Le 16/12/2009 09:55, scott nous fit lire :
>> 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...

finding 100$, you do not know if it is the low or high amount.

The computation is NOT $50*1/2+ $200*1/2 ( = $125)
but either :

 $50*1/2 + $100*1/2 = $75
or
 $100*1/2 + $200*1/2 = $150

Which turn out $75*1/2 + $150*1/2 around $112.5... if that was random.

Finding $100, you have no idea yet if you have a good value or a bad one.
It's a one time game, no long run (once you know the best value... that
you learn at the end of the choice), and girl & boy might be coming in.

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