To find the probability of drawing an arbitrary number of LSs in the opening hand (including redraws):
First start with the probability of drawing 0 LSs. This can be done by using the program, putting the deck size as the population size, putting the number of successes as #LS, the sample size as 8, and the number of successes as 0. In MKCs example of a 56 card deck with 7 LS, this is 0.317 (31.7%)
The probability of drawing exactly 1 LS, you start with the probability of drawing exactly 1 LS in the first 8. For this you can use the calculator as MKC mentioned, with number of successes = 0. Then multiply that by the probability that the top card is not an LS (which is going to be 42/48 in the 56 card example), since out of 48 remaining cards, there are 6 LSs, and 42 non-LSs. In the example, 0.423 is the probability that exactly 1 LS is in the first 8, and multiplying that by 42/48 (.875). .423*.875=.370 = 37%
Now it starts tricky. For the probability of drawing exactly 2 LSs, you either have to draw 1 LS in the original 8, have the top card be an LS, and have the next card not be an LS, or you have to draw 2 LS in the original 8, and have both the next two cards not be LSs. For the first case, it will be .423*.125*42/47=.047 (4.7%). In the second case, we start with the calculator and #successes in sample = 2 (.207), and multiply that by 43/48*42/47 (probability that top card is a non-LS AND that next card is a non LS). This is .166 (16.6%) thus the probability of drawing exactly 2 LSs = 4.7+16.6 = 21.3%
Possibilities 3-7 get decidedly more complicated (but also more rare, as based on the previous calculations, the probability of drawing 0, 1 or 2 LSs is ~90%. I could probably come up with an Excel spreadsheet at some point, I suppose, if anyone is interested.