d#= door #, g a/b= goat a/b, c=car r= revealed, P= probability, l= given. In the numberphile video she used x= d1c and y = d2gr. In this example I'm going to show you what it looks like when you use y=d2ga Now what I'm wondering is why this gives P(xly)= 1/2 when essentially P(xly) in this case is the same as saying the probability of the car being in door 1 given that door 2 is a specific goat(goat a).Which is exactly what you want to find out for the problem. So why is my definition of y "incorrect"? 50-50 confirmed? These are the possible scenarios car, goat, reveal after rules.
1. D1ga, d2gbr, d3c 1/18 + 2/18 = 3/18 = 1/6
2. D1ga, d2c, d3gbr 1/6
3. D1gb, d2gar, d3c 1/6
4. D1gb, d2c, d3gar 1/6
5. D1c, d2gar, d3gb 1/18 + 1/36 = 1/12
6. D1c, d2gbr, d3ga 1/12
7. D1c, d2ga, d3gbr 1/12
8. D1c, d2gb, d3gar 1/12
x=d1c y= d2ga
P(X&Y) = P(XlY) * P(Y)
1/6 = 1/2 1/3
P (Y&X) = P(YlX) * P(X)
1/6 = 1/2 1/3
Full list of scenarios and why there are 8 outcomes instead of 6 in the comments, as well as the simplified 6 scenarios.
1. D1ga, d2gbr, d3c 1/18 + 2/18 = 3/18 = 1/6
2. D1ga, d2c, d3gbr 1/6
3. D1gb, d2gar, d3c 1/6
4. D1gb, d2c, d3gar 1/6
5. D1c, d2gar, d3gb 1/18 + 1/36 = 1/12
6. D1c, d2gbr, d3ga 1/12
7. D1c, d2ga, d3gbr 1/12
8. D1c, d2gb, d3gar 1/12
x=d1c y= d2ga
P(X&Y) = P(XlY) * P(Y)
1/6 = 1/2 1/3
P (Y&X) = P(YlX) * P(X)
1/6 = 1/2 1/3
Full list of scenarios and why there are 8 outcomes instead of 6 in the comments, as well as the simplified 6 scenarios.
