How many papers should the news boy stock? Suppose I'm a news boy and my sales of papers every day is given by $X\sim\text{binom}(10,1/3)$. Purchasing a paper for my stock from the supplier costs \$0.10 and selling a paper at market value nets me \$0.15. Since I can't return papers I've stocked, how many papers should I stock to maximize profits on a day-to-day basis?
I decided to let $Y=15X-10N$ with $N$ being the number of papers I stock. Then
$$\mathbb EY=\sum_{k=1}^{10}(15k-10N){}_{10}\text C_k\frac1{3^k}\left(\frac23\right)^{10-k}$$
and some value of $N$ should maximize this number. But obviously the above is wrong, since the term $15k-10N$ is confused and doesn't really mean anything.
What do I do?
 A: This is known as the Newsvendor model. Let $C_p$ be the selling price, $C_v$ the buying price, $D$ the demand, and $y$ the order quantity. The expected profit given that we order $y$ units is
$$(C_p-C_v)\mathbb E[y\wedge D] - C_v\mathbb E[(y-D)^+], $$
where $\wedge$ denotes $\min$ and $(\cdot)^+$ denotes $\max\{\cdot, 0\}$. Let $F$ be the distribution function of $D$. After some computation, it can be shown that the value of $y$ that maximizes the expected profit is
$$y^* = \operatorname{argmin}_y \left\{F(y) \geqslant \frac{C_p-C_v}{C_p}\right\}. $$
This is a well-defined quantity due to the monotonicity of $F$. In this case, $C_p=0.15$ $C_v=0.1$, and $D\sim\operatorname{Binom}\left(10,\frac13\right)$. 
Now, $$\mathbb P(X=k) = \binom{10}k \left(\frac13\right)^k\left(\frac23\right)^{10-k}=\left(\frac23\right)^{10}\binom{10}k\left(\frac12\right)^k,\quad k=0,1,\ldots,10$$
so $$F(k) = \left(\frac23\right)^{10}\sum_{j=0}^k\binom{10}j\left(\frac12\right)^j, \quad k=0,1,\ldots,10.$$
I used Mathematica to compute the CDF at each integer value:
$$\texttt{Function[k, (2/3)^10 * Sum[Binomial[10, j]*(1/2)^j, {j, 0, k}]] /@ 
 Range[0, 10]}$$
which yielded 
$$
\begin{array}
{c|cccccccccccc}k
&0&1&2&3&4&5&6&7&8&9&10
\\\hline
F(k)& \frac{1024}{59049} & \frac{2048}{19683} & \frac{5888}{19683} & \frac{11008}{19683} & \frac{15488}{19683} & \frac{18176}{19683} & \frac{2144}{2187} & \frac{19616}{19683} & \frac{19676}{19683} & \frac{59048}{59049} & 1 \\
\end{array}
$$
By inspection we find that $F(2)<\frac13<F(3)$, and so the optimal order quantity is $3$.
