After trying unsuccessfully to animate my 3d pop-up card, I decided to try again with a door opening animation. The Monty Hall problem presents you with 3 doors. There is one prize hidden behind the doors. After you pick a door, the game show host opens one of the doors that you did not pick. There is no prize behind the door. You are then asked if you would like to switch doors to stick to your original choice.
First you may be tricked into thinking that both the remaining two doors have a 50/50 percent chance of having the prize. In reality, switching doors at this point highly increases your likelihood of winning! When you picked the first door, your door had a 1/3 chance of having the prize. That means the remaining doors had a 2/3 chance. When one of the other doors was eliminated, your chosen door still has a probability of 1/3, that never changes. This means that the other door has a 2/3 chance of having the prize! The correct choice would be to switch doors.
When I talked to a few friends about this problem, some felt like they should switch without knowing why. Others could not understand why the doors had different probabilities. It was an interesting conversation and we talked about the same scenario except with 100 doors. In that case, if you picked a door and then the game show host opened 98 doors without the prize behind them, wouldn’t you switch the the other remaining door?
Looking into the math behind the Monty Hall problem was actually more fun than writing the code!
Day 50 – Monty Hall