MONTY MONTY

[Ghost Post]

"switch!"...this seems logical, albeit countintuative...
still, what's the flaw in this "50/50" argument:

you and another contestant, X, play the same
3 doors simultaneously, neither knowing
what the other does...

obviously, if you both choose losing doors,
the host can't open the winner...so the rules
would have to be modified in this version...

say, for example, play continues ONLY
when one of you has picked the winner,
and the other has picked a loser....

ok...you take A, X takes B, host opens
C, a loser.. don't you both now
want to switch, each eagerly coveting
the door that the other of you equally
eagerly wants to abandon?...

so you trade doors willingly...how can
EACH of you win 2/3rds of the time?...

[Mercuria]

the flaw is that 1/3 of the possibilities have been eliminated. if the prize is in the third, unchosen door, then the switch, in the monty hall problem, would have given them the prize. had they chosen the door that they did in that situation as a single player, they would have switched to the prize. therefore, 1/3, they switch and win, 1/3 they should switch and win, but the game is halted because of "technicalities."

[Andy]

Here's an equivalent of this variation, with only one contestant...

You pick two doors initially.
If you picked both losing doors, Monty opens all three doors and you lose.
If one of the doors you picked is a winner, Monty opens the third (losing) door. Now you must settle on just one of the two doors you picked initially, either winning or losing.

Probabilities:
Losing on first pick: 1/3
Getting to make second pick: 2/3 (duh!)
losing on second pick: 2/3 * 1/2 = 1/3
Winning : 2/3 * 1/2 = 1/3.
Same as just picking one door with no second chance.