Understanding "Variable Change", The Monty Hall Problem
Uploader Comments (telecoustic)
All Comments (73)
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if anyone dosnt get it still watch this vide /watch?v=mhlc7peGlGg
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Good Explanation and I think the best way to think about it is to look at that chart with all outcomes really close and accept that you are wrong 66.7 % of the time when you start out.
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Great explanation
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Nice video. One thing I'm curious about is if the phrase "Variable Change" is a real term, and if so, what does it mean? I understand the Monty Hall problem and why switching wins 2/3 of the time. i understand conditional probability. But outside of the movie 21, I've never heard of the specific phrase "Variable Change" before. I googled it, and only found references to the movie 21. I wonder if they just made it up.
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It makes total sense...the chances of you picking the car on the first pick is 1/3 so it's 2/3 chance you'll get it wrong so the smarter choice would be to switch seeing that more than likely your initial pick is wrong...
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it's not about the probability changing based on switching or not, it's about the probability of you picking a goat at first and either switching or keeping. if you keep every time there's a 66% chance of loosing, but, if you switch every time, there's a 66% chance of winning.
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The chart made all the difference in the world. Bitchin'~
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i'm pretty sure that's the ace of hearts
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Bad quality i can't see and hear. can't you afford a good camera?
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I get it...Think about it you have a 2/3 chance of getting a goat which means your pick will most likely be a goat it doesnt matter if u c 1 of the choices and now you only have 2 left. Its not 50% .. because you started off at 66% which means you most likely chose a goat than than the car but if you switch your basically switching your goat for the car.
this is confusing
babyaiko 3 years ago
Yes, it is! Sorry if I may have made it more confusing... The last 2 minutes are the most helpful; I kind of babble. The chart in my own opinion helps the most. If you are still confused after, let me know!
telecoustic 3 years ago
ahh it makes more sense nowwhen you just look at it in black and white - its all dependant on the host and initially chosing the incorrect card/door. in a nut shell, its like in poker if you are a 3 - 1 underdog to lose, but then the dealer removed a third of the cards that would NOT help you win the hand thus would give you now a 66.6% chance of winning. Thanks
7271422013 3 years ago
Exactly! Glad I could help.
telecoustic 3 years ago
What happens if you pick the correct card/door straight away, does the host still offer the swap or does he reveal the car straight away. If he still offers the swap you would then be left with a 66.6% chance of losing according to this theory.
7271422013 3 years ago
Yes, he still offers a switch; but NO, you would only be left with a 33.3% chance of losing because he already opened a door with a goat; eliminating 33.3%. But... Since you ALWAYS switch, you would lose if you guessed the car. Understanding that you only pick the car 1 out of 3 times is KEY to understanding how this paradox works. Percentages and stats aren't always based on one time recordings. Knowing that may help your understanding.
telecoustic 3 years ago