Assume the standard (better to switch) solution of the Monte Hall problem.   Then there's the 3-card Monte problem Ok, I understand and am convinced by the standard solution of the Monte Hall Problem, i.e. it is better to switch doors after Monte opens one, and improve one's probability of winning from 1/3 to 2/3.    If I had remaining doubts, they were removed by the many computer simulations, e.g. see http://demonstrations.wolfram.com/MontyHallParadox/.     But then I came up with what I call the "three card monte problem".     Assume you are playing a variant of 3 card monte against a blindingly fast but honest dealer, who will also turn over a losing card, before your once and final guess.    Do you have a strategy that improves your probability to better than 1/2?    Well, here's one.   You choose a card, mentally, but do not reveal it.   The dealer then turns over one of the other cards, which he knows is always not the "special winning card".    There is prob = 1/3 that the turned over card is the card that you previously secretly mentally chose.   If that happens, you still can choose again, and therefore have prob = 1/2 of choosing the winner.    If the card turned over by the dealer is not the card you secretly chose mentally, then you are apparently playing the classic Monte Hall game, i.e. you should secretly mentally switch card positions to increase your prob of winning from 1/3 to 2/3.    Your overall prob. of winning is (1/3)(1/2) + (2/3)(2/3) = 1/6 + 4/9 = 33/54 = 11/18 > 1/2     The paradox arises from the fact that another player, who chose the other hidden card which was not turned over also has the same 11/18 probability of winning.   So you both switch?   The difference between this and the classical Monte Hall paradox is that because no player has to declare any card position, you both can switch.   Is this right?
If this gets too argumentative, I'll put it somewhere else.
 A: The following sentence is false:
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If the card turned over by the dealer is not the card you secretly chose mentally, then you are apparently playing the classic Monte Hall game, i.e. you should secretly mentally switch card positions to increase your prob of winning from 1/3 to 2/3.
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In the classic Monty (note spelling) Hall game, you are guaranteed that Monty will choose one of the two "bad" doors at random, and that he won't choose the door that you chose.  In your situation, the second part of the guarantee is missing.  If you run through whatever your favorite argument for the right answer to the usual problem is and look carefully at each step, you should be able to find the place where this (very necessary) assumption is used.
A: Your strategy simply doesn't work. (You mentioned doing some computer simulations in your first paragraph - try some for your proposed strategy).
The simple symmetry argument shows that whatever you do, you have probability 1/2 of winning. Call the cards A, B, C. Assume that each of them is equally likely to be the winning card, and that the dealer turns over one of the two losing cards with equal probability. If the dealer turns over card B, there is nothing that has removed the symmetry between cards A and C. It doesn't matter what you've said to yourself in your head. Whatever you choose to do, you win with probability 1/2.
In the original Monty Hall problem this symmetry is not present. Suppose you choose card A. Suppose the dealer turns over card B. By now we have lost the symmetry between A and C, since the dealer had the choice to turn over one of B and C, but not A. (Here we make the appropriate assumptions about the behaviour of the dealer - of course they are important but I won't spell them out here). The dealer has "rejected" card C, but has not "rejected" card A. Hence one has extra reason to suspect that card C may be the winning card etc etc. 
