Question

If you draw a card with a value of four or less from a standard deck of cards, I will pay you $⁢134.  If not, you pay me $⁢37.  (Aces are considered the highest card in the deck.)

Step 2 of 2 :  

If you played this game 621 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be entered as negative.

Answer 1

Solution:

Total number of cards in a standard deck of cards = 52

Total number of cards with a value of 4 or less = 12

Total number of cards with a value of greater than 4 = 40

Probability of an event E is given by,

P(E) = no. of favourable outcomes/total number of outcome

Hence, probability of drawing a card with a value of 4 or less is given by,

P(4 or less ) = 12/52

It means I can win $134 with a probability of 12/54 in a single game.

Probability of drawing a card with a value of greater than 4 is given by,

P(greater than 4) = 40/52

It means that I can loose $37 with a probability of 40/52 in a single game.

Now for discrete random variable X, expectation is defined as follows:

Hence, expected win or loss in a single game is given by,

[Since, $37 is a amount which I can loose, therefore I have placed a negative sign before 37.]

Since, expected value has positive sign, therefore it is a expected win. So, I would expect to win $2.461539 in a single game.

Total number of times this game is played = 621.

Hence, expected win if I played this game 621 times = (621 × 2.461539) = 1528.615719

On rounding to 2 decimal places we get,

Expected win if I played the game 621 times = $1528.62

Hence, if I played the game 621 times, I would expect to win $1528.62.

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