what is the quickest, best way to explain the MONTY HALL PROBLEM?

Answer 1

Many of the links above are great.

For most people, the stumbling block is understanding why the probability isn't equal (50-50) for both of the remaining doors. I explain it this way. Your initial guess has a 1/3 chance of being right because you have NO information about any of the doors, so the probabilities are equal at that point. Once a door is opened, you have *some* information about the remaining doors. [One could not have been opened; the other could have been opened but wasn't.] And once you have knowledge about the system, it shouldn't be so surprising that you can achieve a probability better than 50-50.

Once somebody accepts that they have enough information to do better than 50-50, it's usually not hard for them to understand why switching is better than staying.

Answer 2

ok so you have 3 doors with 2 goats and 1 car behind them. you choose a door and before what is revealed behind it the host reveals a goat from one of the other two doors and you have the option to switch to the remaining door or not.

suppose you dont switch. you didnt let anything affect your decision so the odds of winning the car is simply 1/3

suppose you switch. the only way you lose is if you chose the car first. the odds that you choose the car at first is 1/3. if the chances of losing is 1/3, then the chances of winning is 2/3 bc the probabilities must add up to 100%.

conclusion:
switching gives you a better odds of winning the car than not switching.

Answer 3

The probability the car is behind the door you picked is 1/3, so prob it's behind one of the others is 2/3. When Monty eliminates one of those others, then the prob the car is behind the remaining one of those others must be 2/3, since the prob the car's behind your original choice hasn't changed.

Or try P(A) = 1/3, P(B or C) = 2/3.
If P(C) = 0, then since P(B or C) = P(B) + P(C), P(B) = 2/3.

Answer 4

You pick door #1; assume this throughout the entire solution. The host opens a door (other than the door #1 you picked) without a car behind it, for example door #2. Yes, it is true that the probability that the car is behind door #1, given that the car is not behind door #2, is 1/2, but this is *not* the conditional probability that we are looking for! Not only do we know that the car is not behind door #2, we also know that the host opened door #2, which gives additional information since the host could have possible opened door #3 if the car is behind neither of doors #2 and #3. Besides, we cannot conclude that doors #1 and #3 are equally likely to have the car behind them since the host's selection of which door to open was partly based on the door #1 that you picked, which eliminates the symmetry between doors #1 and #3. Furthermore, if the car is behind door #1, then the host could possibly have opened door #3 instead of door #2. On the other hand, if the car is behind door #3, then the host definitely had to open door #2. So the host's opening of door #2 is *evidence* in favor of the car being behind door #3 instead of door #1. I will give two possible explanations of why the conditional probability that the car is behind door #1, given that the host opened door #2, is 1/3 (assuming always that you pick door #1). Explanation #1: Clearly P(car behind door #1) = 1/3. Once you pick door #1, the host must open door #2 or door #3 (but not both doors). Clearly P(car behind door #1 given host opens #2) equals P(car behind door #1 given host opens #3). Obviously the conditional probability that the car is behind door #1 does not depend on which of the two doors the host opens. Furthermore, P(car behind door #1) = 1/3 is a weighted average of the conditional probabilities P(car behind door #1 given host opens #2) and P(car behind door #1 given host opens #3). The weighted average of any number and itself must be that same number. Since these two conditional probabilities are equal, we conclude that they each must equal 1/3. So in particular, P(car behind door #1 given host opens #2) = 1/3. Explanation #2: P(car behind door #1 given host opens #2) = P(car behind door #1 and host opens #2)/P(host opens #2) = P(car behind door #1)P(host opens #2 given car behind door #1) / P(host opens #2) = (1/3)(1/2) / (1/2), because i) the overall probability that the car is behind door #1 is 1/3, ii) the host is equally likely to open door #2 or door #3 given that the car is behind door #1, and 3) the host overall is equally likely to open door #2 or door #3 = 1/3. If you still have doubts, think of 1,000,000 doors instead of 3. You pick door #1, and the host opens 999,998 doors (other than door #1) that do not have the car behind them (say doors #2 to #999,999 for example). Think about this: if door #1 has the car, it is exceedingly unlikely that the one door that the host didn't open (out of 999,999) is door #1,000,000, but if door #1,000,000 instead has the car, then the host definitely had to leave door #1,000,000 unopened. So the host's opening of doors #2 through #999,999 is extremely strong evidence that door #1,000,000 has the car! In fact, the probability that the car is behind door #1 would be 1 in a million. If you still have doubts, do many trials of the Monte Hall experiment! Lord bless you today!

Answer 5

A tree diagram

Edit: http://upload.wikimedia.org/wikipedia/commons/9/9e/Monty_tree.svg

Answer 6

Do you want the answer behind door #1?

Check out this link.

http://en.wikipedia.org/wiki/Monty_Hall_problem

Answer 7

http://math.ucsd.edu/~crypto/Monty/monty.html