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How do mutations code sequence to symbols?

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posted on Mar, 6 2021 @ 04:30 PM
Here's' another way of looking at it:

When the contestant (you) chose from the three doors at the beginning, there was a 2-out-of-3 chance that the door you picked was a goat.

Next the host picks a door, and by the rules the host is supposed to intentionally pick a goat (and being the host, he knows where they are). Like I said above, there is a 2-out-of-3 chance that YOU picked a goat at first, so 2/3 of the time the choice of doors that are left for the host to pick will be one with a goat and one with a car....

And since the host is FORCED to avoid the door with the car every time for those 2-out-of-3 situations, that means the door he didn't chose will have the car 2-out-of-3 times.

tl;dr version:
2/3 of the time (2 out of 3 times the game is played), the doors left for the host will be "goat" and "car" -- and he will always reveal a goat in those situations. Therefore, for 2 out of 3 times the game is played, the door he does NOT reveal will have the car.

edit on 3/6/2021 by Soylent Green Is People because: (no reason given)

posted on Mar, 6 2021 @ 05:33 PM
a reply to: Soylent Green Is People

Hes confusing probability with odds of winning hes not going to get it. thats why i stopped responding to him. Many people make the same mistake thats why vegas makes money lol. Bayes theorem can be very confusing

towardsdatascience.com...

posted on Mar, 7 2021 @ 07:10 PM

originally posted by: Soylent Green Is People

tl;dr version:
2/3 of the time (2 out of 3 times the game is played), the doors left for the host will be "goat" and "car" -- and he will always reveal a goat in those situations. Therefore, for 2 out of 3 times the game is played, the door he does NOT reveal will have the car.

I repeat: I understand the problem, and the accepted answer, and the logic behind the accepted answer.

The problem with the accepted answer is explained in the tl;dr summary you wrote above - namely that that answer relies on multiple plays. The contestant is playing exactly one time and has complete freedom to chose between exactly two doors.

When the contestant plays his ONE chance he/she is NOT playing a best out of 3 or anything of the sort.

ONE chance - TWO doors.

End of story.

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