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.

  • $\begingroup$ Calculate the probabilities, use the expression you gave for expectation. If you calculate the probabilities and put them in your question, or in a comment, in readable form, I will give further help. $\endgroup$ Oct 21, 2014 at 5:51
  • $\begingroup$ The problem is, I don't know how put it in a readable way. I am not a member. So I don't know how to type in a nice math way. $\endgroup$
    – user2414
    Oct 21, 2014 at 5:59
  • $\begingroup$ Give your answers in the form the probability of 0 red is 12/30, the probability of $\dots$. $\endgroup$ Oct 21, 2014 at 6:06
  • $\begingroup$ Would you mind to state what shall 4_C_k stand for? Is it $4\cdot C_k$ or something else? $\endgroup$ Oct 21, 2014 at 6:20
  • $\begingroup$ the probability for rule A is 20/15. Based on the expected value. As well as the probability for rule B, that is 20/15. THis is because I am just using the k values, 1,2,3. That is why i get this result for both cases, when I use the hypergeometric random variable formula. I don't now how I can use the dollar values. $\endgroup$
    – user2414
    Oct 21, 2014 at 6:23

1 Answer 1


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.


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