I think the point is that you should only switch if the expected value is greater than the expected value of not switching. Not switching is 5x/4 (or 3x/2 if the problem description is taken literally) and so is switching. So your return is the same either way. Maybe we get lulled into thinking there's a trick because the expected value is greater than x. But that's not important - there's more than 2x on the table, split between two envelopes, so we should expect an expected value greater than x.

This is different to the Monty Hall problem in that the expected value of switching is *different* to the expected value of sticking with the original choice, and only because the situation has changed (a door is opened). That's what you can't keep applying it.

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