There are three doors from which to choose, only one of which holds a nice price. The odds are 1 in 3 that you will pick the right door. You pick, say, door #1.

Monty Hall, who knows what is behind all three doors then opens up, say, door #2 to reveal something wicked and disgusting, like a politician. As the politician is escorted out to their limousine by lobbyists, that leaves you with your original choice of door #1 and the other door you didn't choose, door #3. Monty asks if you would like to switch. Should you?

Intuitively the answer is it doesn't make any difference. The odds were equally 1 in 3 of choosing the right door, so you have just as much chance of the prize being behind door #1 as you do for door #3, right?

Apparently not. In the book I am now reading, the author says the odds are better if you do switch to door #3, and here is the explanation:

Let's say that there are 100 doors to choose from. You pick, say, door #1. The odds of your having chosen the prize door are 1 in 100 (no matter which door you picked, the odds would be the same).

Now Monty, who knows which door has the prize, starts up his bit of showmanship and begins to reveal what is behind the other doors. Naturally to keep the suspense going, if you chose the wrong door he isn't going to reveal this right away and ruin the game. Instead he keeps opening doors that he knows contain booby prizes.

Finally, he gets to the point where there are only two doors remaining, the one you chose, door #1, and the last door among the 99 others, let's say door #78. Now the question is, if he were to offer to let you switch your choice from door #1 to door #78, would you do it?

If you had any brains you would switch! Why? Because when you made your initial choice, the odds of you picking the right door were 100-to-1. That means that the odds of the prize being behind one of the other 99 doors was 99 out of 100. There was a 1% chance of you being right, and a 99% chance of you being wrong.

Now that Monty opened all but one of those doors, keeping in mind that he knows which one is the prize door, that means that the odds are much greater that the prize is behind that last remaining door, the one you didn't choose. Right? Because the unchosen door, #78, came from a population of choices that had a 99% chance of containing the prize, whereas the door you chose came from a population that had only a 1% chance of containing the prize.

And so it is with the question/problem I posed to you yesterday, the difference being there are only 3 doors instead of 100 doors. When you made the choice of door #1 for the 3-door problem, you had a 1 in 3 (33%) chance of being right. The chances of the prize being behind the other two doors was twice that, or 66% (actually 66 2/3%). Now you have the opportunity to move from a choice that had a 1/3 probability of being right to one that had a 2/3 probability of being right.

Confused? Well I'll leave it at that. And I promise, no more stuff like this, at least for a while! I thought that was very interesting so I decided to share it with you.

## 1 comment:

haha that was in the curious case of the dog in the nighttime

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