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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Essentially, the first door you pick is your bet that it is not the correct door. You get to know whether either of the other two doors would be the correct one, so you have 2/3 chance of being correct by choosing the first one to be the one that won’t be the correct door.
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