The Monty Hall problem

A simple game. There is a game master and a player, you!

• There are n doors. Let’s take n = 3.
• Behind one door there is a price, only the game master knows where it is.
• You must pick a door.
• The game master will open one door. He has two rules:
  1. do not open the winning door
  2. do not open the door picked by the player
• When n > 3 the game master will continue opening doors, given the rules, untill there are 2 doors left.
• Now you have a choice again and this is the main question of the game: do you want to switch? Or: will switching give you a better change of winning?

Analysis

How does it work? How do we solve it?

• Let’s take 3 doors.
• ‘P’ means the door with price and ‘Y’ means that you picked that door.

This are all the possible scenarios.

Now the game master must open a door.

When we check each scenario in the table above, we see there are only 2 options. Option 1 is you picked the winning door and option 2 is you did not choose the winning door.

When we count both options we get this table.

So now it is trivial to see that switching gives you a 2/3 change of winning.

What!?

For me this is really counterintuitive. When you take a large number for n, switching gives you a near 100% change of winning the price! Because I couldn’t believe my own analysis I decided to make a small C# console program to validate this result and to play a little with different game master strategies.

See the code on Github.