# the Monty Hall Problem, explained with memes

Submitted by succtales_backup in lobby

# surreal wrote at July 16, 2018 at 6:51 AM

are you downvoting your own posts?

Yes

# surreal wrote at July 16, 2018 at 9:43 PM

would you prefer if you couldn't vote yourself up?

# buzz wrote at July 16, 2018 at 2:05 PM

i still dont get it. they video suggests a sort of symetricality to picking options, so if one was to RESELECT the same door, wouldnt that be the same as SELECTING the other door? and both new probabilities would be 1/2?

is there any data processing that shows the 76% I can look at?

# u_1f4a9 wrote at July 16, 2018 at 5:33 PM

Think about the strategy from beginning to end: suppose you decide from the get-go to either 1) never switch or 2) always switch.

1. If you never switch, it doesn't matter if the host does anything. Your chance of winning is one out of three.

2. If you always switch, you will win if your first pick was wrong, and lose if your first pick was right. Suppose you were right initially (1/3 chance). Then you will lose because your strategy switches you off the winning door. Now suppose you were wrong initially (2/3 chance). In this case, the host will open a losing door, and you have a losing door, so the only other door has to be a winner. By switching in this case, you win. All in all, the "always switch" strategy means that you will win whenever you pick a losing door to start with, which will happen 2/3rds of the time.

I think people get hung up on not thinking about the problem in its entirety. If the host just opened a door and you hadn't picked first, then yeah, you've got a 50/50 chance of winning.

# indi wrote at July 17, 2018 at 12:38 PM

Very nice and clear way of looking at it.