From Chapter 1, The Monty Hall Problem
This example is subtle, and the answer is often incorrectly guessed by people who should know better. There are sill ongoing debates about this puzzle on various Probability-Puzzle websites because the correct answer seems to be so counter-intuitive to some people that they just won't accept the analysis. It is know by mane names, perhaps most commonly the “Monty Hall” problem, named after the host of a TV game show.
You are a participant in a TV game show. There are three doors (let's call them doors A, B and C). Behind one of these doors is a substantial prize, behind the other two doors is nothing. You have to take a guess. So far this is very straightforward: Your probability of guessing correctly and winning the prize is exactly 1/3.
You take (and announce) your guess. Before the three doors are opened to reveal the location of the prize, the game show host goes to one of the two door that you didn't choose, opens it, and shows you that the prize is not behind this door. The prize, therefore, must either be behind the unopened door that you chose or behind the unopened door that you did not choose. You are now given the option of staying with your original choice or switching to the unopened door that you did not choose. What should you do?
Almost everybody's first response to this puzzle is to shrug – after all, the are now two unopened doors and the prize is behind one of them. Shouldn't there simply be a 0.5 (50%) probability of the prize being behind either of these doors, and therefore it doesn't matter whether you stay with your original choice or switch?
Let's look at all the possible scenarios. Assume that your first guess is door B. (It doesn't matter which door you guess first, the answer always comes out the same).
- If the prize is behind door A, the the host must tell you that the prize is not behind door C.
- If the prize is behind door B, the the host can tell you either that the prize is not behind door A or that the prize is not behind door C.
- If the prize is behind door C, the host must tell you that the prize is not behind door A.
Since each of the above three situations is equally likely, they each have a probability of 1/3.
In situation 1, if you stay with your first choice (door B), you lose. You have the option of switching to door A. If you switch to door A, you win.
In situation 2, if you stay with your first choice you win. If you switch, therefore, your lose.
In situation 3, if you stay with your first choice, you lose. If you switch to door C, you win.
At this point, a table is in order. Remember, your first choice was door B.
|B||A, C||A & C||Win||Lose|
It appears that if you stay with your first choice, you only win in one of three equally likely situations, and therefore you probability of winning is exactly 1/3. This shouldn't really surprise you. The probability of correct guessing one door out of three is 1/3, and there's not much more that you can say about it.
On the other hand, if your only options are to stay with your first choice or to switch to the other unopened door, then your probability of winning if you switch must be 1 – 1/3 = 2/3. There's no getting around this: Either you win or you lose, and the probability of winning plus the probability of losing must add up to the certain event – that is, to a probability of one.
What just happened? What has happened that's different from having just flipped a coin five times, having gotten five heads, and wondering about the sixth flip?
In the coin flipping example, neither the probabilities of the different possibilities or your knowledge of these probabilities changed after five coin flips. In other words, you neither changed the situation nor learned more about the situation... In this game show example, your knowledge of the probabilities changed; you learned that the probability of the prize being behind one specific door was zero. This is enough, however, to make it possible that the expected results of different actions on your part will also change.