Talk:Monty Hall problem/Arguments/Archive 8: Difference between revisions

Revision as of 23:50, 7 July 2017

Revision as of 17:35, 1 May 2022

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:* N = the number of doors at the outset,

:* P<sub><small>Contestant</subsmall></smallsub> = the contestant's probability of choosing right the first time

:* P<sub><small>Monty</subsmall></smallsub> = Monty's probability of choosing the right door, i.e. of leaving closed the door with the car behind it

:Then it's clear that

:* N = 100

:* P<sub><small>Contestant</subsmall></smallsub> = 1 / N = 1/100 = .01 = 1%

:* P<sub><small>Monty</subsmall></smallsub> = 1 - ( 1/N ) = .99 = 99%

:* P<sub><small>Monty</subsmall></smallsub> = 1 - P<sub><small>Contestant</subsmall></smallsub>

:This last statement or formula isn't strictly necessary to solve the problem, of course, but it might help you understand it better: It says that Monty's probability of choosing correctly is 1 minus the contestant's probability of choosing correctly, because the contestant might have got lucky and picked the correct door before he could, thus eliminating Monty's ability to do so.