View Full Version : Expectation and Fair price
aburchett
02022011, 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
02022011, 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?
Twothirds of the time I lose $6.
The other onethird 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
02022011, 06:29 PM
I'm so sorry, the spinner looks like this:
[attachment=0:15s6zp88]spinner.png[/attachment:15s6zp88]
soroban
02022011, 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.
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