# SoullessAutomaton comments on The Monty Maul Problem

• One way to look at ver­sions of this prob­lem that in­volve the host “ran­domly” open­ing doors, none of which hap­pen to con­tain the car, is that this is im­plic­itly elimi­nat­ing prob­a­bil­ity mass by dis­card­ing cases where one of the ran­domly opened doors does con­tain the car.

In the 3 door case, if you pick door 1, there are three pos­si­ble lo­ca­tions for the car (C1, C2, C3). The host then ran­domly opens one of the two re­main­ing doors (H2, H3). Each of the six pos­si­ble com­bi­na­tions is equally likely, but C2+H2 and C3+H3 are im­plic­itly elimi­nated be­cause in those com­bi­na­tions, the car is re­vealed and the game ends.

The re­main­ing four cases are C1+H2, C1+H3, C2+H3, and C3+H2. Ob­vi­ously in this case your odds are 5050 re­gard­less. More sig­nifi­cantly, your over­all odds re­main 13, be­cause of the elimi­nated cases. The same logic ap­plies to cases with more doors, where all com­bi­na­tions of C1+H? re­main un­der con­sid­er­a­tion, while only one com­bi­na­tion in­volv­ing each other car po­si­tion re­mains in play.

This differs from the clas­sic state­ment of the prob­lem be­cause when the host never opens a door with the car, he does not elimi­nate any of the prob­a­bil­ity mass for “you picked the wrong door”, thus con­cen­trat­ing the like­li­hood of “you picked the wrong door” onto a sin­gle door, which gives the usual coun­ter­in­tu­itive re­sult.

But yes, if you know that the host always opens a door and offers to let the player switch, switch­ing will never make you do worse than not switch­ing. If you don’t know this, there is in­suffi­cient in­for­ma­tion to de­cide on a strat­egy.