Choosing new teammates My sister gave me a combinatorical riddle. It doesn't appear to be hard, but I ask you if my thoughts are right, just for certainty. Here it is.

Assume you belong to a group of $100$ people, and you are in a team of $10$ people, being a subset of those $100$ people. Assume that you break up, and you randomly choose people among the the $99$ persons to make a new team of $10$ persons. How big is the chance that you end up with a team without any old team mates?

I thought the chance would be
$$
\frac{90}{99} \cdot \frac{89}{98} \cdot \frac{88}{97} \cdots\frac{82}{91}
$$
Since you have to pick out $9$ persons, and every time you pick someone from A, the chance that the person is "new" is
$$
\frac{|A| \ - \ |\{\text{old team mates in } A\}|}{|A|}
$$
Multiplying the chances gives my guess. Do you think this is right?
 A: Your result is correct numerically, but the argument leading to it is incomprehensible due to the fact that you don't explain what $A$ is (it seems to be changing during the selection process).
I'm assuming that you will be on the new team as a member; and the other members will be selected randomly from the population. For a success $9$ individuals out of the $99$ remaining are definitively forbidden.  There are ${99\choose 9}$ equally probable ways to choose $9$ people from $99$, and ${90\choose 9}$ of these are allowed. Therefore the probability $p$ you are asking for is given by
$$p={{90\choose 9}\over{99\choose 9}}={90\cdot89\cdot\ldots\cdot82\over 99\cdot98\cdot\ldots\cdot 91}\doteq0.408\ .$$
A: It seems to me you have the right idea. I will give a full discussion
for the your benefit and anticipating other students of combinatorics
may read this later.
This problem can be done using either ordered or unordered sample points.
If ordered, then both numerator and denominator of the probability fraction
must count ordered outcomes. If unordered, both numerator and denominator must count unordered ones. To begin, remove yourself from consideration because you are
already chosen. As you say, that leaves 99 people from which you might
choose 9 in random sampling, but only 90 of which are eligible to be
chosen if you are to get an entirely fresh team (other than yourself). 
Ordered: Your answer is P(90, 9)/P(99, 9) =  0.4080 t0 four places,
where P(n, r) denotes the permutations of n things taken r at a time.
Unordered: An equivalent method is C(90, 9)/C(99, 9) = 0.4080,
where C(n, r) denotes a combination (or binomial coefficient).
Hypergeometric: You have an urn with 99 balls of which 90 are red (fresh
teammates) and 9 are black (former teammates). You withdraw 9 balls from
the urn without replacement, and X is the number of red balls drawn.
You seek $P\{X = 9\} = C(90,0)C(9,0)/C(99,9) = 0.4080.$ This is
obtained by plugging into the standard formula for a hypergeometric
PDF (or PMF). Note that C(9,0) = 1, 
Simulation: The brief R program below simulates 100,000 performances
of your experiment. One run gave the approximate value 0.40758, where
the simulation error might be as much as
three digits in the third decimal place. (I always try to check my
combinatorial work, done before coffee on weekends. with a simulation.)


grp = c(rep(0, 9), rep(1, 90));  m = 10^5;  new = numeric(1)
for (i in 1:m) { new[i]=sum(sample(grp,9)) }
mean(new==9)


A: Suppose you are not allowed to choose any of your old-teammates. The number of ways to form your team now would be the number of ways to choose $9$ people from $90$ people, which is given by $90 \choose 9$. 
Without any restrictions the number of ways to do so is $99 \choose 9$. 
Therefore the prob that you don't have any of your team-mates  = $\dfrac{90 \choose 9}{99 \choose 9}$. 
This is the same as the expression you have got. 
