
The Monty Hall Problem - redxblood
https://titan.red/articles/monty-hall-problem
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jdironman
Doesn't choosing between two choices equal a 50% chance instead of 66.6%
chance? You've already eliminated an option.

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Nadya
Let's make the fact there are two goats more explicit: One is a Doe`D` and the
other is a Buck `B`.

If you chose `D` then `B` must be removed leaving you between staying `D` or
choosing `C`.

If you chose `B` then `D` must be removed leaving you between staying `B` or
choosing `C`.

If you chose `C` then `D` or `B` must be removed leaving you between staying
`C` or choosing `D or B`.

Because your original choice was a 33% chance - by removing a goat - it makes
the remaining option a 66% chance of not being a goat. The opened door is
functionally equivalent as _opening both doors that you didn 't choose._

So let's pretend the host doesn't open the door in a separate action. Instead
you're given the choice to open _both doors_ that you didn't pick and if _one
of them_ has the car you win. Would you choose to open your selected door or
both other doors?

Your selection is a 33% chance - but opening both other doors would be a 66%
chance!

~~~
redxblood
That's correct! :) Thanks for the assist.

