Here is a trivia quesion I heard recently:
There are three doors. One door has a prize, the other 2 doors have lemons. You choose a door (but don't open it) and there are two doors left. A friend comes along and opens one of the remaining doors revealing a lemon. What is the probability that the remaining unchosen door has the prize? What is the probability that the door you first chose has the prize? Do you stick with the door you first picked or do you pick the remaining door? Demonstrate the answer mathematically.
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Mike GoldPosted Aug 3, 2007, 12:16 AM
The Answer:
That is correct. The probability is 1/n that you chose the correct door and (n-1)/n that the remaining door contains the prize. So for the case of 3 doors:
When you picked at random one of 3 doors, your probability of choosing the correct door was 1/3 and will always be 1/3 for your initial pick.
The probability is also 1/3 for each of the remaining doors. When your friend opens a door containing a lemon, the remaining door now has a probability of 2/3 that it contains the prize.
As unintuitive as this may seem, you actually have a 2X better chance of guessing the prize is behind the remaining door rather than the prize that you initially picked.
ManuelPosted Jul 31, 2007, 2:47 AM
1. The possibility that the other remaining door is that one for a prize is 1/n if all doors can be elimintated, including that one with the prize.
2. If the searched one won't be eliminated the possibility is (n-1)/n.
This is because the chance to choose the right one with first decision is 1/n and to choose a wrong one is (n-1)/n. And causing only wrong doors will be eliminated the chance that the other remaining door is the right one is equal to the chance that you choosed a wrong door.
Mike GoldPosted Jul 31, 2007, 2:14 AM
Jan MontanoPosted Jul 30, 2007, 10:03 PM
1%?
1% because we're not talking about closed or open doors. we're talking about doors. And it doesn't matter whether someone opens it or not and whether we see lemons or not inside the opened door. It will always be 1/x where x = total # of doors.
I can't think of any answers anymore aside from this (^_^)
Mahesh ChandPosted Jul 30, 2007, 8:07 PM
Mike GoldPosted Jul 30, 2007, 3:37 PM
-Mike
Mahesh ChandPosted Jul 30, 2007, 1:12 PM
100% and 50%. I'm just a lucky guy.
Mike GoldPosted Jul 30, 2007, 11:49 AM
There are 100 doors and only one contains a prize. You go ahead and choose one of the 100 doors, but don't open it. Your friend comes along and opens 98 doors all containing lemons. What is the probability that you contain the prize? What is the probability that the remaining door contains the prize?
Jan MontanoPosted Jul 30, 2007, 3:42 AM
The probability that the remaining unchosen door has the prize is 50%. The probability that the door I first chose has the prize now is 50%. It doesn't matter whether I stick with my door or not. Because the two closed doors have the same probability now that one lemon door has been opened.
Unless I'm missing something. I can't think of anything anymore. But I'm still thinking. wahh! hehehe.
*EDIT
Assuming the lemons actually refer to two or more (^_^) Then the probability would be zero, because the prize is a lemon which has been opened already by my friend. and each door has lemons. hehehe.