The scenario is the following: You are on a game show and you are faced with three doors. Behind one of the doors is your dream prize (my very own lightsaber) while behind the other two doors are dud prizes (a homeopathic contraceptive). You are given the opportunity to select one of the doors and attempt to win your dream prize. After you select a door but before revealing the location of the prize, the host reveals to you one of the doors that did NOT contain the prize, leaving the door you selected and one other. The host then asks you whether you wish to stick with the door you originally selected or change your decision and choose the other remaining door.
This problem is referred to as the 'Monty Hall' problem and the question is whether a) you should stick with your original decision, b) change your decision and selected the other remaining door or c) it does not make a difference either way.
So the answer is that if you wish to increase your chances of winning your dream prize you proceed with option b). That is you should abandon the door you originally chose and select the other door. By doing this you increase your chances of winning from 33% to 66%.
There are a number of places where people can go wrong with this puzzle, however in my experience the most common response is c). That it makes no difference whether you switch or stay. This response suggests that there is an equal probability of the prize being behind each door or that you have a 50% chance of winning regardless of the door you choose. This is in fact incorrect for reasons I'll now attempt to explain.
Hopefully you can all see that to begin with, the probability of guessing the door which contains the prizes is 33% or 1 in 3. This also means that there is a 66% chance that the door you chose does not contain the prize. That is, there is a 66% chance that the prize resides behind one of the other two doors.
When the host reveals one of the doors that does not contain the prize the chance of your door containing the prize does not change, it remains at 33%. Importantly though we know that the probability that the prize is behind one of the doors is 100%. So given there are two doors remaining and the probability of your door containing the prize is 33%, the probability of the other door containing the prize must be 66%. There was a 66% chance that one of the other two doors contained the prize. Now however we know that one of those doors does not contain the prize, therefore the other door must account for the remaining 66%.
If you're not yet convinced by my solution or you're having trouble getting your head around it, imagine the exact same scenario only instead of 3 doors to guess from there are now 100 doors. Your chance of guessing the correct door is now 1% or 1 in 100. So you guess a door and then I take away 98 of the doors that did not contain the prize leaving just 2 doors. The one you chose and one other. I now ask you whether you'd like to stick with your original choice or switch doors, what do you do? You switch obviously because you know that the chance of getting it right the first time was 1 in 100 (1%). The probability in this case that the other door has the prize is 99%.
No one emailed me the correct answer which means I hold onto the used handkerchief. The good news though is that this means the prize pool doubles for the next time I post a problem. So check back in if you're eager to win the handkerchief plus an additional prize of equal or lesser value.
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