability is $0.6-q$;
- In Sunny weather, the probability $0.6+q$.
The prize also depends on the weather:
- In Rainy weather, the prize is R;
- In Sunny weather, the prize is S.
$R$ and $S$ can be negative.
The chance of each weather (Rainy or Sunny) is 0.5. What value of $q$ should you choose?
The expected profit is:
$$E(q) = 0.5 R (0.6 - q) + 0.5 S (0.6 + q) $$
$$E'(q) = 0.5 (S-R)$$
$E'$ never equals zero, so the optimal value of $q$ is on an endpoint. This means that the best strategy is:
- If $S>R$, select the maximum possible value: $q_{opt}=0.3$;
- if $S<R$, then $q_{opt}=-0.3$;
- if $S=R$, select any $q$ you want (it doesn't matter).
This makes sense: it's best to select a horse whose expertise is in winning the higher prize.
Now, suppose that in order to get the prize, your horse has to win twice instead of once. Intuitively, I would expect to get the same result for $q$, but this is not the case!
The expected profit now is:
$$E(q) = 0.5 R (0.6 - q)^2 + 0.5 S (0.6 + q)^2 $$
$$E'(q) = - R (0.6 - q) + S (0.6 + q) = 0.6(S-R)+q(S+R) $$
$$E''(q) = S+R$$
Equating $E'(q)$ to 0 gives:
$$q^* = 0.6 \frac{S-R}{-(S+R)}$$
When $S+R>0$ (the second derivative is positive), $q^*$ is a minimum point, and the optimal value of $q$ (the maximum point) is on one of the endpoints - similar the 1-win case.
But, when $S+R<0$, $q^*$ is a maximum point. This means that the optimal value of $q$ is usually NOT on an endpoint. How can this be explained intuitively? In particular, why is it different than the case of having to win once? Why not select $q=+0.3$ or $q=-0.3$ as before?
When $R+S$ is positive, $E(q)$ is a quadratic in $q$ with $q^2$ having positive coefficient. So $E'(q_0) = 0$ implies $q_0$ minimizes the expectation, not maximizes it. The optimal values will still be $q = \pm 0.3$ (but with the slight change that if $R=S$ you still want to pick $q = \pm 0.3$).
When one of $R$ and $S$ is positive and the other is negative, but $R+S$ is negative, then $|\frac{S-R}{-(R+S)}| \geq 1$, and so our "optimum" value of $q$ is outside the allowed range (and indeed outside any possible allowed range). So this is still maximized at an endpoint.
When both $R$ and $S$ are negative, you're no longer trying to win both races, instead you are trying to lose at least one. Picking an endpoint maximises the chance of $LL$, but also increases the chances of $WW$. Whereas picking somewhere in the middle minimizes the chances of $WW$ by maximizing the chances of $WL$. (This happens because the weather stays the same between the two races, so now you want a horse that is about equally good at losing on either surface, rather than putting all your eggs in one weather basket).