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The Monte Hall problem
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[QUOTE="Forloveoforange, post: 5081523, member: 1015"] What makes the odds stay the same, and what makes it the same game, is that the host knows. If it were random guesses on the hosts part, then the doors would keep the same face value (in relation to the other unopened doors). The host doesn't have the option of picking random doors, so that is what keeps it the same game, and what keeps Door 1 from never having more than a 1 in 3 chance. Otherwise it would actively get better, because his luck-driven guesses would narrow the choices down equally. Face value. The doors look the same, but the non-randomness changes everything. If there were a bunch of doors, his luck in finding the car would equal the luck of your first guess, and you could keep getting better and better odds on your guess. This is why your guess stays the same, and why the other doors don't change in relation to your door. Your side 1 in 3 and the other side 2 in 3, or if 4 door game, 1 in 4 and 3 in 4, etc. Switching is unimportant in a random game. You would keep comparing equally to what is left. Every door would have the same face value, so no reason to switch, or not to switch. You take away doors in a random way, that makes the odds go up for what is left, including your original choice. So that's it. I guess that is the key that is mind-boggling. I think some people think that switching automatically freshens a game, but if its a RANDOM game, then I don't see it. The luck factor should keep your guess increasing in value. If you have 100 doors and you're down to 2 equal chances, you have gotten yourself to 50-50. But even then, you really had only a 2% chance at getting to this "50-50" spot in the game. But remember that is if the game is RANDOM. RANDOM can be a bit mind-boggling too. [/QUOTE]
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