In this post, we'll examine the Monty Hall problem intuitively. First, we'll start with the problem in it's basic formulation:

You're on a game show. Here are the rules of the game: There are three doors, and behind one of them there's a brand new car. Behind the two other doors, there are goats. You'll choose one of the doors, and subsequently, the game show host, who knows where the car is, will open one of the other doors, revealing a goat. Then you are given a choice: Will you stay with your initial choice or will you switch to the other unopened door?

Many answer initially that it doesn't matter. They reason that, as there are only two doors left, one of which include the car and one that does not, the chances that any one of them veils the car is 1/2. Sensible as this may seem, it's incorrect. In fact, the probability that the other door veils the car is 2/3, so you should switch.

How can this be? Of course, when we first picked a door, there was a 1/3 chance that we picked the right one. But when the game show hosts opens another door, doesn't that additional information change the situation? In fact, it doesn't. We don't gain any additional information from his opening of a door, since he'd open a door to reveal a goat whether or not we picked the right door to start with. We knew from the beginning that the probability was 1/3, and we haven't gained any additional information to change things, so we're still looking at a 1/3 probability.

Here's another way of looking at it: 1/3 of times, you'll pick the right door to start with. Staying will grant you the car, and switching gives nothing. But 2/3 of the times, you'll pick the wrong door, and the host will open the other door that also veils a goat. Those times, staying gives nothing, and switching grants you the car. So switching gives you the car in 2/3 of instances.

In the next part of this series, we'll have a look at the mathematics of this game, but for now, let's leave it at this. As a side note, if the host didn't know where the car is, but he still opened a door at random and just happened to reveal a goat, then the chances would, indeed, be 1/2 and you'd be indifferent to switching.

In the alternative formulation that I gave in my first entry, I said that the correct decision is to stay with your initial choice. The formulation was slightly different, with only one significant difference. But what? I've had a few friends asking me about this on MSN, and one of them actually came up with the correct answer by himself. Can anyone figure out what that difference is, and why it changes things so drastically?Please comment.

You're on a game show. Here are the rules of the game: There are three doors, and behind one of them there's a brand new car. Behind the two other doors, there are goats. You'll choose one of the doors, and subsequently, the game show host, who knows where the car is, will open one of the other doors, revealing a goat. Then you are given a choice: Will you stay with your initial choice or will you switch to the other unopened door?

Many answer initially that it doesn't matter. They reason that, as there are only two doors left, one of which include the car and one that does not, the chances that any one of them veils the car is 1/2. Sensible as this may seem, it's incorrect. In fact, the probability that the other door veils the car is 2/3, so you should switch.

How can this be? Of course, when we first picked a door, there was a 1/3 chance that we picked the right one. But when the game show hosts opens another door, doesn't that additional information change the situation? In fact, it doesn't. We don't gain any additional information from his opening of a door, since he'd open a door to reveal a goat whether or not we picked the right door to start with. We knew from the beginning that the probability was 1/3, and we haven't gained any additional information to change things, so we're still looking at a 1/3 probability.

Here's another way of looking at it: 1/3 of times, you'll pick the right door to start with. Staying will grant you the car, and switching gives nothing. But 2/3 of the times, you'll pick the wrong door, and the host will open the other door that also veils a goat. Those times, staying gives nothing, and switching grants you the car. So switching gives you the car in 2/3 of instances.

In the next part of this series, we'll have a look at the mathematics of this game, but for now, let's leave it at this. As a side note, if the host didn't know where the car is, but he still opened a door at random and just happened to reveal a goat, then the chances would, indeed, be 1/2 and you'd be indifferent to switching.

In the alternative formulation that I gave in my first entry, I said that the correct decision is to stay with your initial choice. The formulation was slightly different, with only one significant difference. But what? I've had a few friends asking me about this on MSN, and one of them actually came up with the correct answer by himself. Can anyone figure out what that difference is, and why it changes things so drastically?Please comment.

## 0 comments:

## Post a Comment