The Monty Hall Problem I just dont get it.

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Why does the probability “lock in” and not change if you remove a variable but doubles for the other one?

I read it multiple times and it still doesnt make any logical sense to me.

In: Mathematics

33 Answers

Anonymous 0 Comments

You have three doors: P (prize), E1 (Empty) and E2 (empty again). It’s important to keep track of the difference between E1 and E2. So now we draw up a table of all possible scenarios, and mark it up with the odds of you and monty choosing each option, from which we can calculate the probability for what the third door has inside

|You|Probability|Monty|Probability|Remaining Door|Probability|
|:-|:-|:-|:-|:-|:-|
|P|1/3|E1|1/2|E2|1/6|
|P|1/3|E2|1/2|E1|1/6|
|E1|1/3|E2|1/1|P|1/3|
|E2|1/3|E1|1/1|P|1/3|

There’s four possible scenarios, two of which (with total probability 1/3) are bad, and two of which (total probability 2/3) that are good.

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