
monty hall solution  I don't believe so.
When Monty opens a door, I will be left with two doors,
and so probability of any of the two doors having the $$ is 1/2, isn't it?
Consider a different angle: In the list of cases you had given, if monty chooses door 3, it can happen in (the first) two cases only, and either case has an equal chance, ie., probability of 1/2, isn't it?
Raghuram
Thursday, January 02, 2003
I realized this should be in google as it was noted elsewhere in the discussions.
Found a very detailed explanation in another site. It finally gave this argument:
"Imagine that there were a million doors. Also, after you have chosen your door; Monty opens all but one of the remaining doors, showing you that they are "losers." It's obvious that your first choice is wildly unlikely to have been right. And isn't it obvious that of the other 999,999 doors that you didn't choose, the one that he didn't open is wildly likely to be the one with the prize? "
But let me reverse the question: What is the probability that the door he didn't open has the $$? Or, would you switch if there were only two doors to begin with, and after you chose a door, Monty added 999,999 doors with goats? I think that the confusion is between the logical mind and mathematical mind. No wonder this is a hotly discussed topic.
Raghuram
Thursday, January 02, 2003
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