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There are three doors. Behind the doors, there are 2 goats and 1 car. But we don't know which thing is present behind which door. Our target is to get the car by guessing. Afterwards, we select a door (let the 1st door). Keeping the selected door closed, another door (let the 2nd door) is open which has a goat.Now, is it beneficial to change the selection so that we can get a car?

I have known about an answer where it is said that it is beneficial to change/switch the answer.It suggested to use probability saying that, the probability of the unselected door (3rd door) increases after opening the 2nd door.But the probability of the selected door (1st door) remains unchanged. But how does it happen?

Nehal Samee
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    The non-switcher will win if his original choice was correct. Probability on that: $\frac13$. The switcher will win if his original choice was incorrect. Probability on that: $\frac23$. This works under condition that the opened door is opened by someone who deliberately opens a door behind which there is no car. – drhab Feb 03 '18 at 09:50
  • But why does the "deliberate opening" causes the probability of the unselected door to increase? – Nehal Samee Feb 03 '18 at 09:53

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It's because the second door is defined posterior to knowing which one contains the goat, unlike the first door which is fixed.

Akababa
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  • Does it mean that if you define something after discarding any outcome , then its probability of occurring increases? – Nehal Samee Feb 03 '18 at 09:52
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    Yeah, because if I tell you the second door is not losing then it makes it more likely to be winning. – Akababa Feb 03 '18 at 09:54