K <- c("X",0) L <- c("-X",0) q <- c(K,L) w <- sample(q, 10000, replace = TRUE) sum(w == "X") sum(w == "-X") sum(w == 0)
In the event of L our expected gain loss sample space is [-X, 0]
In the event of K our expected gain loss sample space is [X,0] — Jeremiah
Let's try running a simulation on that structure.
K <- c("X",0)
L <- c("-X",0)
q <- c(K,L)
w <- sample(q, 10000, replace = TRUE)
sum(w == "X")
sum(w == "-X")
sum(w == 0)
The Result are:
x: 2528
-x: 2510
0: 4962
``` — Jeremiah
Might as well just not even bother looking in the first envelope then, and pick the second one from the start. Looking at the contents seems not to affect this decision. This tells me it doesn't matter if you switch.If we open the envelope and see £10 then we know that the other envelope contains either £5 or £20. By switching there's a 50% chance of losing £5 and a 50% chance of gaining £10. Switching seems like the better option. — Michael
I don't know R, but from what I can gather that's just selecting an outcome at random and showing that the number of times you win is equal to the number of times you lose?. — Michael
We've established that where X is gained, X = 10 and where X is lost, X = 5. — Michael
You are letting your examples confuse you once more. What we have establish is that -X and X are equally likely to occur. If X = 5 then that is -5 or 5. If X = 10 then that is -10 or 10. If X = 15456783134.1346854654 then that is -15456783134.1346854654 or 15456783134.1346854654. That is what it means to be an unknown variable. What you are suggesting is that for each X there are two possible outcomes, and well that does not fit the definition of a function. You are not considering that L and K are independent events and that -X and X are conditional events. — Jeremiah
Let's call case one L and case two K. Where L is the event you start with 2X and K is the event you start with X. These are two different events.
You open it up and see you have 10 bucks:
In the event of L if you switch then you gain -X, which in this case is a loss of 5 bucks. If you don't switch then you gain 0.
In the event of K if you switch then you gain +X, which is gain of 10. If you don't switch then you gain 0. — Jeremiah
Let X=10.
If you choose Envelope X then you get 10 and switching gains you ten. If you choose Envelope 2X then you get 20 and switching loses you 10. — Baden
No 20 and 5 are incompatible possibilities given you've already chosen an envelope. 20=4(5). There is no 4X. — Baden
There is no point in this scenario after an envelope is chosen where X( £5) and 4X(£20) are compatible possibilities. — Baden
It depends on my utility function U. Utility functions map one's wealth to levels of 'utility', ie satisfaction. They are generally assumed to be monotonically increasing and concave down.What should you do? — Jeremiah
What should you do? — Jeremiah
So, you have 10. But only one scenario applies to your choice, either the one where you lose by switching in which case X is five and 2X is 10, or you gain by switching in which X is 10 and 2X is 20. You can't be in both scenarios at the same time and X has a different value in both, so your calculation fails. — Baden
We know the chosen envelope has £10 in your scenario and we also know that one of the envelopes contained an amount X and one of the envelopes contained an amount 2X, and therefore we know we can only move from X to 2X or from 2X to X by switching. Therefore we know we have a 50% chance of gaining X by switching and a 50% chance of losing X by switching and that therefore there is no point in switching. — Baden
It looks like you're assuming the player's utility function is the identity function, which would be unusual and unrealistic. Even if we assume that, the quoted calculation doesn't take account of all the available information. There is new information, which is the known dollar value in the opened envelope. That changes the utility calculation.The expected value of each envelope is $(X + 2X)/2 = $3X/2 = $1.5X. You should be indifferent between switching or not. — Andrew M
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