Probability hand has 4 aces, given 5-card hand and ....

[fred2028]

3. A 5-card poker hand is dealt and the last card is turned face up. Determine the probability that you have been dealt 4 aces, given that the card turned over is an ace.

My work:

Since this is conditional:
P(A|B) = P(A n B) / P(B) A = Dealt 4 aces B = Turned card is ace
And here is where I'm stuck. I dunno how to go about calculating the probability that of the remaining 4 cards, 3 are aces. I tried 3/4 * 13^4, and I tried 13^3 and 13^4. The answer in the back of the book is 1/20825.

Any help?

 

[galactus]

\(\displaystyle \L\\(\frac{3}{51})(\frac{2}{50})(\frac{1}{49})\)

 

[fred2028]

\(\displaystyle \L\\(\frac{3}{51})(\frac{2}{50})(\frac{1}{49})\)

Oh I see. Aren't you supposed to take into account that it's conditional though?