> Unless I am vastly misreading this, it seems there is a simple way to
> set up this problem. Two cups, arbitrarily labeled A and B.
>
> A contains $X, B contains $2X
> or
> A contains $2X, B contains $X
>
> I don't see an obvious way to turn this into the Boy or Girl, or Monty
> Hall, paradox, so I would have to guess that yes, it would be a good bet
> to take the switch.

How can a tactic where you *always* switch the cup be more profitable than 
never switching?  In both cases, in the long run you're going to end up 50% 
of the time choosing the higher amount and 50% the lower amount.

I'm still trying to figure out where the mistake is in the OP's reasoning...

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