In one box, 2 times more money than in another. We opened one of them, there is $ 100, should we open the second one? How to maximize EV ?
I first thought box 1 has 100$. EV to stay is 100$.
Box 2 will have either 50 or 200$.
(50+200)/2=125
On average, our EV to switch will be 125$, 25$ more than staying.
This didn't seem correct. We know someone opening their first boxes and another person always opening the second one will have the same amount in the end after many boxes. I had to Google to find out what the answer was. I'll not spoil it.
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Last edited by Bubbles; 02-18-2018 at 07:32 AM.
It's a math teaser. Start here: https://en.m.wikipedia.org/wiki/Two_envelopes_problem
Kind of like Monty Hall, but with two choices instead of three.
You really don't have any information that is relevant to the problem after opening the first box. One box has a smaller amount, S, and one the larger amount, L. The expected value of either box, assuming each has an equal chance of containing either amount, is:
EV = 1/2*(S +L)
Since L =2*S:
EV = 1/2(S + 2*S) = 3*S/2
Knowing the value of one box doesn't change the EV's of each box. It just gives the EV of both boxes two possible values that are still equal and equally likely. One if the revealed amount ends up being the smaller amount and another if it ends up being the larger amount. Only one is reality and you don't know which one. You can't say I like this EV better so I will or won't switch. You only know the amount in the box you chose, A. These two events have the same probability:
If A = S: in which case you gain S by switching
If A = 2*S: in which case you lose S by switching.
Since the two events listed above have equal probability so switching has as an effect of zero (1/2(S + -S)).
Since there is no EV gained in switching, it becomes a problem like Don's coin flip scenario. A preference as to whether you prefer a $100 sure thing or switching to the other box that either has $50 or $200. Since $100 is a trivial amount I would probably take the unrevealed amount in the other box. If the revealed box contained a life changing amount I would keep it.
Interesting....I linked over to the wikipedia on this and it was too much to take in before morning coffee.
While not identical, does this bear any relation to the double up option offered on some VP games? I suppose that is more like you have one envelope in hand with $X and are offered the option to exchange to one of two others, one with $2X and one with $0X. If you take the offer you have a random shot at double or nothing.
You are right. If the amounts are unknown in both boxes the values of S and L are undefined variables twice or half of each other. Once you know the value of one box that changes since S is either 1/2 that particular value or twice that particular value. So basically S no longer equals S in each of the two possibilities. So the other box would on average contain 1.25 times the value of the known box. That was bothering me when I did my analysis but I didn't understand why. Thank you for getting me to finish my line of reasoning.
Last edited by Three; 02-18-2018 at 12:14 PM.
For utility, everyone will have a different answer. For EV, however, there is no difference. This is a classic high school math problem... Maybe this link will be easier to understand.
https://brilliant.org/wiki/two-envelope-paradox/
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