Econ HW - Expected Value

[ItalianFX]

For some reason this isn't making sense to me, so hopefully I can get the right answer here.

Question: What is the expected value of a random toss of a die? (Fair and six-sided.)

This next question I just have no idea how to do it.

Suppose your current wealth, M, is 100 and your utility function is U = M^2. You have a lottery ticket that pays $10 with a probability of 0.25 and $0 with a probability of 0.75. What is the minimum amount for which you would be willing to sell this ticket?

Thanks.

[relativity_x]

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Question: What is the expected value of a random toss of a die? (Fair and six-sided.)

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I would say it's 1/6

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Suppose your current wealth, M, is 100 and your utility function is U = M^2. You have a lottery ticket that pays $10 with a probability of 0.25 and $0 with a probability of 0.75. What is the minimum amount for which you would be willing to sell this ticket?

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I'm not sure how the utility function and current wealth plays into this question, but I'd assume the minimum price you'd sell the ticket for is P=0.25(10)+.75*(0)=$2.50.

It's possible since the prize is 10 dollars your utility from the prize is 10^2 or 100. If this is the case, then the minimum price would be Pmin=0.25(100)+.75(0)=25.


disclaimer: these may/may not be correct.

[buttonpusher]

isnt the dice answer 3.5?

I also think the minimum price you should sell the ticket for is $2.50.

[jman3232]

EV= probability*value
1/6*1+1/6*2+1/6*3....+1/6*6

so yes it is 3.5

The second part... hmmm EV of U without selling ticket is 1/4*(110)^2 + 3/4*(100)^2= 3025+7500= 10525= U

so... because U=m^2, m= 102.59...
Sell it for at least $2.59. It probably doesn't make that much sense, but its what I see.

[tabako]

Dice: EV = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 3.5

If you sell the ticket for price P, you now have (100 + P) dollars. Your utility is now (100 + P)^2.

If you keep the ticket, 25% of the time you hit for $10 and now have $110. 75% of the time you stay at $100.

Your utility here is: (.25)(110)^2 + (.75)(100)^2

You are willing to sell the ticket if your utility from selling is >= to the expected utility from holding on to the ticket. The least you will sell it for is when the two are exactly the same. So solve:

(100 + P)^2 = (.25)(110)^2 + (.75)(100)^2
10000 + 200P + P*P = 10525
P^2 + 200P - 525 = 0

P = $2.59

This is slightly larger than the solution of $2.50, which makes sense if you think about it.

[ItalianFX]

What I don't understand about the dice problem is, why are we weighting the sides of the die? Assume that we only roll the die once, then each side has an equal chance of coming up, one out of 6 or 1/6. Why weight them 1, 2, 3, 4, 5, 6?

[tabako]

How much would you pay to play this game with me?:

You pay me some amount of dollars. You roll a 6-sided die. I pay you the number dollars equal to the number that comes up.



You should be willing to pay no more than the expected value of the die. Do you still think this number is 1?

[ItalianFX]

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(100 + P)^2 = (.25)(110)^2 + (.75)(100)^2
10000 + 200P + P*P = 10525
P^2 + 200P - 525 = 0

P = $2.59



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I'm a freakin' idiot. How do you solve for P in that last step?

[relativity_x]

a=1
b=200
c=-525

b=(-b+/-sqrt(b^2-4ac))/(2a)

[tabako]

Quadratic equation. It looks like it would be a pain to factor it, but you might be able to (the second solution you get with the quadratic equation is negative and obviously doesn't apply here).

For the coin question: To calculate EV, you need to multiply two things together. 1) probability of an event happening. 2) utility when that event happens (some will be positive, some negative, in this coin problem)

Do this for all the possible events in the problem and then add the products together.

The coin problem should be very simple.