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Post Closed as "Not suitable for this site" by Xi'an, Shawn Hemelstrand, utobi
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edit approved Oct 8 at 5:33
Graham Bornholt
edit approved Oct 8 at 5:33
Graham Bornholt
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User1865345
User1865345
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I saw this solution someone gave to the Monty Hall Problem and was wondering if the notation was wrong:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car. Then if the contestant switches, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 2 3$.
If they don't, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 1 3$.

I know this is problematic and this is my attempt to fix it:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car, and $S$ the event that the contestant changes their mind after being shown another door. Then if the contestant switches, $P(W|S) = P(C \cap W|S) + P(C^c \cap W|S) = \frac 2 3$.
If they don't, $P(W|S^c) = P(C \cap W|S^c) + P(C^c \cap W|S^c) = \frac 1 3$.

Does this make sense? Is there a way to justify the solution I saw?

Thank you!

I saw this solution someone gave to the Monty Hall Problem and was wondering if the notation was wrong:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car. Then if the contestant switches, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 2 3$.
If they don't, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 1 3$.

I know this is problematic and this is my attempt to fix it:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car, and $S$ the event that the contestant changes their mind after being shown another door. Then if the contestant switches, $P(W|S) = P(C \cap W|S) + P(C^c \cap W|S) = \frac 2 3$.
If they don't, $P(W|S^c) = P(C \cap W|S^c) + P(C^c \cap W|S^c) = \frac 1 3$.

Does this make sense? Is there a way to justify the solution I saw?

Thank you!

I saw this solution someone gave to the Monty Hall Problem and was wondering if the notation was wrong:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car. Then if the contestant switches, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 2 3$.
If they don't, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 1 3$.

I know this is problematic and this is my attempt to fix it:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car, and $S$ the event that the contestant changes their mind after being shown another door. Then if the contestant switches, $P(W|S) = P(C \cap W|S) + P(C^c \cap W|S) = \frac 2 3$.
If they don't, $P(W|S^c) = P(C \cap W|S^c) + P(C^c \cap W|S^c) = \frac 1 3$.

Does this make sense? Is there a way to justify the solution I saw?

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WinnieXi
WinnieXi
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Notations regarding the Monty Hall Problem

I saw this solution someone gave to the Monty Hall Problem and was wondering if the notation was wrong:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car. Then if the contestant switches, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 2 3$.
If they don't, $P(W) = P(C \cap W) + P(C^c \cap W) = \frac 1 3$.

I know this is problematic and this is my attempt to fix it:

Let $C$ be the event of making a correct pick in the first place and $W$ the event of winning a car, and $S$ the event that the contestant changes their mind after being shown another door. Then if the contestant switches, $P(W|S) = P(C \cap W|S) + P(C^c \cap W|S) = \frac 2 3$.
If they don't, $P(W|S^c) = P(C \cap W|S^c) + P(C^c \cap W|S^c) = \frac 1 3$.

Does this make sense? Is there a way to justify the solution I saw?

Thank you!