Econ HW - Expected Value
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. |
Re: Econ HW - Expected Value
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Question: What is the expected value of a random toss of a die? (Fair and six-sided.) [/ QUOTE ] I would say it's 1/6 [ QUOTE ] 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? [/ QUOTE ] 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. |
Re: Econ HW - Expected Value
isnt the dice answer 3.5?
I also think the minimum price you should sell the ticket for is $2.50. |
Re: Econ HW - Expected Value
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. |
Re: Econ HW - Expected Value
[ QUOTE ]
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. [/ QUOTE ] Both of the answers so far for the second question are wrong...the 3.5 for the dice answer is correct. One answer came close...your utility function is M^2, so what you want to do is find the point where selling the ticket for a fixed amount is equal to the expected utility from the ticket. I'll show you the easier one, you'll have to do the lottery ticket, figure the expected util, and then some math. From selling the ticket (for X dollars, say), your wealth after is (100+X), so your utility form selling the ticket is (100+X)^2. Figure out the expected utility from selling the ticket, and find out where X makes you indifferent between the two options. Shane |
Re: Econ HW - Expected Value
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. |
Re: Econ HW - Expected Value
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?
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Re: Econ HW - Expected Value
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? |
Re: Econ HW - Expected Value
Think about it like this: you get the number of dollars as the dice rolls (i.e. a roll of 4= $4).
What is the expected value of one roll (in dollars)? |
Re: Econ HW - Expected Value
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Think about it like this: you get the number of dollars as the dice rolls (i.e. a roll of 4= $4). What is the expected value of one roll (in dollars)? [/ QUOTE ] Ok, that makes sense. Maybe I'm thinking too much in probabilities? How would this relate to the EV of a flip of a coin? Could you show me that? I know it's .5, but I'd like to see how we do that too. |
Re: Econ HW - Expected Value
How about this question:
A fair coin is flipped twice and the following payoffs are assigned to each of the four possible outcomes: H-H: win 20, H-T: win 9, T-H: lose 7, T-T: lose 16. What is the expected value of this gamble? I'm thinking I know how to do this one, but I'm also thinking it's harder than I am figuring it to be. |
Re: Econ HW - Expected Value
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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 [/ QUOTE ] I'm a freakin' idiot. How do you solve for P in that last step? |
Re: Econ HW - Expected Value
a=1
b=200 c=-525 b=(-b+/-sqrt(b^2-4ac))/(2a) |
Re: Econ HW - Expected Value
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. |
Re: Econ HW - Expected Value
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a=1 b=200 c=-525 b=(-b+/-sqrt(b^2-4ac))/(2a) [/ QUOTE ] Ok, that's what I thought. I kept thinking there had to be a different way. I'm really not as dumb as I appear! [img]/images/graemlins/blush.gif[/img] |
Re: Econ HW - Expected Value
[ QUOTE ]
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. [/ QUOTE ] Yeah I thought so. (.5)(.5)(20)+(.5)(.5)(9)+(.5)(.5)(-7)+(.5)(.5)(-16) = 1.5 |
Re: Econ HW - Expected Value
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Ok, that makes sense. Maybe I'm thinking too much in probabilities? How would this relate to the EV of a flip of a coin? Could you show me that? I know it's .5, but I'd like to see how we do that too. [/ QUOTE ] EV= probability times value: the probability is .5, but there is no true value, like a dice has a set value of 1,2,3... etc. but there is no set values for a coin. If you set heads=1 and tails=0, then the EV=.5 for a fair coin, but for a normal, unspecified situation stating the EV is kinda strange... if you know what I mean. |
Re: Econ HW - Expected Value
Say we have a two-sided disc with a 1 and 2. If we flip the disc, the probability of getting a 1 or 2 is 1/2. Taking the example of the die, then the EV of flipping the disc would be 1(.5)+2(.5) = 1.5, correct?
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Re: Econ HW - Expected Value
[ QUOTE ]
Say we have a two-sided disc with a 1 and 2. If we flip the disc, the probability of getting a 1 or 2 is 1/2. Taking the example of the die, then the EV of flipping the disc would be 1(.5)+2(.5) = 1.5, correct? [/ QUOTE ] exactly. |
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