Using two fair dice, one dark green and the other light green, and determine whether two events are independent. Say you toss the dark green one first.
P(sum = 7) = 6/36 = 1/6 P(sum = 6) = 5/36 P(dark green = 1) = 1/6
Since, P(sum=7 | dark green=1) = P(dark green=1), events are independent Since, P(sum=6 | dark green=1) \= P(dark green = 1), events are dependent
So, I get how to compare, but I don't understand this intuitively, the way I understand how it would work with a pack of cards or a bag of marbles w/o replacement.
I read another post similar, but I did not get the reasoning behind the explanation. If you know the outcome of one of the two dice, then you can only have one outcome with the second one that will total the required sum: 6 in the case of P(sum=7) and 5 in the case of P(sum=6). Since the dark green's 1 locks the success of the next roll, shouldn't both events be dependent?
With an example you will see it clear. If you throw two dice, as there is no physical connection between both dice we say that both throws are independent. Now suppose another case, we retrieve balls from a bag. There are balls of two colors inside, say red and blue. If we take out one blue for the first time and don't replace it, then we have changed the likelihood of getting a ball of a particular color next time. I have changed the 'system' by retrieving a ball. Now the system has more balls of one color than another. The 'system' remembers the change. So one event is dependent on the other. But dice have no memory.
The fact of you knowing the outcome of one die, does not change how probability is distributed on the other die. So $$P(A \text{ and } B)=P(A)\cdot P(B)$$
Thinking that the past dice casts affect the future ones is called 'The Gambler's Fallacy'. And It has made loose lots of money to some people at the casinos.
Hope it helps.