Two Pair Outs... A Sklansky question

[General Sal]

I have a question that I"m hoping someone could answer. It's an "outs" question. I was looking in Sklansky's Advanced Hold "em on the probability section. In it, it shows the percentages for four outs. Obviously, I understand how four outs on an inside straight draw works. But, it also lists "two pair" as having four outs, and I don't get it. Can someone explain it to me?

With an inside, let's say I have a 10 and J. Board shows A K 2. I know that I need a queen to make my draw. And there are four queens. Therefore, four outs.

With a two pair scenario, let's assume I have a 10 and J. The flop comes J 9 3. There are only three 10s left in the deck, so how is it that I have four outs? I'm confused!

I'd appreciate the answer.

[Dave B]

If you have j 10 and the flop come j 10 3-then you have 2 pairs and 4 outs. There are 2 Js and 2 10s.

In your scenario of J 10 w/ J on the flop, you only have one pair. In your case there are 5 outs, 2 Js and 3 10s.

[General Sal]

It makes sense now... he was talking about two pair drawing to a full house! I see said the blind man. Thank you, Dave.

[Dave B]

No problem-I can see where the confusion came from.