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This is the regular Monty Hall Problem where you choose a door and then Monty opens a door with a goat behind it.

Don't understand why the conditional probability is 2/3?

WLOG assume you pick door 1 and C be the event get car and $M_i$ be the event of Monty opening door $i.$

$P(C) = P(C | M_2)P(M_2) + P(C | M_3)P(M_3)$ by LOTP.

Then $P(M_2)=P(M_3)=1/2$ because of these 3 cases (G,G,C) (G,C,G) and (C,G,G). Then to find $P(C|M_2)$ means you can remove the case (G,C,G) and out of the remaining 2 you get a car in only half of them if you switch. How do you get 2/3?

srm26
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  • "conditional" on what? What are $M_2,M_3$? Are you switching your selection or not? – lulu Apr 18 '24 at 15:00
  • $M_i$ is Monty opening door $i.$ – srm26 Apr 18 '24 at 15:01
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    How is you asking this different than the hundreds of other times people have asked for clarification on the monty hall problem? Why should we not close this as a duplicate immediately? – JMoravitz Apr 18 '24 at 15:02
  • Please edit your post to include all the necessary definitions, don't leave them in the comments. And, have you gone through any of the standard computations for Monty Hall? They should answer your question. – lulu Apr 18 '24 at 15:02
  • @JMoravitz All I could find is unconditionally switching to show the probability is 2/3. This is isn't unconditionally switching, it's what's the probability of getting it after Monty opens a specific door. – srm26 Apr 18 '24 at 15:03
  • Your probability of success (whether you switch or not, which isn't clear from your post) is independent of which door Monty opens. – lulu Apr 18 '24 at 15:05
  • "Then to find $P(C|M_2)$ means you can remove the case (G,C,G) and out of the remaining 2 you get a car in only half of them if you switch." You assume those two cases have equal probability when $M_2$ occurs. They don't. If you had put enough detail in your probabilities to actually calculate them rather than hand-waving, you might be able to discover this. – David K Apr 18 '24 at 16:26

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