somebody wrote:
> You are on a game show. The host says he has hidden money under two cups,
> one amount being double the other one. You pick one at random, he turns it
> over, it's $100. He gives you a second chance (that he would do so was
> determined before the game and not dependent on which cup you'd turn over):
> You can switch, or take the $100.
> 
> You figure: If I switch, I will either get $50 or $200 with equal
> probability. Expected return is thus $50*1/2 + $200*1/2 = $125, which is
> larger than the $100 I'd spend to "play" this second game. Looks like a good
> deal. Is it?
> 
> 

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.

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