The Monty Hall Problem

titan.red

1 pointredxblood3 comments

Doesn't choosing between two choices equal a 50% chance instead of 66.6% chance? You've already eliminated an option.

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!

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

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