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

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

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

Expert verified

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

Step by step solution

01

Given Information

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

02

Explanation

Now, For Wager 1,

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

role="math" localid="1650512570990" P=no.offavorableoutcomesTotalnumberofoutcomes

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

which gives, role="math" localid="1650512629767" P(win)=452

and role="math" localid="1650513005603" P(lose)=1-452=4852

Then put it in the expected value formula,

which gives $10452+(-$1)4852

On solving it we get, -$0.15.

Similarly For Wager 2,

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

P(win)=1352and

P(lose)=1-1352=3952

Then put these values in expected value formula,

which gives,$21352+(-$1)3952

On solving this we get,-$0.25.

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

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