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.