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Let’s Have Data Fun: the Monty Hall Problem

by | Jul 14, 2020 | Uncategorized | 0 comments

The Monty Hall problem is named after the original host of the game show Let’s Make a Deal, Monty Hall. The problem is this:

Let’s pretend you’re on a game show, and you’re given the choice to pick one of three doors. Behind one door is a car (a winning prize) and behind the other doors are a goat (considered a losing “prize”). You pick a door, say No. 1, which isn’t opened yet. Then the host, who knows what’s behind the doors, opens a door that has a goat, say No. 2. Now they ask you, “Do you  want to pick door No. 3?”. Is it to your advantage to switch your choice?

Instinctively you might think you’re chances of winning a car went from 33% (1 door out of 3) to 50% (1 door out of 2). Which is how this puzzle is often explained in TV, movies, etc. Or they’ll say you’re chances went from 33% to 66%, if you plan to switch doors. Or another scenario where the host opening the door changed your odds of winning the car.

When in fact your odds are always 66%, when you plan to switch doors. The host opening the door doesn’t change your odds, as that is part of the puzzle/game. This can be very counter intuitive and I’ll explain why this is.

You need to think of the big picture. Picking your first door, the host opening a door with a goat/losing “prize”, then you switching is the full picture.

There are 3 different scenarios. For simplicity, let’s pretend door 1 has a goat, door 2 has a goat, and door 3 has the car.

First scenario. You pick door 1, host opens door 2 and reveals a goat. You switch to door 3 and win the car. 😃

Second scenario. You pick door 2, host opens door 1 and reveals a goat. Then you switch to door 3 and win the car. 😃

Third scenario. You pick door 3, host opens door 1 (or door 2) and reveals a goat. Now you switch to door 2 (or door 1) and get the goat/losing “prize”.  😥 

In 2 out of the 3 scenarios you win the car, when you plan to switch. So your odds of winning the car from the start is 66% ( 2 out of 3) and remains 66% throughout the puzzle/game.

Basically your strategy is, at the start pick a door that has a goat, then the host eliminates the other door with a goat and you switch to the door with the car. From the outset you want to pick a door with a goat. 2 of the 3 doors contain a goat. Hence, your odds of winning the car are 66% ( 2 out of 3, or 2/3).

Until next time, continue having data fun.

Looking for more fun with data? You might enjoy some of our other posts!

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