In the three door game why do you first have a 33 percent chance of guessing the correct door but after the host opens one of the doors do you have a 66 percent not a 50 percent chance?

Answer 1

This is a very famous problem in probability and the answer goes against the intuition of most people including mathematicians. So picture 3 doors and a prize behind 1 of the doors. Let's say it is the WikiAnswers gold badge, worth lots of time and work Now since there are 3 doors and 1 prize, the chance of picking the gold badge is 1/3 or 33%. So I could predict that one time out of three, you pick the door with the prize. Now here is where it gets fun. In any probability problem, including this, the sum of the probabilities must equal 1 or 100%. If we pick one door the chance of finding the gold badge, is 1/3 as we said. If we look at all 3 doors, the chance of finding the prize is 1 or 100% since we know the badge is behind one of them and if we look at 2 doors, the chance of finding the prize is 2/3. So now you pick a door. There are only two choices. Either you picked the winning door or you did not. No other choices! Now Mathdoc, ok it was really Monty Hall, opens up one of the doors and reveals that there is no prize there. Monty has not just eliminated one of the three doors, he has eliminated one of the doors with a prize! This detail is important. Now say your initial choice was correct, you picked the door with the gold WikiAnswers badge. Now since you picked the right one, the other two are wrong ones. Now Monty opens one of them, does not really matter which. So if you stay with your first choice you are right 1/3 of the time. Now, let's say you picked wrong to start with. The door you pick does NOT have the gold badge. That means 2/3 times your initial choice is wrong as we said above. We also explained above that if we look at 2 doors, 2/3 times we can pick the prize. So now we have the prize behind one of those 2 doors. So since you picked the wrong door to start with this means that 2/3 of the time the prize is behind one of other two doors. So looking at those two doors that you did not pick as one probability unit, there is a 2/3 chance of picking the prize, but Monty is kind and has shown you where it is not which means it must be behind the other door. Remember there are only two doors you did not pick and one of the two has the gold and Monty showed you the one that does not so the other one does! Now think of the remaining door as inheriting the probability that the two doors had. That is to say, the remaining door now has a 2/3 probability of having the prize. Your initial chance of being right is 1/3 and the door remaining has 2/3 chance of being right.. Should you switch? Hmm I think so! So why is it not 50% you asked? That is a good question and important to understand in order to grasp the problem. The idea is that Monty shows you a losing door, so one door is right and one is wrong; so there should be a 50 % chance of picking the correct door. It seems that it does not matter if you switch or not. Here is why the intuition fails. Monty always has to open a losing door. So one losing door is always eliminated. Because this is a probability problem, the probability of your initial choice being correct and your remaining choices must equal 1 or 100%. That means if your initial choice being correct probability is 1/3, the probability of the other two being correct is 2/3 ( 1-1/3) If a door is not opened then there are two choices to switch to, it is kind of like just changing your mind. But this is not what happens. If it did you would just multiply 1/2x 2/3. Since Monty opens one door, you only have 1 to pick from, so instead of your odds being 1/2 you now need to multiply by 1. (you can pick form two doors, 1/2, you pick from one door 1 choice) So you must multiply 2/3 x1 NOT 1/2. That gives us a probability of 2/3 or 66% So it is always better to switch! To review in the more classic Monty method. The car is behind door C, so A and B have a goat. You get a car if you win and a goat if you lose. Say you pick the door with the car. You are then shown either door A or B with the goat. If you change you lose, if you stay you win. Now say you pick door A. Now I show you door B with a goat. If you switch you win, if you stay you lose. Third possible scenario, you pick door B and you are shown the goat behind door A. Same thing, if you change doors you win if you don't you lose. Of course each of the three options above has a 1/3 chance of happening. Since you are equally likely of picking any of the three. Now in the above scenarios, you win 2/3 times if you change. In 1/3 scenarios you win if you don't change. So when you switch 2/3 times you win the car that is 2/3 or 66%. YES THIS IS NOT OBVIOUS, that is why so many people became goat herders on the show.