Monty Hall question, but varied. I haven’t seen the option that no matter what you have a 40% chance overall to win. You always pick 1 of 3 doors, then 1 is discarded & you pick 1 of 2. So don’t you always have a 2/5 chance of being right no matter what?
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I can see why you’re saying 2/5, but simply put that’s just not how probability works. Along with the other explanations here it might help to look at this demonstrated in a diagram. [This](https://brilliant-staff-media.s3-us-west-2.amazonaws.com/tiffany-wang/zWt5C1YlGm.png) is the best one I can see (from [this explanation](https://brilliant.org/wiki/monty-hall-problem/)).
As you can see from that, there are six possible combinations of choice & outcome: 3 where you stick and 3 where you change.
(One thing that’s hard to get your head around here is that probability depends on our *knowledge*. The location of the car doesn’t change, but opening a door changes our knowledge and our assessment of the probability.)
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