Two cards are randomly drawn without replacement from a common 52 card deck, P(Jack)=? Q: Two cards are randomly drawn without replacement from a 52 card deck of 
   common playing cards. What is the probability of drawing a Jack?
I have found a few ways to answer this problem, one the ways I prefer but don't fully understand. Here is my solution:
$$P(J_1 \cup J_2)=P(J_1)+P(J_2)-P(J_1\cap J_2)=1/13+1/13-(1/13)\cdot (3/51)=33/221$$
$33/221$ is the correct final answer but I'm not sure how these inputs can be correct, for instance how is P(J1)=
P(J2)? 
 A: I think this is an intuitive way:
$P(Jack)=1-P(Jack^\complement)=1$-${48}\choose{2}$/${52}\choose{2}$
I use combinations to calculate $P(Jack^\complement)$:


*

*There are ${52}\choose{2}$ ways to draw two cards without replacement

*There are ${48}\choose{2}$ ways to do the same thing without Jacks

A: It doesn’t matter whether you draw the jacks simultaneously or one at a time. The probability that the first card is a jack is the same as that of the second card being a jack. You can verify this by computing these probabilities in two different ways. 
First, drawing sequentially, the probability that the first draw is a jack is clearly $\frac1{13}$. For the second card, we have two cases: if you’ve already drawn a jack, the probability that the next card drawn is a jack is $\frac3{51}$. On the other hand, if you haven’t already drawn a jack, then it’s $\frac4{51}$. Putting these together using the Law of Total Probability, $$\Pr(\text{second card is a jack})=\frac1{13}\cdot\frac3{51}+\frac{12}{13}\cdot\frac4{51}=\frac1{13}.$$  
If you’re drawing the cards simultaneously, then there are really two steps to the process: first you draw the two cards and then you place them into slots #1 and #2. There are four possible outcomes: no jacks in either slot, a jack in both slots, a jack in slot #1 only, or a jack in slot #2 only. We’re only interested in the last three, the probabilities of which are $$\begin{align}\Pr(\text{jack in slot #1 and jack in slot #2}) &= \frac4{52}\cdot\frac3{51} \\ \Pr(\text{jack in slot #1 only}) &= \frac4{52}\cdot\frac{48}{51} \\  \Pr(\text{jack in slot #2 only}) &=\frac{48}{52}\cdot\frac4{51}.\end{align}$$ The last two probabilities are obviously equal. These events are disjoint, so the overall probability of having a jack in slot #1 is the sum of the first and second values, as is the overall probability of having a jack in slot #2 (and is again equal to $\frac1{13}$).
