Card picking dependant probability without replacement - P(6,6,Red)

Hi All,

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've made some questions up myself to try and work out the answer and check it 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.




Active Member
How about splitting it up and finding the probabilities of red6 red6 red, red6 black6 red, black6 red6 red, and black6 black6 red