SOLUTION: I draw a card from a standard 52-card deck. If I draw an Ace, I win 1 dollar. If I draw a 2 through 10, I win a number of dollars equal to the value of the card. If I draw a face c

Question 1161243: I draw a card from a standard 52-card deck. If I draw an Ace, I win 1 dollar. If I draw a 2 through 10, I win a number of dollars equal to the value of the card. If I draw a face card (Jack, Queen, or King), I win 20 dollars. If I draw a $\clubsuit$, my winnings are doubled, and if I draw a $\spadesuit$, my winnings are tripled. (For example, if I draw the $8\clubsuit$, then I win 16 dollars.) What would be a fair price to pay to play the game? Express your answer as a dollar value rounded to the nearest cent.
Your answer should be a number with two digits after the decimal point, like $21.43$.
EXPECTED VALUE QUESTION
I would appreciate any help
This is expected value. So for the ace, it would be E = (1) 4/52 as a decimal(.923076923) and for the face cards it would be (12/52 ) as a decimal ??(20) so it would be E= (1)(923076923) + (20)(0.230769231) the rest i don't know, please help

Found 2 solutions by greenestamps, solver91311:
Answer by greenestamps(13347) (Show Source): You can put this solution on YOUR website!

You clearly know HOW to calculate an expected value. But in the work you show you are oversimplifying the problem.

A couple of comments before I start discussing the approach to solving the problem: don't convert each part of the answer to decimal form. Leave each part of the answer in the form n/52; add all those fractions, and wait until you have that total fraction before converting to decimal.

Better yet, since every one of those fractions has denominator 52, ignore the denominator until the very end and divide by the 52 once to get your final decimal expected value.

For example, for the aces, the calculation (ignoring the "52" denominator) is not simply (1)(4), because the ace of clubs is worth double and the ace of spades is worth triple. So the calculation is 1(1+1+2+3) = 7(1)) = 7.

Then you need to consider the suits when doing the calculations for the other cards.

For the face cards, there are 3 each in hearts and diamonds, 3 in clubs, and 3 in spades; the calculation is 20(3+3+3(2)+3(3)) = 7(20+20+20) = 7(60).

Then you will have similar calculations for the 2 through 9 cards:
2's: 2(1+1+2+3) = 7(2)
3's: ... = 7(3)
...
9's: ... = 7(9)
10's: ... = 7(10)

Add all those pieces together and divide by 52; that is the expected value of your winnings.

Then of course if the game is EXACTLY "fair", that expected value should be the cost of playing the game. However, it is unlikely that the expected value will turn out to be a whole number of dollars, or even a whole number of cents -- so there probably won't be a price for playing the game that is EXACTLY fair.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

If you draw a red Ace, you get $1. There are 2 red aces in the deck, so the probability of drawing a red Ace is 1/26. The expected value of drawing a red ace is then (1/26)(1) which is approximately $0.04.

The club Ace gets you $2, 1 club ace, 1/52 times $2 is the same $0.04.

The spade Ace is worth $3, 1/52 times $3 gives E(AS) = $0.06.

For the 2s, you have 2 times 1/26, 4 times 1/52, and 6 times 1/52.

The rest are summarized in the following table:

A♥♦ 1 1/26 $0.04 A ♣ 2 1/52 $0.04 A ♠ 3 1/52 $0.06 2♥♦ 2 1/26 $0.08 2 ♣ 4 1/52 $0.08 2 ♠ 6 1/52 $0.12 3♥♦ 3 1/26 $0.12 3 ♣ 6 1/52 $0.12 3 ♠ 9 1/52 $0.17 4♥♦ 4 1/26 $0.15 4 ♣ 8 1/52 $0.15 4 ♠ 12 1/52 $0.23 5♥♦ 5 1/26 $0.19 5 ♣ 10 1/52 $0.19 5 ♠ 15 1/52 $0.29 6♥♦ 6 1/26 $0.23 6 ♣ 12 1/52 $0.23 6 ♠ 18 1/52 $0.35 7♥♦ 7 1/26 $0.27 7 ♣ 14 1/52 $0.27 7 ♠ 21 1/52 $0.40 8♥♦ 8 1/26 $0.31 8 ♣ 16 1/52 $0.31 8 ♠ 24 1/52 $0.46 9♥♦ 9 1/26 $0.35 9 ♣ 18 1/52 $0.35 9 ♠ 27 1/52 $0.52 10♥♦ 10 1/26 $0.38 10 ♣ 20 1/52 $0.38 10 ♠ 30 1/52 $0.58 F♥♦ 20 3/26 $2.31 F ♣ 40 3/52 $2.31 F ♠ 60 3/52 $3.46 $15.48

The expected value of the game is the sum of the individual expected values. By the way, don't add the numbers in the expected value column -- they are all rounded and you will get an incorrect sum. The correct sum of the unrounded values which is then rounded to two digits is shown.

John

My calculator said it, I believe it, that settles it

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