I really only understood the Monty Hall Problem intuitively when I sat down to write a simulation of it to convince someone else that the math (which had alread

I really only understood the Monty Hall Problem intuitively when I sat down to write a simulation of it to convince someone else that the math (which had already convinced me) was actually correct. Here's Julia code for it: function monty_hall(switch::Bool; n=3) winner, guess = rand(1:n), rand(1:n) reveal = setdiff(1:n,[winner,guess])[rand(1:end)] if switch guess = setdiff(1:n,[guess,reveal])[rand(1:end)] end return guess == winner end This allows you to simulate the switch and don't switch strategies: julia> mean([ monty_hall(false) for _=1:100000 ]) 0.33166 julia> mean([ monty_hall(true) for _=1:100000 ]) 0.6676 But more importantly, you can see that in the don't-switch case, the reveal choice is dead code – it has no effect on wether the original guess was right or not. That's the key insight: since the original guess had 1/3 chance of being right before the reveal, it still has a 1/3 change of being right after the reveal. Before the reveal the other two choices also each had a 1/3 chance of being winners, but after the reveal their combined 2/3 chance is concentrated entirely on the door that wasn't revealed, making that the best option. The code also lets us see what happens with a generalized Monty Hall problem for more doors: julia> mean([ monty_hall(false, n=4) for _=1:100000 ]) 0.24862 julia> mean([ monty_hall(true, n=4) for _=1:100000 ]) 0.37462 julia> mean([ monty_hall(false, n=5) for _=1:100000 ]) 0.19763 julia> mean([ monty_hall(true, n=5) for _=1:100000 ]) 0.2666 Even for higher numbers of doors, it's better to switch.