I was explaining the Monty Hall problem to someone thus:
You have three doors, and you pick one, giving you a $1/3$ chance of being right. The presenter opens one of the other two doors, knowing which door knows has a goat behind it. Then:
- you would not pick the opened door as it contains a goat, and not the expensive prize. So the probability that the opened door is a winner, is now reduced from $1/3$ to $0$.
- the remaining closed door must now have a $2/3$ chance of winning, an increase from the original $1/3$.
The person insisted that I was incorrect in stating that the probabilities shift in 1, 2 above. I do not understand if the opposer is correct or not. In some sense, I understand what they were saying: that the outcome does not change the probability. But at the same time, if we label the doors as "door chosen, door opened by presenter, and door remaining", then the probabilities will always be, respectively, $1/3$, $0$, $2/3$, right? The car will be behind the door that is untouched 2/3 times, no matter what.
Was I wrong to suggest that as the game is played, the probabilities of the two unchosen doors shift from $1/3$ and $1/3$ to $0$ and $2/3$?
Best Answer
$$ P\left({\mbox{Prize is behind chosen door}}\right){}\neq{}P\left({\mbox{Prize is behind chosen door}\,|\,\mbox{Prize is not behind door with goat}}\right)\,, $$ in general.