Card picking dependant probability without replacement - P(6,6,Red) I'm trying to teach myself statistics online @ khan academy and although its a wonderful resource I have no one to turn to for help when I don't understand. I hope someone here might be able to give me some advice. 
I thought I understood it all and thought would test myself so I've made some questions up to try and work out the theoretical probability, then would cross x the answer against a simulation i've written in VBA/Excel to really understand it all.
But.... I'm really stuck on working out the probability of picking card #1 with value 6, #2 with value 6 then #3 a Red card.
Working out P(6,6) is easy for me = 4/52 * 3/51 ≈ 0.4525% 
My issue is that after having picked two 6's in a row, what is the probability of then picking out a red card - I just can't get my head around it. Because you could have already picked 2 red cards which would reduce the probability or you could have picked 1 or none.
I don't know how you work it out from here, and googling it doesn't help as i'm not wording my question right and not getting the results I want.
Can anyone help me / push me in the right direction. 
ANYTHING would be great.
Thanks
Lewis

You got it already. Since you are aware that first two picks being red or not affects the probability of third pick being red, you should consider it case by case:
Case 1: First two 6's are black. First, we can find the probability of having two black 6's in our two picks. This is nothing but $\frac{2}{52}\cdot\frac{1}{51}$. In this case, having a red card on the third pick is $\frac{26}{50}$ since we still have $26$ red cards.
Case 2: One of first two picks is red. In this case, probability of having one black 6 and one red 6 is nothing but $\frac{2}{52}\cdot\frac{2}{51}\cdot2$. Here, we multiplied by $2$ since we can have our first pick a red 6 and second pick a black 6; or first pick a black 6 and second pick a red 6. And these cases are just symmetric so instead of separating them, we just multiplied the result by $2$. Then having a red card on the third pick is $\frac{25}{50}$ because we are left with $25$ red cards after first two picks.
Case 3: I am leaving it to you to find the probability for this case and the answer. I can check it if you write it as a comment.
