Originally Posted by

**OPFOR**
Paymeister, as has been said, you are correct in that any series of random events is random, and does not depend on preceding events (roulette spins, coin flips, etc). This, however, does NOT hold true here: there is only one event,; you are just given more information about it DURING the event. Imagine you spun a roulette wheel after betting on black, then found out (magically) that the ball would NOT land on 10 of the black squares...would you be smart to change your bet to red if you were given the opportunity? Of course - you know have 10 less winning options... (Note that this is NOT analogous to the Monty Hall problem, it's just to show that the MH problem is one event, with information added during the event. The idea that some “random” events do have a memory is the key to “card counting” in Black Jack and the basis of all odds predictions in poker – the odds change DURING each shoe/deal as more information, i.e. more cards, are shown to the player.)

Ok, I thought the explanation (and example) I gave would show the answer, but apparently not. I'll write it here again, and try to expand on it so it's more clear...

At the outset, you have three choices, and NO information about what is behind any of the doors. Picking one at random, you have a 1/3 chance of picking the BP and a 2/3 chance of picking a G. Assume the BP is behind door number one (it doesn’t matter which for the explanation, you can just change the numbers...)

You have three options; you chose 1, 2, or 3.

1) If you chose 1, Monty could show you 2 or 3 (both being goats). If you switch, you lose. If you stay, you win.

2) If you chose 2, Monty **MUST** show you 3 (1 being the BP). If you switch, you win. If you stay, you lose. It is important to note that he cannot show you 1, because that would reveal where the BP is. He MUST show you the other losing door.

3)If you chose 3, Monty **MUST** show you 2 (1 being the BP). If you switch, you win. If you stay, you lose. Same here – he MUST show you the other losing door.

After he has shown the one losing door, you now have another option, with more information. You know that if you picked a goat (which you had a 2/3 chance of doing at the outset), then Monty was just forced to show you where the OTHER goat was. So, if you had guessed wrong (2/3 chance, remember?), you now know where both goats are – behind the door you choice, and behind the one Monty revealed. That means the BP is behind the door neither originally chosen, nor revealed.

If you originally guessed right (1/3 chance), Monty could reveal either door, and you don’t really gain any new information. Remember, though, that you had a 2/3 chance of picking a goat at the outset, and only a 1/3 chance of picking the BP. So, we assume (based on probability) that we originally picked a goat, and now Monty has shown us the other goat, therefore (again based on probability – not certainty) you now know where both goats (and therefore the BP) are.

Check the play by play again – if you stay, you only win if you had originally picked correctly (1/3 chance). If you switch, you win if you had originally picked incorrectly (2/3 chance). Remember there are only three original options – picking 1, 2, or 3….

Clear yet? It’s counterintuitive to some extent, that’s what makes it an interesting problem, but if you simply look at all the possible outcomes you will see that, by switching, you will win in 2 of the 3 possible outcomes and by staying you will win in only 1 of the 3 possible outcomes….