I think you reached the wrong conclusion in your final paragraphs. Can you show how the expected value calculation could be different for “the amount of money contained in the other envelope expressed in terms of the amount of money in this envelope [which is in dollars, BTW]” versus “the amount of money in the other envelope expressed in dollars”?
I see that Wikipedia says that there is no generally accepted solution to the paradox. That almost certainly means people are interpreting it differently. I’ll give my opinion. But let me recast the problem in more pointed terms.
Mr Moneybags writes a check for some real positive amount of money and puts it in an envelope. Then he writes a check for double the amount of the previous check and puts it in another envelope. He shuffles the envelopes. You pick one and open it, and you find $N. He offers to let you switch to the other just this once. Should you?
On first glance it appears that the other envelope could have $N/2 or $N*2, and that you are entirely ignorant about which is the case, so your ignorance priors are 1⁄2 for each possibility. Then the expected value calculation comes up all wrong! Call the hypothesis that you opened the lower value envelope L and the hypothesis that you opened the higher value envelope H.
But that’s nonsense. My opinion as to what went wrong was that it’s false that you are entirely ignorant after opening the envelope, because you can now do an update on your ignorance prior.
We’ll start, of course, with Bayes’ formula P(H|N)/P(L|N) = P(N|H)/P(N|L) * P(H)/P(L) * P(N)/P(N). The last ratio cancels. The second-to-last ratio also cancels because the shuffling of the envelopes made it equally likely you’d open either one. We know basically nothing about the distribution that the lower value was first drawn from, so a convenient ignorance prior on it is that it was some uniform distribution over the interval $A to $B. Consequently, the higher value would be described by a uniform distribution over the interval 2*$A to 2*$B. The distributions may or may not overlap. I’ll take each case separately.
Common start: A=0.
P(N=k|H)={1/3 for 0<k<=B; 1 for B<k<=2B}
P(N=k|L)={2/3 for 0<k<=B; 0 for B<k<=2B}
Integrate each with respect to k from 0 to 2B, then plug in
P(H|N)/P(H|L) = (4/3*B)/(2/3*B) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.
Separate start: A>0 and 2A<B.
P(N=k|H)={0 for A<k<2A; 1/3 for 2A<=k<=B; 1 for B<k<=2B}
P(N=k|L)={1 for A<k<2A; 2/3 for 2A<=k<=B; 0 for B<k<=2B}
Integrate each with respect to k from A to 2B, then plug in
P(H|N)/P(H|L) = P(N|H)/P(N|L) = (4/3*B-2/3*A)/(2/3*B-1/3*A) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.
No overlap: B<2A.
P(N=k|H)={0 for A<=k<=B; 1 for 2A<=k<=2B}
P(N=k|L)={1 for A<=k<=B; 0 for 2A<=k<=2B}
Integrate each with respect to k from A to 2B, then plug in
P(H|N)/P(H|L) = P(N|H)/P(N|L) = (2B-2A)/(B-A) = 2
and convert back to probabilities
P(H|N)=2/3, P(L|N)=1/3.
So viewing your prize, you suddenly feel it’s twice as likely that you got the higher amount. Bizarre, huh? And your expected value calculation becomes
so there’s no point switching after all. Even weirder is that, before you open the envelope, this is a situation where you know for sure what you will believe in the future, but you nevertheless can’t make that update till then.
I think you reached the wrong conclusion in your final paragraphs. Can you show how the expected value calculation could be different for “the amount of money contained in the other envelope expressed in terms of the amount of money in this envelope [which is in dollars, BTW]” versus “the amount of money in the other envelope expressed in dollars”?
I see that Wikipedia says that there is no generally accepted solution to the paradox. That almost certainly means people are interpreting it differently. I’ll give my opinion. But let me recast the problem in more pointed terms.
Mr Moneybags writes a check for some real positive amount of money and puts it in an envelope. Then he writes a check for double the amount of the previous check and puts it in another envelope. He shuffles the envelopes. You pick one and open it, and you find $N. He offers to let you switch to the other just this once. Should you?
On first glance it appears that the other envelope could have $N/2 or $N*2, and that you are entirely ignorant about which is the case, so your ignorance priors are 1⁄2 for each possibility. Then the expected value calculation comes up all wrong! Call the hypothesis that you opened the lower value envelope L and the hypothesis that you opened the higher value envelope H.
But that’s nonsense. My opinion as to what went wrong was that it’s false that you are entirely ignorant after opening the envelope, because you can now do an update on your ignorance prior.
We’ll start, of course, with Bayes’ formula P(H|N)/P(L|N) = P(N|H)/P(N|L) * P(H)/P(L) * P(N)/P(N). The last ratio cancels. The second-to-last ratio also cancels because the shuffling of the envelopes made it equally likely you’d open either one. We know basically nothing about the distribution that the lower value was first drawn from, so a convenient ignorance prior on it is that it was some uniform distribution over the interval $A to $B. Consequently, the higher value would be described by a uniform distribution over the interval 2*$A to 2*$B. The distributions may or may not overlap. I’ll take each case separately.
Common start: A=0.
Separate start: A>0 and 2A<B.
No overlap: B<2A.
So viewing your prize, you suddenly feel it’s twice as likely that you got the higher amount. Bizarre, huh? And your expected value calculation becomes
so there’s no point switching after all. Even weirder is that, before you open the envelope, this is a situation where you know for sure what you will believe in the future, but you nevertheless can’t make that update till then.