Question

If you draw a card with a value of 4 or less from a standard deck of cards, i will pay you $7 if not you pay me $6. (Aces are considered highest cards in the deck.)

Step 2: If you played this game 825 times how much would you expect to win or lose? Round answer two decimal places.

Answer 1

Playing the card-drawing game 825 times, with a standard deck and rules as given, you would expect to lose approximately $13,898.25.

Step-by-step explanation:

To calculate the expected value for playing the game of drawing a card with a value of 4 or less from a standard deck of cards 825 times, we first need to find the probability of drawing such a card and then use the expected value formula.

There are 16 cards with a value of 4 or less in a standard deck (4 suits each of Aces, 2s, 3s, and 4s). With Aces being high, they'll count as high cards, not as ones you want to draw, so that leaves us with 12 cards that are either a 2, 3, or 4.

Therefore, the probability of drawing a card with a value of 4 or less is 12/52, and the probability of not drawing one is 40/52. The expected value (EV) for a single game is calculated as:

EV = (Probability of winning) x (Amount won per win) + (Probability of losing) x (Amount lost per loss)

EV = (12/52 x $7) + (40/52 x -$6)

This simplifies to:

EV = ($21/13) - ($240/13)

EV = -$219/13

EV = -$16.85 (rounded to two decimal places)

Over 825 games, the expected total value would then be:

825 x EV = 825 x -$16.85 = -$13898.25 (rounded to two decimal places)

Playing this game 825 times, you would expect to lose $13,898.25.