>>11914746The only way I can make it be 1/2 is by not shuffling the balls, so that we treat the balls as distinguishable and always pick the first ball. To make this a little clearer
Suppose in each box one ball is heavier than the other, and you always pull the heavier ball out of the box first.
let G mean "the heavier ball is gold", and g,G mean "the lighter and the heavier ball are gold".
let s denote a light silver ball, and S a heavy silver ball.
suppose also that the box with a gold and silver ball has the gold ball the heavy ball and the silver ball the light ball, so that the boxes are
[g,G], [s,G], [s,S]
then, since we always pull out the heavy ball first,
P(G) = 2/3
P(gG|G) = P(G|gG) P(gG) / P(G) = 1 * (1/3) / (2/3) = 1/2
Thing is, OP clearly states that you take a ball from the chosen box at random, so it's not 1/2 as in this case, it's 1/3 because you shuffle the balls.
>>11914756>pickED,pick = 2/3.explain what you mean by that
can you define it in C#? is it some kind of array?
