Two envelopes labelled A & B
Two dudes (Harry & Tom)
Both know that one envelope contains exactly TWICE as much as the other.
Flip coin and Harry wins and chooses A. Tom gets B.
Both are offered swap.
Both consider options and logically decide that the other envelope has a better probability of maximizing their money. They both swap.
How can that be?
That is the paradox.
Originally posted by David113But if your envelope contains $100 you know the other envelope contains eitherr $50 or $200.
"logically decide that the other envelope has a better probability of maximizing their money"?
Obviously, both envelopes have the SAME probability of maximizing their money - 0.5.
Logically you decide to swap
(50%chance of +100 and 50% chance of -50).
I am still aware that there are only two envelopes out there, say x and 2x and that by swapping I am either losing or gaining x. The opening of the envelope has not determined the value of X (although limits it to two values).
Whilst if I was offered to double of half my money in isolation then obviously the odds are in my favour by taking the gamble, where there are only two envelopes an equation needs to balanced with the owner of the other envelope and the supposed probability in my favour is merely not correctly stating the equation by using all facts within my knowledge.
Originally posted by wolfgang59That all depends on what you believe the probability distribution of the possible values is.
But if your envelope contains $100 you know the other envelope contains eitherr $50 or $200.
Logically you decide to swap
(50%chance of +100 and 50% chance of -50).