The Monty Hall Problem

The Misconception: Random chance can not be altered.

The Truth: Sometimes, it can.

Here is the Monty Hall Problem:

Suppose you are on a game show, and you’re given the choice of three doors. You get whatever is behind the door you pick.

Behind one door is a car. Behind the other two, goats.

You pick a door. Let’s say it was number one.

Monty Hall knows what’s behind all the doors, and he opens another door instead. Let’s say he picks number three. When the door opens, a goat is revealed.

Monty asks if you would like to change your choice. Two doors remain closed. You originally picked door number one. Nothing has changed behind the doors.

So, here is the question. Would changing your answer improve your chances of getting the car?

The insane answer is – yes.

Changing your answer to door number two improves your odds from about 33 percent, to about 66 percent.

Even though nothing has changed, changing your answer will now double your odds of winning.

You don’t believe me, but that’s because you are not so smart.

Read on for the explanation, try it for yourself here: Link

This is going to fry your brain, so pay attention.

At first, you have a one in three chance of winning, and a two in three chance of losing.

Once the host picks door a door, the original chances don’t change. The chances of the car being behind the door you first picked is 1/3, but the chances of the car being behind either door number two or three is now 2/3.

Yes, I know, this doesn’t seem to make any sense, but that’s because your brain doesn’t compute fractions very well.

There are three possible outcomes if you switch.

1. You pick number one. The car is behind door number one. Monty reveals door number three. It’s a goat. You switch. You lose.
2. You pick door number two. The car is behind door number one. Monty reveals door number three. It’s a goat. You switch. You win.
3. You pick door number three. The car is behind door number one. Monty reveals door number two. It’s a goat. You switch. You win.

See? Two out of three times switching wins.

There are three possible solutions if you don’t switch.

1. You pick number one. The car is behind door number one. Monty reveals door number three. It’s a goat. You don’t switch. You win.
2. You pick door number two. The car is behind door number one. Monty reveals door number three. It’s a goat. You don’t switch. You lose.
3. You pick door number three. The car is behind door number one. Monty reveals door number two. You don’t switch. You lose.

See? If you don’t switch, two out of three times you lose.

Don’t freak out if this didn’t make sense at first. You were just thinking in concrete terms. You thought the odds didn’t change when Monty revealed the goat, but they did.

In a study by Mueser and Granberg in 1999, 0nly 13 percent of 228 people presented with this problem chose to switch.

You see two unopened doors and think the odds are equal. Two doors, two chances.

You think your odds are evenly distributed across all unknowns, but they aren’t.

If you pretend Monty didn’t know there was going to be a goat behind one of the doors, it starts to make sense intuitively. Pretend he said, “You pick a door, and I’ll pick a door. If either one has a car behind it, you win.”

By switching, you get to win even if one of the two doors opened has a goat behind it. By not switching, you have to let the original odds determine your fate.

No matter how smart or educated, most people still don’t switch when they first see this problem. You make choices like this based on intuition, and intuition isn’t good with huge fractional equations, and as you can see, this one is a doozy:

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