The Monty Hall Problem The problem : You are presented with 3 doors: one of the door is a winning one (we will call it the correct door), the other 2 doors are losing (we will call them the false doors). You are to pick one of the doors. Then one door that is false and that you didnβt pick gets revealed. You get to choose whether to stick with the door you previously picked or to switch to the other unrevealed door. The q uestion is: Is it advantageous to switch doors, to sti ck with the first one , or does it not matter? In the solution I will be showcasing 3 different approaches to this problem, all of which lead to the same conclusion. Approaches: 1) After choosing a door, every other door (except for one) that is false gets revealed 2) After choosing a door, one other door that is false gets revealed 3) After choosing a door, one third of the total amount of doors other doors that are false get revealed Used format: π ... the total amount of doors π ( π 1 ) ... probability that the door picked at first is the correct door π ( π 2 ) ... probability that the door to which you switch from firstly picked door is the correct door π· = π ( π 2 ) π ( π 1 ) ... ratio of the two probabilities π· < 1 β it β s better to not switch the door π· > 1 β it β s better to switch π· = 1 β it doesnβt matter whether you switch or not First approach X is a whole number greater than two π ( π 1 ) is calculated as picking 1 favored outcome out of π outcomes: π ( π 1 ) = 1 π Since every door except for the one picked at first other random door was revealed to be false, one of the two not revealed doors must be the correct door, meaning: π ( π 1 ) + π ( π 2 ) = 1 β π ( π 2 ) = 1 β π ( π 1 ) = 1 β 1 π = π π β 1 π = π β 1 π π· = π ( π 2 ) π ( π 1 ) = π β 1 π 1 π = π β π β 1 π = π β 1 Graph of π· = π ( π ) lim π β β ( π· ) = lim π β β ( π β 1 ) = β R esult For 3 doors it is 2 times more likely that youβll win if you switch With increasing amount of total doors the more advantageous itβd be to switch doors (for π = 120 doors π· = 119 ) Second approach π is a whole number greater than two π ( π 1 ) is calculated as picking 1 favored outcome out of π outcomes: π ( π 1 ) = 1 π After choosing a door only one other false door gets revealed, meaning that when switching there is ( π β 1 ) / π probability that one of the doors to which weβre switching is a correct one, and out of these weβre only able to pick 1 out of ( π β 2 ) available do ors left ( unavailab le : one door we're switched from, second was revealed as false) π ( π 2 ) = π β 1 π β 1 π β 2 = π β 1 π β ( π β 2 ) = π β 1 π 2 β 2 π π· = π ( π 2 ) π ( π 1 ) = π β 1 π β ( π β 2 ) 1 π = π β π β 1 π β ( π β 2 ) = π β 1 π β 2 Graph of π· = π ( π ) lim π β β ( π· ) = lim π β β ( π β 1 π β 2 ) = 1 R esult For 3 doors it is 2 times more likely that youβll win if you switch With increasing amount of total doors the less it would matter whether you switched or not, but it would still be favorable to do so (for 1 20 doors π· = 1 , 008 ). Third approach π is a multiple of 3, meaning π = 3 β π ( π = 1 , 2 , 3 , ... ) π ( π 1 ) is calculated as picking 1 favored outcome out of π outcomes: π ( π 1 ) = 1 π After choosing a door, X/3 other false doors get revealed to be false, meaning that when switching, there is ( π β 1 ) π β pro bability that one of the doors to which weβre switching is a correct one, and out of these weβre only able to pick 1 out of ( 2 3 β ) β π β 1 available doors (1 door weβre switching from, 1/3 of doors were revealed as false) π ( π 2 ) = π β 1 π β 1 2 3 β β π β 1 = π β 1 π β ( 2 3 β β π β 1 ) = π β 1 2 3 β β π 2 β π π· = π ( π 2 ) π ( π 1 ) = π β 1 π β ( 2 3 β β π β 1 ) 1 π = π β π β 1 π β ( 2 3 β β π β 1 ) = π β 1 2 3 β β π β 1 Graph of π· = π ( π ) lim π β β ( π· ) = lim π β β ( π β 1 2 3 β β π β 1 ) = 1 , 5 R esult For 3 doors it is 2 times more likely that youβll win if you switch. With increasing amount of total doors itβd still be advantageous to switch doors (for 120 doors π· = 1 , 506 ) Final Conclusion For each of the approaches, it is always 2 times more likely youβll win if you switch to the other door left. Created by bored STEM student tired of a youtube comment discussion