Probability of card combinations I am trying to improve my understanding of combination probabilities. I stumbled upon the Wikipedia site for Texas Holdem probabilities and am currently stuck in calculating the probabilities of various single hands that are listed in the table.
Hand                                            Probability     Odds
...
Connected cards (cards of consecutive rank)     0.157           5.38 : 1
Any 2 cards with rank at least 9                0.208           3.81 : 1
Not connected nor suited, at least one 2-9      0.534           0.873 : 1

I managed to calculate all probabilities except for the last one. I cannot figure out how they calculated it. 
Here is what I calculated:
If at least one card has to be within range 2-9 then there are 8 possible hands for a pair. With all possible suit combinations this gives 48 overall possibilities for a pair.
The possible hands for unsuited, not connected cards are:
If one card is a 2 then possible hands are
2,4 2,5 2,6 2,7 2,8 2,9 2,10 2,J 2,Q 2,K 2,A

If one card is a 3 then possible hands are
3,5 3,6 3,7 3,8 3,9 3,10 3,J 3,Q 3,K 3,A

and so on until one card is a 9
9,J 9,Q 9,K 9,A

This gives a total of hands of
$\sum_{n=4}^{11}n = 60$
With respect to all possible suit combinations the number of total combinations is $12*60=720$. Together with the combinations for pairs I get
$720 + 48 = 768$
The number of all possible combinations for the first two cards, as stated in the article, is 1326. This results in a probability of 
$768 / 1326 = 0.579$
which is clearly different from the value in the table.
Where is the error when calculating the possible combinations?

The quoted odds of $0.534$, assuming proper rounding, translate to $708$ combinations.  I think the author does not allow pairs in "Not connected or suited, one card $2-9$".  This would reduce your count to $720$.  Then there are $12$ hands with $A2$ nonsuited that you can consider connected because they can form a straight.  These are less valuable than other connected cards because they can only form one straight, while $78$ can form four. This gets the count to $708$ as required.  You have correctly computed what you were trying to compute, the issue is understanding what was to be computed.
