Ok, here goes... sorry for the annotation, if it is kind of vague.

Assume you want to take

**k **items out of

**n** items, without caring about the order. Then the possible ways to do this are, according to probability theory, n!/(k!*(n-k)!).

So, let us assume you need to be dealt

*specifically *3 spades out of 156 cards. Since there are 39 spades in total, there are 39!/(3!*36!)=9139 ways to do this. Note that I used the number of spades, not total number of cards, as

**n**. Now, the sample space (the possible ways to pull any 3 cards from a pool of 156 cards) is 156!/(3!*153!)=620620. Therefore, the probability to be dealt 3 spades is: P("3 spades")= 9139/620620 = 0.014725597, which is, by the way, quite low.

However, we do not specifically need spades, but any 3 cards that are the same suit. Since the possibilities of being dealt 3 spades, or 3 diamonds, or 3 clubs, or 3 hearts are all independent, we can simply add them:

P("3 of same suit")=P("3 spades")+P("3 diamonds")+P("3 clubs")+P("hearts")=4*P("3 spades")=0.058902388, because all suits have the same probability to be dealt.

Now, for the second part, things get a little more interesting, but still easy to calculate.

Instead of using 39!/(3!*36!) and 156!/(3!*153!), use the numbers you define. In your example, where 30 cards are missing, of which 9 are spades, the probability to be dealt 3 spades is: (30!/(3!*27!))/(126!/(3!*123!)). Note that both the total nmber of spades

*and *the total number of cards has to be modified.

You can calculate the probabilities for the rest of the suits in the same way. I hope this helps