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> 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 e
by lisper 4y ago
> 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 4y agoImpossible. 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 4y ago> 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."