Min/Max Expectation problem (very difficult) Problem: The manager of a fish market speculates that the number of requests for salmon on any given day is a random variable $X$ with the probability function
$$f(x)=\begin{cases}
\frac14,&\text{if }x=0\\\\
\frac12,&\text{if }x=1\\\\
\frac14,&\text{if }x=2\;.
\end{cases}$$
There is a profit of $\$2$ on each salmon he sells and a loss of $\$1$ on each salmon he does not sell. Assuming that each salmon can be sold only on the day it is up for sale, that each request is for a single salmon, and that the managers's speculation is correct, find the number of salmons the market should have per day to maximize profit.
Attempt at a solution: So, what I think this is saying is that there is a $.25$ chance that no salmon will be requested, a $.50$ chance that $1$ salmon will be requested, and a $.25$ chance that $2$ salmon will be requested. This adds up to $1$, so it makes sense. In the second part, it states that each request is for a single salmon, but each the probability function of that only gives $.50$ (not sure if this is right) 
Anyway, I attempt trial and error. So if we start with $10$ salmon, then he sells only $5$, and the other $5$ go to waste. So he loses $\$5$. If he starts with $8$ salmon, he loses $\$4$ ($8 *.5 *2 - 8*.5 * \$1$), gives us $4$... Continue this to $0$, he loses none, but has no salmon.
I'm lost
I think I need help modeling and solving , thanks in advance
 A: The statement that each request is for a single salmon just means that no customer ever asks for more than one salmon at a time. Since $f(0)+f(1)+f(2)=1$, the demand on any given day is always $0,1$, or $2$; there is no chance that he will sell more than two salmon, so he certainly should not order more than two. This is a small enough problem that we can easily tabulate the possible outcomes:
$$\begin{array}{c}
\qquad\qquad\qquad\qquad\qquad\text{Orders:}\\
\begin{array}{rc|ccc}
&&0&1&2\\ \hline
&0&0&-1&-2\\
\text{Gets Requests For:}&1&0&2&1\\
&2&0&2&4
\end{array}
\end{array}$$
Now we calculate the expected profit for each of his three possible orders. (Well, okay: technically an order of $10$ salmon is possible, but we know that he loses money on it, so we rule it out.)
Clearly he is certain to have a profit of $\$0$ if he orders $0$ salmon.
If he orders $1$ salmon, his expected profit is $\frac14(-1)+\frac12\cdot2+\frac14\cdot2=\$2.25$.
And if he orders $2$ salmon, his expected profit is $\frac14(-2)+\frac12\cdot1+\frac14\cdot4=\$1$.
A: What is the range of salmon he can sell? It appears from the question that x can only equal 0, 1, or 2.
