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InitialsDiceBearhttps://github.com/dicebear/dicebearhttps://creativecommons.org/publicdomain/zero/1.0/„Initials” (https://github.com/dicebear/dicebear) by „DiceBear”, licensed under „CC0 1.0” (https://creativecommons.org/publicdomain/zero/1.0/)KE
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40
Joined
5 yr. ago

  • Just count the number of possibilities. If you change there there two possible first choices to win + if you do not change 1 possible choice to win = 3. If you change there is one possible first choice to lose + if you do not change there two possible first choices to lose=3 P(x1)=P(x1') = 3/6

  • First, fuck you! I couldn't sleep. The possibility to win the car when you change is the possibility of your first choice to be goat, which is 2/3, because you only win when your first choice is goat when you always change.

    x1: you win

    x2: you change

    x3: you pick goat at first choice

    P(x1|x2,x3)=1 P(x1)=1/2 P(x3)=2/3 P(x2)=1/2

    P(x1|x2) =?

    Chain theory of probability:

    P(x1,x2,x3)=P(x3|x1,x2)P(x1|x2)P(x2)=P(x1|x2,x3)P(x2|x3)P(x3)

    From Bayes theorem: P(x3|x1,x2)= P(x1|x2,x3)P(x2)/P(x1) =1

    x2 and x3 are independent P(x2|x3)=P(x2)

    P(x1| x2)=P(x3)=2/3 P(x2|x1)=P(x1|x2)P(x2)/P(X1)=P(x1|x2)

    P(x1=1|x2=0) = 1- P(x1=1|x2=1) = 1\3 is the probability to win if u do not change.