Econ HW - Expected Value

[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?

[jman3232]

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)?

[ItalianFX]

[ QUOTE ]
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.

[ItalianFX]

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.

[jman3232]

[ 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.

[/ 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.

[ItalianFX]

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?

[ItalianFX]

[ QUOTE ]


(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?

[relativity_x]

a=1
b=200
c=-525

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