20 sats \ 8 replies \ @Bell_curve 1 Jul \ parent \ on: What's the best of best 'Game Show' of all time? gaming
3 doors πͺ, one prize π
I pick door πͺ 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 πͺ I didnβt select is 2/3
Now Monty Hall opens door πͺ 3 which has no prize ; this means p = 0 for door πͺ 3
That means door πͺ 1 has p = 2/3
The combined probability of door 1 and door 3 is always 2/3
Since p for door πͺ 3 is 0 then p for door πͺ 1 is 2/3
Monty Hall asks me do I want to stay or choose door πͺ 1
I tell Monty Hall I want to switch , give me door πͺ 1
I double my probability by switching
Monty did us a favor by showing us one empty door
Hope this is clear. Itβs easier to explain with pen ποΈ and paper π
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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 itβs 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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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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The probability of each door has to add up to 1 or 3/3
Eliminating one of the goat π doors doesnβt change the original probabilities
Original probabilities are 1/3 π and 2/3 goat
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