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The famous "let's make a deal" riddle from Marilyn vos Savant's column "Ask Marilyn," got a lot of media attention when it appeared years ago. It became one of the most widely-debated probability problems of the past decade. Why? Because so many logicians and others thought she was wrong. They even wrote nasty letters to editors of newspapers. In the end, they wrote apologies - Marilyn was right all along.
In the original riddle, you are to suppose that you are a contestant on the old television game show "Let's Make A Deal." The emcee, Monty Hall, has told you that behind one of the large doors an the stage is a new car. The other doors have either a small gift or a goat. You get to choose a door, and keep whatever is behind it.
After you choose, Monty reveals what is behind one of the doors you didn't choose. It isn't the car. The car, then, is behind the door you chose or the other one. Mr. Hall gives you the option of switching to the other one if you wish. The question, then, is whether you should switch or not, or doesn't it make any difference (assuming you want the car more than the goat or small gift).
Note: Perhaps one reason people had difficulty with this problem is that they weren't familiar with the show. Monty Hall never opened the door with the car after a contestant made her initial choice. In other words, if the car was behind one of the two doors, he would open the other one, and then give you the choice of switching. Most people would assume this, even if they hadn't seen the show, but it is worth pointing out.
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To increase the probability that you'll get the car, you should switch doors. This seems counter-intuitive to some, because it seems that no matter what Monty has revealed, you are left with two doors and no way to know which one has the car. The probability would seem to be 50% for either. It isn't.
Think of it this way: If you went through this routine 300 times, you would choose a door without the car about 200 of those times, right? (2 out of three doors have no car, so 200 out of 300 times - on average - you would choose a door without the car) Now, each of those times Monty will eliminated the other non-car door for you, meaning if you switch you will get the car, right?
Of course the other 100 times you will have chosen the right door from the start, and will lose the car when you switch. However, winning 200 out of 300 times is better than 50%, right? With a probability of 2/3, or 66.7%, you should always switch.
There are other ways in which people have explained this. To me, the interesting thing about the "let's make a deal," riddle, is that it is an example of the relevance of knowing how to figure probabilities. This is how the game show was really played, after all. I wonder what the statistics are on how many contestants switched or stayed with their first choice?