Question

A standard deck of playing cards contains $52$cards, of which $4$are aces and $13$are hearts. You are offered a choice of the following two wagers:

I. Draw one card at random from the deck. You win $\backslash (10$if the card drawn is an ace. Otherwise, you lose $\backslash )1$.

II. Draw one card at random from the deck. If the card drawn is a heart, you win $\backslash (2$. Otherwise, you lose $\backslash )1$.

Which of the two wagers should you prefer?

(a) Wager $1$, because it has a higher expected value

(b) Wager $2$, because it has a higher expected value

(c) Wager $1$, because it has a higher probability of winning

(d) Wager $2$, because it has a higher probability of winning

(e) Both wagers are equally favorable.

Short answer

I would prefer Wager $1$, because it has a higher expected value so option (a) is correct answer.

Step-by-step solution

Given Information

We are given that there are $52$cards in which $4$are ace and $13$are hearts we are given the choices for wages and we have to choose the correct wages from given options.

Explanation

Now, For Wager $1$,

if it is an ace then we will win $\$10$otherwise we will lose $-\$1$

role="math" localid="1650512570990" $P=\frac{no.offavorableoutcomes}{Totalnumberofoutcomes}$

no. of favorable outcome =$4$and total number of outcomes=$52$

which gives, role="math" localid="1650512629767" $P\left(win\right)=\frac{4}{52}$

and role="math" localid="1650513005603" $P\left(lose\right)=1-\frac{4}{52}=\frac{48}{52}$

Then put it in the expected value formula,

which gives $\$10\times \frac{4}{52}+(-\$1)\times \frac{48}{52}$

On solving it we get, $-\$0.15$.

Similarly For Wager 2,

for every hearts win=$\$2$and otherwise lose=$\$1$

P(win)=$\frac{13}{52}$and

$P\left(lose\right)=1-\frac{13}{52}=\frac{39}{52}$

Then put these values in expected value formula,

which gives,$\$2\times \frac{13}{52}+(-\$1)\times \frac{39}{52}$

On solving this we get,$-\$0.25$.

As wager $1$is expecting higher value, So, preference will be option (a).