What is the expected value of your bet? Round your answer to two decimal places. Losses must be expressed as negative values.

Question

Suppose that you and a friend are playing cards and you decide to make a friendly wager. The bet is that you will draw two cards without replacement from a standard deck. If both cards are clubs, your friend will pay you $12. Otherwise, you have to pay your friend $3.

Step 1 of 2 :  

What is the expected value of your bet? Round your answer to two decimal places. Losses must be expressed as negative values.

Answer 1

Probability of drawing two clubs without replacement= P(Club1)*P(Club2).

Now, there are 13 clubs in a deck of 52 cards.

So P(Club1)=13/52, P(Club2)=12/51

P(Club1)*P(Club2)=(13/52)*(12/51)=1/17

Expected value= x1*P1+x2*P2=$12*(1/17)+(-$3*(16/17))=-$36/17= -$2.12

Answer 2

both clubs without replacement

has P(2C) = P(1st club) x P(2nd club) = (4/52)(3/51) = (1/4)(1/17)= (1/(4)(17) =1/68

P(not 2C) =1-1/68 =68/68-1/68 = 67/68

expected value of the bet= 12(1/68) - 3(67/68

=12/68- 3(67/68) = (12-3(67))/68= (12-201)/68= -189/68=about -$2.78

it's a losing bet, but slightly better than losing $3 per bet, due to the small chance of winning the "jackpot" of $12. $2.78 as a loss will never happen, except as the average loss over time from repeated bets. It's the mean loss or mean negative "win"

Answer 3

What's the probability of getting two clubs? ₁₃C₂/₅₂C₂ = 1/17

So the expected value of your bet would be 12(1/17) - 3(16/17) = -36/17 or -$2.12.