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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Easiest way to think about the problem: Your first choice *constrains* Monty Hall. If you pick the prize (1/3 of the time), he can show you one of two goats. Switching always loses. If you pick a goat (2/3 of the time), he *has* to show you the other goat. Switching always wins.
It’s that simple.
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