Conditional Probability P(A ∩ B) So I found myself in a infinite loop while trying to do some probability. If A and B are independent, calculating P(A ∩ B) is as simple as P(A)P(B). However, how do I calculate P(A ∩ B) if they are dependent? I know P(A∣B)⋅P(B)=P(A∩B).
example:
Charlie and Doug take turns rolling two dice, Charlie goes first. If Charlie rolls a sum of 6 before Doug rolls a sum of 7, Charlie wins. What is the probability Charlie wins.
Solution attempt:
P(Charlie wins) = P(Charlie rolls 6, 1st turn) + P(Charlie rolls 6 | Charlie and Doug don't win previous rounds)
        = 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)*P(Charlie and Doug don't win previous rounds)
        = 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)*P(Charlie doesn't win previous)P(Doug doesn't win previous)
    = 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)(31/36)*(30/36)    
Now this is where I'm stuck, as I don't know how to calculate P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds). Can anyone give me some pointers?

Let $p$ be the required probability that Charlie wins the game. Charlie can win in two ways, fast or slow:   (i) Fast: Charlie wins immediately, probability $\frac{5}{36}$; (ii) Slow: Charlie doesn't win immediately, neither does Doug on his first try, but Charlie ultimately wins, probability $\left(\frac{31}{36}\cdot \frac{30}{36}\right)p$.  Thus
$$p=\frac{5}{36}+\left(\frac{31}{36}\cdot\frac{30}{36}\right)p.$$
Solve for this linear equation for $p$.
Remark: Without explicitly saying so, we have used conditional probability. The probability that Charlie (ultimately) wins given that neither player wins on his first try is just $p$.  
We can instead use your general strategy. The probability that Charlie wins on his first try is $\frac{5}{36}$. The probability that Charlie wins on his second try is $\left(\frac{31}{36}\cdot\frac{30}{36}\right)\frac{5}{36}$. The probability Charlie wins on his third try is  $\left(\frac{31}{36}\cdot\frac{30}{36}\right)^2\frac{5}{36}$. And so on. For the probability Charlie wins, we add up the infinite geometric series with first term $\frac{5}{36}$ and common ratio $\frac{31}{36}\cdot\frac{30}{36}$. We can find the sum using the ordinary formula for the sum of an infinite geometric series.
