[lisper]

> Let A = 50 Envelope 1 is 100 Envelope 2 is 25

No. The problem specifies that E1 is twice E2, but here you have E1 = 4 x E2.

A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time.

[andrewshawcare]

Impossible. Schrodinger’s envelope, then. You can’t reflect the probability to be 2A AND 1/2A if A represents the value of an envelope.

The value of the other envelope must include the probability that it is the original A (not some sleight of hand new A’ that, itself, is based on an expected value).

[lisper]

> You can’t reflect the probability to be 2A AND 1/2A if A represents the value of an envelope.

Yes, that's right. That's exactly what I said: "A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time."