Question

A game advertises, EVERYONE IS A WINNER! It costs $1 to play. You get to choose a card from a standard deck of 52

playing cards. If you choose a red card, you win $3. If you choose a black card, you get $2. What is the expected value of

playing this game?

$1.50

$1.80

-$0.50

$2

Answer 1

Answer:

$X_{red} = 3, p(red) =\frac{1}{2}$

$X_{black} = 2, p(red) =\frac{1}{2}$

And the cost of play would be 1 with probability 1 for any given game. Then we can find the expected value like this:

$E(X) = 3 *\frac{1}{2} +2 \frac{1}{2} -1$

And solving we got:

$E(X) = 1.50$

And then the best answer for this case would be:

$1.50

Step-by-step explanation:

For this case we can calculate the expected value with this formula:

$E(X) =\sum_{i=1}^n X_i P(X_i)$

We assume that the standard deck is formed just with red and black cards. For this case we have the following info:

$X_{red} = 3, p(red) =\frac{1}{2}$

$X_{black} = 2, p(red) =\frac{1}{2}$

And the cost of play would be -1 with probability 1 for any given game. Then we can find the expected value like this:

$E(X) = 3 *\frac{1}{2} +2 \frac{1}{2} -1$

And solving we got:

$E(X) = 1.50$

And then the best answer for this case would be:

$1.50