Question

If you draw a card with a value of five or less from a standard deck of cards, I will pay you $7$ 7. If not, you pay me $7$ 7. (Aces are considered the highest card in the deck.) Step 2 of 2 : If you played this game 543543 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

`Amount = -\$1461.9`

Explanation:

We have the following:

`Total = 52` --- cards in a standard deck

Represent cards with a value of 5 or less with x

So:
`x = \{2, 3, 4, 5\}` i.e 4 cards

However, each of these has a frequency of 4.

So:
`n(x) = 4\ cards* 4`

`n(x) = 16`

The probability is:

`p(x) = (16)/(52)`

The cost of this is:

`C(x) = +\$7`

Represent cards with a value of above 5 with y

`n(y) = 52 - 16`

`n(y) = 36`

The probability is:

`p(y) = (36)/(52)`

The cost of this is:

`C(y) = -\$7` --- It is negative because you lost

In a game, your expected amount is:

`E = n(x) * C(x) + n(y) * C(y)`

`E = (16)/(52) * (+\$7) + (36)/(52) * (-\$7)`

`E = (16* \$7)/(52) - (36* \$7)/(52)`

`E = (\$112)/(52) - (\$252)/(52)`

`E = (\$112 - \$252)/(52)`

`E = -(\$140)/(52)`

When you play 543 times, the expected amount is:

`Amount = E * 543`

`Amount = -(\$140)/(52) *543`

`Amount = -(\$140 *543)/(52)`

`Amount = -(\$76020)/(52)`

`Amount = -\$1461.9`