Pot Odds

[oldvarmint]

In chapter 5 (Pot Odds)of Sklanskys "The Theory of Poker". He gives an example of counting all the "seen" cards. He starts out with a 7 card stud open ended straight draw (6,7,8,9)after 4 cards have been dealt. The number of "other" seen cards are 8. He then gives the odds if none of the seen cards are a 5 or a 10; to hit the straight as 49.8 percent. Could someone tell me where I'm missing this. I see a total of seen cards (counting my 4) as 12, total cards left unseen 40. So I see it 8 cards can help me hit the straight and 32 won't. Which means to me you have a 25% chance of hitting it (I know I'm probably doint this all wrong). How does Sklanskly come up with 49.8?

His next example deals with a razz game. I have a 4,3,2,A and the number of seen cards as 10. He sayes you have a 81.8% chance to hit a 8 or lower, if none of the seen cards are a 8,7,6,5. Once again if I count the seen cards (14) and unseen cards (38). I come up with 16 cards to make the 8 lower and 22 won't make it. At least this is closer with a difference of Sklansky's 81 to my 72 (once again I know I'm probably doing the math wrong, thats why I'm ask this question).

So I'm all screwed up, could someone help me here on what I'm doing wrong.

Oldvarmint