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 diamonds, your friend will pay you $29. Otherwise, you have to pay your friend $4.Step 2 of 2 : If this same bet is made 891 times, how much would you expect to win or lose? Round your answer to two decimal places. Losses must be expressed as negative values.

Answer 1

Answer:

Ok, the probability of winning is:

For the first draw, we have 13 diamonds in a 52 card deck, the probability of drawing a diamond is:

p1 = 13/52

for the second draw we have 12 diamonds and 51 cards in the deck, now the probability is:

p2 = 12/51

the total probability is equal to the product of the individual probabilities, this is:

P = p1*p2= (13/52)*(12/51) = 0.059

now, the expected value can be calculated as follows.

EV = (P1*X1 + P2*X2)

where P1 is the probabilty of event 1, and X1 is the event 1,

In this case X1 is the event of winning $29, then P1 = 0.059

X2 will be the event of loosing $4, so P2 = 1 - 0.059 = 0.941

Then the expected value for 1 round of the game is:

EV = (0.059*$29 - 0.941*$4) = -$2.05

So if you play 891 times, expected value of 891 times the expected value for 1 round, this is:

891*EV = 891*(-$2.05) = -$1826.55

So you loss around  $1826.55