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Expected Value - Carnival Game - Need Help Please

AngelaS

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Apr 22, 2012
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I need to find the expected value of a carnival game that I made up. Here is my game:
For this game, the only supply that you will need is a spinner with two sections; one section is red and one section is blue. The red section is 75% of the entire spinner, while the blue section takes up the remaining 25% of the spinner. The object of the game is for the participant to spin the spinner two times and land on each color once; red and blue. It does not matter the order of the colors on which the participant lands. If the participant is able to land on each color once, during their two spins, they win $4.00. The cost to play “Spinner Winner” is only $3.00.

Would the expected value (or average) that the charity would win per participant be $2.25? The probability of winning is 3/16 and losing is 13/16. So, wouldn't I just take the net profit of $36 and divide that by 16 to get the $2.25. The expected value also has to show a losing value for each player....I don't even know where to start for that.
 

batman350z

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Apr 19, 2012
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Let me give a stab at this problem since it is right down my alley for my final exam.

I declare X = the winnings of a random customer
Customer either wins $4 or $0, so
P(X=0) = \(\displaystyle \frac{13}{16}\)
P(X=4) = \(\displaystyle \frac{3}{16}\)

Expected winnings for a random customer = E(X) = 0(\(\displaystyle \frac{13}{16}\)) + 4(\(\displaystyle \frac{3}{16}\)) = $0.75
Since a random customer has a expected winning of $0.75, then the expected losing would be $3 - $0.75 = $2.25

Lastly, the expected profit made by the charity for a random customer = 3 - E(X) = 3 - 0.75 = $2.25

[edit] fixed the winning amount [/edit]
 
Last edited:

AngelaS

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Apr 22, 2012
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Thank you. That helps a lot! I think it's "clear as mud" now. :p
 

AngelaS

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Apr 22, 2012
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Ugh... now I'm more confused than I was when I started! :( So my probability of 3/16 (win) and 13/16 (lose) isn't even right?
 

pka

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Jan 29, 2005
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7,688
Ugh... now I'm more confused than I was when I started! :( So my probability of 3/16 (win) and 13/16 (lose) isn't even right?
I think that it is clearer to use decimals.
Here is how you loose: \(\displaystyle (.75)^2+(.25)^2=0.625\)
You get two blues or two reds.

Now \(\displaystyle 1-0.625=0.375\) that is the probability of winning: one of each color.
\(\displaystyle $1(0.372)-$3(0.625)=-$1.50\)
 

AngelaS

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Apr 22, 2012
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I had come up with the 3/16 by multiplying 1/4*3/4=3/16. 1/4 being the "blue" and 3/4 being the "red". I don't think you misunderstood the rules..I'm just not grasping this whole probability and expected value subject. How you both (JeffM and pka) spelled it out makes sense... but I really thought I was on the right track. I guess not.
 

batman350z

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Apr 19, 2012
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Ugh... now I'm more confused than I was when I started! :( So my probability of 3/16 (win) and 13/16 (lose) isn't even right?
Well, it sure looks like that to me, but maybe I do not fully understand the game. How did you come up with 3/16?

Did you understand my point about FOUR possibilities? That my be where you went wrong or where I misunderstood the rules.
Yes, JeffM is right, you need to consider all the possibilities of winning and losing. 3/16 = 3/4*1/4 which is spinning red and then blue, but you can also win by spinning blue first and red second 1/4 * 3/4. So you'll need get all the permutation of winning and losing to calculate your probability.
 

AngelaS

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Apr 22, 2012
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Thanks! I'll go and rework my problem before it is due tomorrow.
 

soroban

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Jan 28, 2005
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5,588
Hello, AngelaS!

I need to find the expected value of a carnival game that I made up.

We have a spinner with two sections; one section is red and one section is blue.
The red section is 75% of the entire spinner; the blue section takes up the remaining 25% of the spinner.
The object of the game is for a player to spin the spinner twice and land on each color once: red and blue.
The order of the two colors does not matter.
If the participant is able to land on each color once, during their two spins, they win $4.00.
The cost to play “Spinner Winner” is only $3.00.

Would the expected value (or average) that the charity would win per participant be $2.25?
The probability of winning is 3/16 and losing is 13/16.
So, wouldn't I just take the net profit of $36 and divide that by 16 to get the $2.25.
The expected value also has to show a losing value for each player.
I don't even know where to start for that.

JeffM explained the four cases and their probabilties.
Please digest that first . . . or further explanations will be useless.

Note that the player pays $3 for every game.

Then \(\displaystyle \frac{3}{8}\) of the time he wins $4 . . . so he has a gain of $1.


We have:

. . \(\displaystyle \begin{array}{cccc} \text{Event} & \text{Prob.} & \text{Result} \\ \hline \text{Win} & \frac{3}{8} & +\,\$1 \\ \text{Lose} & \frac{5}{8} & -\,\$3 \end{array}\)


Hence: .\(\displaystyle E \;=\;\left(\frac{3}{8}\right)(1) + \left(\frac{5}{8}\right)(-3) \;=\;\frac{3}{8} - \frac{15}{8} \:=\:-\frac{12}{8} \:=\:-\frac{3}{2}\)


On the average, the player can expect to lose $1.50 per game.

On the average, the charity can expect to win $1.50 per game.
 

AngelaS

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Apr 22, 2012
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Thank you all!!!! I just redid my homework project.... it's three pages long, so hopefully I explained all my calculations correctly. I can't thank you enough as you saved me from an "E"!
 
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