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The issue as I see it is that the probability exists that the other envelope contains Z/2 or 2Z, but not both - ie. the probability that one of those exists neg
by dalai 13y ago
The issue as I see it is that the probability exists that the other envelope contains Z/2 or 2Z, but not both - ie. the probability that one of those exists negates the other from existing, because in order for both to be possible you're introducing a 3rd value where only 2 existed at the start.
So there is 50% probability that the other one contains Z/2 and 50% that it has 2Z. Do you disagree with that statement?
I have two values A and B. I ask you to select one. What is the probability that the remaining value is C? Obviously it's zero.
No it doesn't. According to the problem, both 2Z and Z/2 are possible and you don't have the information to decide either way. Think of a simple coin toss, before I show you the result it is 50-50 to be heads or tails, even though it is just one of the two.
- dools 13y agoSo there is 50% probability that the other one contains Z/2 and 50% that it has 2Z. Do you disagree with that statement? Yes! Well, in some circumstances. Unless you know how the values were placed into the envelope to begin with, how could you possibly assign a probability to it? I just found this great link today: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... which basically resolves the problem I'm having by taking into account how the values get into the envelopes in the first place. If you have absolutely no knowledge of how the values got into the envelope in the first place then there's a 50/50 chance you got the higher value on your first pick, and a 50/50 chance you'll get the higher value on your second pick. You can't make a more informed decision unless you know how the values were initially chosen.