Money problem

• Jan 11th 2009, 09:20 PM
Money problem
A bag contains ten \$1 bills, five \$2 bills, four \$5 bills, one \$10 bill, one \$20 bill, and one \$100 bill. In order to play the game, which consists of drawing on bill randomly out of the bag, you must pay \$20.

Q: A construct a probability distribution for this situation.
A: ????

Q: Find the expected value
A: I think, (1x10)+(2x5)+(5x4)+(10x1)+(20x1)+(100x1) = \$7.27

Q:Is this game fair? if it is not fair, determine how much someone should have to pay in order for the game to be fair.
A: Well it should cost \$7.27 for the game to be considered fair.

I dont know what to do on the first one, nor if i have done the 2nd one right, which therefore would make me do the last one wrong.
A:
• Jan 11th 2009, 09:59 PM
mr fantastic
Quote:

A bag contains ten \$1 bills, five \$2 bills, four \$5 bills, one \$10 bill, one \$20 bill, and one \$100 bill. In order to play the game, which consists of drawing on bill randomly out of the bag, you must pay \$20.

Q: A construct a probability distribution for this situation.
A: ????

Q: Find the expected value Mr F says: The expected value of the amount of money pulled from the bag, I assume ....?

A: I think, (1x10)+(2x5)+(5x4)+(10x1)+(20x1)+(100x1) = \$7.27 Mr F says: How did you get \$7.27!? The left hand side adds up to 170 ....!

The answer (use the probability distribution - see how to get it below) is found using the definition: 170/22 = \$7.73.

Q:Is this game fair? if it is not fair, determine how much someone should have to pay in order for the game to be fair.
A: Well it should cost \$7.27 for the game to be considered fair.

I dont know what to do on the first one, nor if i have done the 2nd one right, which therefore would make me do the last one wrong.
A:

There are 22 bills. Ten of them are \$1 bills. So Pr(X = \$1) = 10/22. Calculate Pr(X = \$2), Pr(X = \$5) etc. in a similar way. The exhaustive set of probabilities is the probability distribution for the amount of money pulled from the hat.
• Jan 11th 2009, 09:59 PM
(10/22)*(-19)+(5/22)*(-18)+(4/22)*(-15)+(1/22)*(-10)+(1/22)*(0)+(1/22)*(80) = \$-12.27

so -12.27 is the expected value?

if so how do i do problem A?
• Jan 11th 2009, 10:06 PM
mr fantastic
Quote:

(10/22)*(-19)+(5/22)*(-18)+(4/22)*(-15)+(1/22)*(-10)+(1/22)*(0)+(1/22)*(80) = \$-12.27

so -12.27 is the expected value?

if so how do i do problem A?

Expected value of what??

The expected value of money pulled from the bag is \$7.73.

The expected value of money you win is 7.27 - 20 = - \$12.27.

I have already said how to get the probability distribution. What part of what I said don't you understand?
• Jan 11th 2009, 10:41 PM
so we know the expected value is -12.27

i got 3.86 for how much the game should cost for the game to be fair.

right?

i did (10/22)(x-1)+(5/22)(x-2)... so on and plugged that equation into y1. and x into y2, the intersected at 3.86,3.86
• Jan 11th 2009, 10:57 PM
mr fantastic
Quote:

so we know the expected value is -12.27

i got 3.86 for how much the game should cost for the game to be fair.

right?

i did (10/22)(x-1)+(5/22)(x-2)... so on and plugged that equation into y1. and x into y2, the intersected at 3.86,3.86

No.

For the game to be fair you require the expected value of the money you win to be zero.

Let the cost per game be C. Then you require 0 = 7.73 - C.
• Jan 11th 2009, 10:59 PM