Game of Red balls two drawings are made, which rule would you choose if playing the game, rule A or rule B? In the game of redball two drawings are made without replacements from a bowl that has four white ping pong balls and two red ping pong balls. The amount won is determined by how many ping-pong balls. The amount won is determined by how many of the red balls are selected. For a $5 bet, a player can opt to be paid under either Rule A or Rule B, as shown. If you were playing the game, which would you choose? Why?
For A: # of Red ball drawns 0,1,2 and their payoff in dollars 0 , $2, $10 respectively.
Similarly for B: # of Red ball drawns  0,1,2, and their payoffs in dollars 0, $1, $20.
Solution: Let X be a discrete variable with probability p_x(k). The expected value of X, denoted by E(x) is given by E(x) = (Sum of for all k) k*p_x(k).
For Rule A: we have from the hypergeometric random variable P(X=k) = p_x(k) = [2_C_k * 4_C_(2-k)]/6_C_2, for k = 0,1,2.
THus, E(x) = (Sum from 0 to 2) K*[2_C_k * 4_C_(2-k)]/6_C_2 = 0* [2_C_0 * 4_C_(2-0)]/6_C_2 + 1* [2_C_1 * 4_C_(2-1)]/6_C_2 + 2* [2_C_2 * 4_C_(2-2)]/6_C_2 = 0 + 2*4/15 + 2*6/15 = 20/15.
Can someone please help me? I don't know how to us the payoffs. I don't know where would I plug the payoffs in the equations. Anything would be appreciate it.
Thank you. 
 A: Let us find the probability of $0$ red, of $1$ red, or $2$ red. We will not use the "hypergeometric formula," though we do in a remark at the end.
For $0$ red, imagine picking one at a time. The probability the first is non-red is $\frac{4}{6}$. Given this happened, the probability the second is non-red is $\frac{3}{5}$, so the probability of $0$ red is $\frac{4}{6}\cdot\frac{3}{5}$. This is $\frac{12}{30}$. 
Similarly, the probability of $2$ red is $\frac{2}{6}\cdot\frac{1}{5}$. This is $\frac{2}{30}$. 
For $1$ red, that's the rest, so it has probability $1-\frac{12}{30}-\frac{2}{30}$. This is $\frac{16}{30}$. 
Rule  A: Let random variable $X$ be the gross amount you win (not counting the price of playing).
Then $X$ has value $0$ with probability $\frac{12}{30}$. It has value $2$ with probability $\frac{16}{30}$. And it has value $10$ with probability $\frac{2}{30}$. Thus the mean gross win is given by
$$E(X)=0\cdot\frac{12}{30}+2\cdot\frac{16}{30}+10\cdot\frac{2}{30}.$$
Thus $E(X)=\frac{52}{30}$. By the way, your expected net win is $\frac{52}{30}-5$, negative. don't play!
Rule B: Let random variable $Y$ be our gross winnings under Rule B. Now it's your turn. 
Remark: Let us find the probability of say $1$ red using the formula. We get $\frac{\binom{2}{1}\binom{4}{1}}{\binom{6}{2}}$. Calculate. After a while we get $\frac{8}{15}$, which is the same as the $\frac{16}{30}$ I obtained earlier. You can do the others in this style. 
