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Nice one!! Similar to the three doors problem. If I am correct, the point is the implication: if the mean value is bigger than one -> switch envelop.

Before choosing any of the envelop, according to how the amount in the envelop is parametrised we can have mean value of 3/4x (envelop1 = x, envelop2 = 0.5 x) or 3/2x (envelop1 = x, envelop2 = 2 x). Using the same implication above, in the first case is not convenient to play the game (pessimistic parametrisation!), in the second it is convenient (optimistic parametrisation!).

This probably makes sense only if x is the quantity that we put in the game from our pocket, and in the game A we can receive x or 0.5x while in the game B we can receive x or 2x.

If, after "normalising", there are two quantities that we can win: 1 dollar or 2 dollars and the probability of winning one or the other is 0.5, then the mean value is always 3/2, either when we start the game choosing an envelop for the first time, or if we are asked to change it during the game, as no new information are acquired from one situation to the other.

So, I would not waste time changing envelop an infinite amount of time; instead I would play as many time as possible :-)

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