How to approach this conditional probability question The problem is as follows:

Let X be the winnings of a gambler. Let $p(i)=P(X=i)$ and suppose that 
  $p(0)=1/3;\\ p(1)=p(-1)=13/55;\\p(2)=p(-2)=1/11;\\p(3)=p(-3)=1/165$.
  
  Compute the conditional probability that the gambler wins $i$, $i=1,2,3$ given that he wins a positive amount. 

I am not looking for a solution, but just be pointed in the right direction to begin solving this question. 
I don't think this has been posted before, at least I was unable to find this question. If it has, I apologise.
 A: Since you seem to have figured it out from the prod in the comments earlier, here is a fuller explanation.
Here we note two important things.  First, the definition of conditional probability:  the probability of event $A$ occurring given that we know that event $B$ occurs is
$$Pr(A\mid B) = \frac{Pr(A\cap B)}{Pr(B)}$$
Second, we remember one of the axioms of probability: 
$$\text{If}~E\cap F = \emptyset~~\text{then}~~Pr(E\cup F) = Pr(E)+Pr(F)$$
In this problem, our sample space can be partitioned into the possible outcomes of the amount won by the gambler.  We can describe it as $\Omega = \{-3,-2,-1,0,1,2,3\}$ with $Pr(X=i)$ given in the problem statement.
Letting $B$ be the event that the outcome is positive, we recognize that $B=\{1,2,3\}=\{1\}\cup \{2\}\cup \{3\}$ where the final representation is as a union of disjoint sets, allowing us to use the aforementioned axiom.
Thus, $Pr(B)=Pr(\{1\}\cup\{2\}\cup \{3\}) = Pr(\{1\})+Pr(\{2\})+Pr(\{3\})=\frac{13}{55}+\frac{1}{11}+\frac{1}{165} = \frac{1}{3}$
Now, for each of the outcomes in question, $Pr(X=1\mid X>0) = Pr(\{1\}\mid \{1,2,3\}) = \frac{Pr(\{1\}\cap \{1,2,3\})}{Pr(\{1,2,3\})} = \frac{Pr(\{1\})}{Pr(\{1,2,3\})}$ since $\{1\}\subseteq \{1,2,3\}$.  Plugging in the values given in the question gives $Pr(X=1\mid X>0) = \frac{13/55}{1/3} = \frac{39}{55}$
Similarly one can plug in the remaining values for the other outcomes to find those probabilities.
A: Intuitively, you're looking for the probability that you win a certain positive amount, say 1, given that you win a positive amount.
We are given that
$$p(1)=13/55$$
if you pick 1 from $\{-3,-2,-1,0,1,2,3\}$.
What if we do away with nonpositive values?
It's like asking for the probability you pick 1 from $\{1,2,3\}$.
Then we have:
$$p(1 | 1,2,3)=\frac{13/55}{13/55+1/11+1/165}$$
Similarly, we have
$$\\p(2 | 1,2,3)=\frac{1/11}{13/55+1/11+1/165};\\p(3 | 1,2,3)=\frac{1/165}{13/55+1/11+1/165}$$
