Calculating probability of prizes being distributed at random? There are 10 people and 12 prizes. There are 6 x 100 dollars, 3 x 150 dollars and 3 x 200 dollars. The prizes are distributed at random.
a) What is the probability of Person A getting $100?
b) Will Person A have a higher expected return if he colludes with Person B and Person C, such that they will share their prizes equally?
I've calculated that the worse case scenario is all 3 of them getting 100 dollars only, the probability being 6/12 * 5/11 * 4/10 = 9%. But there seems to be something missing.
Do I have to consider them being further down the line, for example 6/11 * 5/10 * 4/9?
 A: Consider the single player $A$ first. He get $\$100$ with probability $\frac12$, $\$150$ with probability $\frac14$ and $\$200$ with probability $\frac14$. Expected return is 
$$
\mathbb E[A] = 100\cdot \frac12+150\cdot \frac14+200 \cdot\frac14 = 137.5.
$$
If playes $A$, $B$, $C$ cooperate, they can get three $\$100$ prizes, or three $\$150$ prizes,  or three $\$200$ prizes, or two $\$100$ and one $\$150$ prizes, or ... 
Sure, we can calculate the expected return directly. Note that there are $10$ cases. 
Say, 
$$\mathbb P(\$100\times 3) = \frac{\binom{6}{3}}{\binom{12}{3}}=\frac1{11},$$
$$\mathbb P(\$100\times 2+\$150) = \frac{\binom{6}{2}\binom{3}{1}}{\binom{12}{3}}=\frac9{44},$$ 
and so on. And then expected return is 
$$
\mathbb E[A_2]=\frac13\left(3\cdot 100 \cdot\frac{1}{11}+(2\cdot 100+150)\cdot\frac{1}{44}+\ldots\right)
$$
Here $A_2$ denotes total return of $A$ if he cooperates.
This is not the best way. 
Denote by $B$ and $C$ return of players $B$ and $C$ in the game. $A$, $B$ and $C$ are identically distributed (but dependent). So
$$
\mathbb E[A_2] = \mathbb E\left[\frac{A+B+C}{3}\right] = \frac{\mathbb E[A]+\mathbb E[B]+\mathbb E[C]}{3} = \frac{3\mathbb E[A]}{3} = \mathbb E[A]
$$
So the cooperation results the same expected value as in the single play. 
