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There are 3 doors. Behind one door is a prize. I know which one that is but you do not. You choose a door. I then open one of the two remaining doors that does not contain the prize, and I offer to allow you to change your choice to the remaining door. The question is; is it better to stick with your original choice, to switch to the remaining one, or does it not make any difference?
Since there are 2 doors left, shouldn't it be 50%/50%, so it should not matter if you switch?
You are twice as likely to win if you switch.
Your initial choice stood a 1 in 3 chance (33.33%) of being right. The fact that I have opened a door does not change that. The two doors you did not choose stood a combined chance of 2 in 3 (66.66%) of having the prize. That has also not changed. Since we know the open door has a 0 in 3 chance of winning, that means the remaining door has inherited a 2 in 3 chance of having the prize; twice the chance of your original choice.
The usual mistake is to assume that since I opened a random door, the two remaining doors stand an equal chance of winning, and the initial choice has no bearing on the end result. However, I did not open a random door. 2 times out of 3, your initial choice was wrong, and as a result, I was left with only one door that I could open. Your initial choice forced me to open the only alternative door, meaning that your initial choice does influence what is available for the second choice. On those times, switching will win.
If you still need proof, that is what this demonstration is for. If you need more explanation, see Wikipedia.
There are 3 doors. Behind one door is a prize. I know which one that is but you do not. You and another player choose a door (not the same door as each other). I then open one of the doors you chose that does not contain the prize, and eliminate that player. I now offer to allow the remaining player to change their choice to the remaining door. The question is; is it better to stick with their original choice, to switch to the remaining one, or does it not make any difference?
They are twice as likely to win if they stick.
Once you have understood the original Monty Hall problem, it can be confusing that the answer to this one is different. However, the principle is the same. Once again I was only given two doors to open. However, this time, those doors have already been chosen by the players, meaning that there is a 1 in 3 chance that neither player will win. One player is eliminated, and the door is opened, meaning that the remaining player now has the 2 in 3 chance of winning.
2 out of 3 times, I am forced to choose one specific player to be eliminated, as the other player has chosen the winning door. It is those times that the player will win if they stick.
If you need more explanation, see Wikipedia.
The Monty Hall problem, 28 October 2006.
Version 1.1.2, 6 September 2007.