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3 doors πŸšͺ, one prize πŸ†
I pick door πŸšͺ 2 which has p = 1/3 where p is probability
That means my chance of getting nothing is 2/3
The combined probability of the two doors πŸšͺ I didn’t select is 2/3
Now Monty Hall opens door πŸšͺ 3 which has no prize ; this means p = 0 for door πŸšͺ 3
That means door πŸšͺ 1 has p = 2/3
The combined probability of door 1 and door 3 is always 2/3 Since p for door πŸšͺ 3 is 0 then p for door πŸšͺ 1 is 2/3
Monty Hall asks me do I want to stay or choose door πŸšͺ 1
I tell Monty Hall I want to switch , give me door πŸšͺ 1
I double my probability by switching
Monty did us a favor by showing us one empty door
Hope this is clear. It’s easier to explain with pen πŸ–ŠοΈ and paper πŸ“
I get the concept but I am struck on the idea that odds go from 2/3 to 1/2 once you eliminate a door.
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Let me start over and be more concise
Imagine there are 100 doors You pick 1 door Probability is 1 percent
Monty opens 98 doors with goats
There is a 99 percent chance it’s not the door you selected
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The combined probability of two doors is always 2/3
We start with 1/3 + 1/3 + 1/3 = 1
We eliminate a door 1/3 + 0 + 2/3 = 1
The key is your initial selection probability never changes
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Yes but then the combined probability of my door and the one already opened is also 2/3 so if my door plus opened and the door I didn’t pick plus the opened one both are 2/3 why is it beneficial to switch?
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The combined probability of your door and the door already open is 1/3 + 0
The probability of the open door is 0
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If you switch p is 2/3
If you stay p is 1/3
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19 sats \ 1 reply \ @grayruby 2 Jul
I get what you are saying but still doesn’t seem right.
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The probability of each door has to add up to 1 or 3/3
Eliminating one of the goat 🐐 doors doesn’t change the original probabilities
Original probabilities are 1/3 πŸ† and 2/3 goat
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