The Monty Hall Problem (Cult of the Lamb Animatic) 

Also subtitles looking good man!

​@@ghoulish_GuyGHOUL SPOTTED Hell yea!! Ty fam glad the sub works

JAIL. NOW.

in the words of Raman halt "BBBBBBBBBBBBBBOOOOOOOOOOOOOONNNNNNNNNNEEEEEEEEEEEEEEE"

​@@ghoulish_Guy i trust everyone with a Ralsei pfp, there always so wholsom-

@@MxIbatomik red crown is that you

oh… oh, Oh OH *OH* God! NO. Why!?!

​​​​@@justareguralcitizen9492*YYYEEESSSS! YYYYYYEEEEESSSSSS!*

He’s also wearing Narinder’s cloak at end too 😅

She doesn't?

It got PTSD form that night!

Alone at the edge of a universe humming a tune For merely dreaming we were snow

OHhhhhhhhh thank you cause honestly I wanted to know the math

the crux is that your choice isnt retroactively made into a better choice by knowing one of the answers is wrong. if you had a one in a million chance of getting it right, then you havent been retroactively been made into a genius that picked a door with a 50/50 chance of being correct, its still the same door. youre effectively changing the question, its not "do you want to switch" youre betting on the fact that your first guess was probably wrong

That’s correct but Another way of looking at it (probably the less straightforward way but eh) is as follows. Let’s say you got it right the first time, which happens 1/3 of the time. The host opens a door, then tells you you can open the other door. given that you’ve already picked the right door, the other two doors are the same. In this 1/3 of the time, switching will make you lose. Now alternatively, you got it wrong the first time, which happens 2/3 of the time. Then the host rules out the other wrong door, and lets you switch to the correct door. In this scenario which occurs 2/3 of the time, if you switch, you get it right. 1/3 of the time, you’re in a scenario where switching is wrong, and 2/3 of the time you’re in a scenario where switching is right. It has nothing to do with the current choice between 2 doors, as is commonly and intuitively assumed, but instead to do with the probability you already got it right. As the comment before me said you’re banking on your first guess being wrong

@@Lylybeebee wouldn't it be 50/50? No matter your first choice, one of the wrong ones will always be revealed. Your new choice is like choosing between a completely unrelated 50/50. Of the three doors, let's say 2 is correct. If you pick door 1, door 3 will be opened. Door 3 is completely redundant as you didn't pick it originally and it was revealed before you could switch to it. Same problem for 3, just swap 3 n 1. If you pick 2, either 1 or 3 will be opened. Doesn't matter, you can't (or at least, shouldn't) pick the opened one. The door opened is never your original pick or the right one, so that door will always be redundant (I think, I'd love to be proven wrong, this is making my 'tisms fire)

@@billgatestrappedinthe7year909 but that’s the thing, your original choice does matter. If your original choice was wrong, then it becomes right to switch, if your original choice was right then it becomes wrong to switch. Your original choice being wrong happens 2/3 of the time, and your original choice being right happens 1/3 of the time. This means that in the switch, 2/3 of the time switching gets it right, while 1/3 of the time switching gets it wrong. That is to say, the odds of getting it right by switching is 2/3

Okay, thank you! I’ve heard of this before (honestly it might’ve been on game theory lol) but I don’t think they actually explained how it’s a 2/3 chance if you swap, just that it is.

Statistics says that, but I've seen deal or no deal and it hardly works our for the person :p

So... who said that the chance resets after elimination of one of the doors? And shouldn't it be a variant case of Schroedinger's box anyway if you talk about probability determination?

@@TheArklyte you have a 2/3 chance to win if you choose both the two other doors, and if one of the doors is eliminated, the other one is still 2/3

@@ThatOne6uy that doesn't work as the probability value of ALL doors resets and chance value is determined only when YOUR door opens.

You explained it simply. But also wrong. If the host is guaranteed to remove an incorrect door, then one of those 3 doors was never an option for you to begin with. It doesn't matter if it's 3 or 300, if the host knowingly eliminates a wrong door every time you choose to keep or switch then it will always boil down to the last 2 doors. Thus, before you even pick a door, the game is 50/50.

​@@vigorouslethargy It's not a math problem. It's psychology. There are two "groups" of doors. Winning and losing. Reducing the number of doors to two results in a choice of "Keep your initial group, or take the other group." So if the odds were against your previous choice being the winning option, you should switch groups. The actual doors don't matter, only which "Group" you are most likely to currently be in.

@@KefkeWren But the winning group and the losing group have equal odds as soon as one of the losing doors is removed from play. It's only psychology for people who don't understand very simple probability. If there were 10 doors, you have a 10% chance of guessing it right the first shot. With every losing door removed, the probability of each remaing door being a winner, including the one you chose, increases. After it dwindles to 5 doors, they all have a 20% chance of being correct. Try it yourself. Have someone put a quarter under one of 10 random cups, go through the motions, and at the very end when there's only 2 cups left switch your choice. According to your description, you should have a 90% chance to win by swapping. Repeat enough times and you'll see your actual win rate will be around 50%.

@@vigorouslethargy You didn't choose when they had equal odds. You chose when they had different odds, and now you're being offered the chance to take what you didn't pick then.

@@vigorouslethargy There's not three outcomes, so don't think of it as choosing between three options. Think of it as flipping a coin that's been weighted on one side, then being asked before you've seen the result if you want to turn it over or reveal it as it lies. Even if the weight gets taken off, the coin is already flipped. It doesn't get flipped again.