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aburchett
02-02-2011, 05:13 PM
A person spins the pointer and is awarded the amount indicated by the pointer.



It costs $8 to play the game. Determine:


The expectation of a person who plays the game.


The fair price to play the game.

My work:
E=P(wins)(amount won)+P(loss)(amount lost)
E=1/3(-6)+2/3(-8)
E=1/3*-6/1 + 2/3*-8/1
E=-2/1+-16/3
E=-6/3+-16/3
E=-22/3=7.333
E=$7.33

Fair price=expectation + cost of play
Fair price= 7.33 +8.00
Fair price= $15.33

This just doesn't seem right to me, can someone tell me where I went wrong?

soroban
02-02-2011, 06:11 PM
Hello, aburchett!

You left out a lot of details . . . like:
. . how is the spinner divided?
. . what are the payoff?


A person spins the pointer and is awarded the amount indicated by the pointer.

It costs $8 to play the game.

Determine:

. . (a) The expectation of a person who plays the game.
. . (b) The fair price to play the game.

My work:

E\:=\:P(\text{win})\cdot (\text{amount won}) \,+\,P(\text{lose})\cdot(\text{amount lost})

E \:=\:\left(\tfrac{1}{3}\right)(\text{-}6) \,+\,\left(\tfrac{2}{3}\right)(\text{-}8) . ??
Do I understand the situation?

Two-thirds of the time I lose $6.
The other one-third of the time, I lose $8.

\text{My expectation is: }\:-2-\frac{16}{3} \:=\:-\frac{22}{3}

\text{I can expect to }lose\text{ an average of }\$7.33\text{ per game.}


And you think I'll pay $8 to play this game . . . I don't think so!

aburchett
02-02-2011, 06:29 PM
I'm so sorry, the spinner looks like this:
[attachment=0:15s6zp88]spinner.png[/attachment:15s6zp88]

soroban
02-02-2011, 10:44 PM
Hello again, aburchett!

Okay, much better . . .


A person spins the pointer and is awarded the amount indicated by the pointer.



* * *
* | *
* | *
* | $8 *
|
* | *
* $2 * - - - - *
* | *
|
* | $12 *
* | *
* | *
* * *

It costs $8 to play the game.

Determine:

(a) The expectation of a person who plays the game.

(b) The fair price to play the game.
We can expect the following:

. . \begin{array}{c}\text{win \$2 with probability }\tfrac{1}{2} \\ \\[-3mm] \text{win \$8 with probability }\tfrac{1}{4} \\ \\[-3mm] \text{win \$12 with probability }\tfrac{1}{4} \end{array}


If we played for free, our expected value would be:

. . E \;=\;\left(\tfrac{1}{2}\right)(2) + \left(\tfrac{1}{4}\right)(8) + \left(\tfrac{1}{4}\right)(12) \;=\; 6

We could expect to win an average of $6 per game.

(a) Since we pay $8 to play each game, we have an average loss of $2 per game.


(b) The fair price to pay is $6 per game.