I have read through around 10 articles and webpages on this problem, and still don’t understand. I’ve run simulations and yes, switching does get you better odds, but why?
In: Mathematics
Lots of good ways of thinking here. Here’s a way of visualising it.
You’ve a dartboard split into three equal sections. Two sections lose, one section wins.
You’re blindfold and the board is spun, then you throw darts until you hit the board, completely randomly.
Then you are given the choice – do you want the outcome you landed on, or the other one?
You’re still blindfolded, but you know you were twice as likely to land in a losing segment than a winning one even though *now* you are being offered the 50/50 of ‘the one you chose, or the one that’s left’
Ok. You pick a goat in round one. Host says do you want to change your mind? You say no- the game is over and you lose. You say yes, he opens 1/2 of the remaining doors and shows you where 1/2 goats are. Because the host shows you where the car is NOT, the chances of winning increased from 1/3 to 1/2. This is conditional probability. probability of A|B (A given B) 👍
You have to know you’ll have the switch option everytime for this to apply, right? Because if the game show runners had the option and a desire to limit your winnings, then the option would be influenced by other factors right?
So the problem assumes you always get the option to switch?
At the start of the scenario there was a 66% chance that the prize was behind a door you didn’t pick. Once they remove one of the doors that you didn’t pick from the choice the odds didn’t change, just how much information you have about what was behind each of the doors you didn’t pick.
Opening one of the doors doesn’t make that door disappear. There are still three doors – one that you *know* doesn’t have the prize, & two that are unknown.
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