While working on simulating the Birthday problem, I already knew what I was going to do next: apply the same approach to the Monty Hall problem. This problem is even more bizarre than the birthday problem.

It is based on a classic TV game show. The setup is straightforward: a player is presented with three doors, one hiding a car and the other two hiding goats and then player has to guess the correct door to win the car. The player initially picks one door. Then, the host, who knows what’s behind each door, reveals a goat behind one of the two unchosen doors. At this point, the player is given a choice: stick with their original door or switch to the other unopened door.

At first glance, it might seem like the probability of winning the car is 1/3, regardless of the player's choice. However, if the player switches to the other door, the probability apparently jumps to 2/3. Yes, switching actually doubles your chances of winning. Its also probably better described here

Just like my previous blog post, I've written another script to simulate the Monty Hall problem with N trials. The script creates two players who play the games N times. One player always sticks with their initial door choice, while the other player always switches their choice after the host reveals a goat. The script then plots the average cumulative probabilities of both strategies on a graph. And interestingly the probabilities for the "keep" and "switch" strategies converge to 1/3 (33.33%) and 2/3 (66.66%), respectively, as the sample size increases.