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19 sats \ 7 replies \ @grayruby 1 Jul \ parent \ on: What's the best of best 'Game Show' of all time? gaming
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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30 sats \ 0 replies \ @Bell_curve 1 Jul
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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20 sats \ 5 replies \ @Bell_curve 1 Jul
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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19 sats \ 4 replies \ @grayruby 1 Jul
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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50 sats \ 0 replies \ @Bell_curve 2 Jul
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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20 sats \ 2 replies \ @Bell_curve 1 Jul
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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50 sats \ 0 replies \ @Bell_curve 2 Jul
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 🏆 and 2/3 goat
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