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alexfierroislife:

alexfierroislife:

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the monty hall saga

please watch brooklyn nine-nine

[Audio transcript below the read more.]

Keep reading

…but its a 50% chance

Its not. Its 2/3 if you switch. But its complicated.
Think of it like this: you pick door 1. There is a 1/3 chance door 1 wins. There is a 2/3 chance that doors 2 OR 3 win. The host opens a door he KNOWS is empty. But that set (2 or 3) is still 2/3 chance, but is now only 1 door. By switching doors, you select the choice with the higher chance. The key is that the host will only open an empty door. He has more info than you.

Also this show is great

Why does that change anything though?

Because we have new information.
Ok, so, which has a higher chance: door 1 or doors 2 AND 3? Doors 2 AND 3 have a 2/3 chance combined. When the host opens door 2, door 2 now has a 0 chance, while door 3 has a 2/3 probability now (because the set of 2 and 3 is still 2/3, but we know 2 is 0%) . Door 1 still has only a 1/3 chance. The host will only open an empty door. That’s key.

https://i.stack.imgur.com/AO5z6.png

im in mobile, but that link has a visual that shows this more clearly.

I honestly don’t understand though. if its a 1/3 chance in each of the three doors. but then one door doesn’t have it. Doesn’t that just put it at a ½ chance in each of the remaining doors? Which means regardless of which door you picked, there’s a 50% chance.

Like even if the set of 2 and 3 remains at a 2/3 chance, that doesn’t mean that door 2’s probability changes from 1/3 to 2/3.
Because if door 3 doesnt’ have it, then we can’t change its chance to zero without discounting it completely and switching to a ½ chance. Because the idea of a chance being in 3 is when we have 3 doors that haven’t been opened. Whether one door has it or not doesn’t influence the probability of the other doors having it (when not discluding the opened door), because the car is already set behind one door.

Or to phrase it maybe better. Door 3’s probability wouldnt become 0 just because it doesnt have the car. The probability being x/3 is based on the state of the doors being unopened, and the chance of the car being there would remain 1/3 whether its there or not.

Door 3s probability goes to 0 because we opened the door and saw it had no car in it. We have more information now, and that information (and knowing that the host knows even more than us) is what causes the switch.

The combined probability of “doors you didnt choose” is still 2/3, but you know door 3 is empty. You are choosing between the 1/3 door, or the 2/3 “doors you didnt pick”

Imagine if there were 10 doors. You pick door 1 (1/10 chance). The host then opens door 2, 3, 4, 5, 6, leaves door 7, open door 8, 9, and 10. Because the host KNOWS Which doors are empty, their knowledge affects what doors get opened and impacts the probability.

MY TIME HAS COME

Here’s a simple explanation I totally didn’t spend way too many hours coming up with!

This tactic works if you always switch after you get the second door opened. Because if you do that, it means that initially you aren’t banking on picking the RIGHT door, you are banking on picking the WRONG one. And the probability of THAT is 2/3 (because two out of three doors are wrong ones). And if you pick that first door right (that is, it’s the empty door), then the third door (the one you didn’t pick and the one that wasn’t opened) is the winning one. You only lose if you guessed wrong initially and the totally random door you picked was the winning door. The chance of which is just 1/3. Bam!

Here’s a visual illustration:

Note how it’s the two guys who picked the WRONG door on their first guess who win!

(and, y’know, how there’s two of them)

@alexfierroislife

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