the monty hall saga

please watch brooklyn nine-nine

Can we just applaud b99 for addressing gay sex as casually as it would heterosexual sex because it’s disturbingly rare?


But what’s the answer though?!

You are twice as likely to win if you switch doors.


When you select door 1 there is a 1/3 chance you win and a 2/3 chance you lose. The host, who knows which door is right, will always eliminate an empty door (this only works because the host has more information than you)
So now you get to choose between the 1/3 chance you were right the first time, or the 2/3 chance you were originally wrong.

To think of it another way, the set of doors 2 and 3 have a probability of 2/3. Because tje host eliminates an empty door (making the open door a 0 probability now), that 2/3 chance “collects” in the remaining door. Your original door is still only 1/3 because you made that choice before you had the new information.

when you select your door initially, it DOESN’T MATTER AT ALL
what it boils down to, is a decision between keeping the door you initially picked (with a 50% chance of being the right one at this point) or picking the OTHER door (which ALSO has a 50% chance of being correct)

To put in other terms, when you initially select one door you have a 1/3 chance of it being correct. The host eliminates a door that you did not pick, a door that would have been a losing choice. You now have a choice between two doors. You have received no additional information about these doors. You have a 50/50 shot with either of them. Because probability isn’t affected by prior choices.

But you did initially have a 1/3
The host can only choose between the doors you didnt choose. So when he eliminates all but 1 remaining door, he isnt affecting the probability of your door, only the door he is leaving.

No, once one of the doors is revealed the odds slip from 1/3 to ½ for each unopened door

No. The host can only remove a door ypu didnt choose, so your doors probability isnt affected by the new information. Opening an empty door narrows the “you were originally wrong” choice to just 1 door, but still with the 2/3 probability.

Again, you can actually test this in simulation.


If you follow the algorythm, you win if you chose the wrong door at first. You have a 2/3 chance of doing that.

Like, you assume that you’ve chosen the wrong door and one of the remaining doors is the right one. AND THEN YOU LITERALLY HAVE IT POINTED OUT TO YOU WHICH ONE. Boom. Win.


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