Probability of getting 3rd club on 21st draw without replacement How can I obtain the probability of getting the 3rd club on the 21st draw? It makes sense that there must be exactly 2 clubs in the first 20 draws and that the 21st must be the third club. 
As there are 13 clubs in a standard card deck, would I be right to say that there are 20 * 13C2 ways of picking 2 clubs for the first 20 draws? This is the point I am getting stuck at… 
 A: Answer:
It is equivalent to drawing a 20 card poker hand with exactly 2 clubs and 18 nonclubs and the 21st card is a club
The way 2 clubs could be chosen is ${13\choose2}$
The way 18 Non-clubs could be chosen is ${39\choose18}$
The total number of ways 20 cards could be drawn out of a deck is ${52\choose20}$
Thus the probability is $$\frac{{13\choose2}.{39\choose18}}{{52\choose20}}.\frac{11}{32}$$
A: There are 13 clubs in the deck. This means that the are also 39 non-clubs. If two of the first twenty cards drawn are clubs, that means that 18 must be non-clubs. 
There are 13C2 ways to pick two club cards and 39C18 ways to choose the eighteen non-club cards.
A: P(3rd club on 21st card) = 
P(exactly 2 clubs in first twenty cards) x P(21st card is club | exactly 2 clubs in first twenty cards) =
C(13,2) x C(39,18) / C(52,20)  x  11/32
A: I think it will be.... ${13\choose2}$* ${39\choose18}$*${11\choose1}$/${52\choose21}$
A: If you're just counting the clubs and where they're at, then there are $\binom{20}{2}$ ways for there to be two clubs in the first 20 draws.
You could observe that there are $\binom{20}{2} \cdot 13 \cdot 12$ ways to pick specific clubs to go into the first 20 cards... but if you're simply analyzing where the clubs are in the deck, it's much easier to simply ignore the ranks of every card and just look at the suits.
But it's hard to compute the probability you want just looking at it this way: the only reasonable way I see to compute a probability is to compute
$$ \frac{ \text{number of wanted club distributions} }
{\text{number of all club distributions}} $$
which means you also have to decide how to arrange the remaining 10 clubs in the remaining 31 cards to compute the numerator.

Formally: each of the $\binom{52}{13}$ ways to pick which cards in the deck are to be clubs is equally probable. So we can compute the desired probability if we identify how many of these are of the form we want.
Each desired distribution can be completely and uniquely specified by choosing


*

*Which 2 cards of the first 20 are clubs

*Which 10 cards of the last 31 are clubs


and each such choice gives us a valid distribution of clubs, so we have correctly reformulated the problem. In this form, we can count easily, and compute the desired probability as
$$ \frac{ \binom{20}{2} \binom{31}{10}}{ \binom{52}{13} } $$
A: This is the probability that an arrangement of the full 52-card deck contains a club in position 21, 2 clubs in the first 20 positions, and 10 clubs in the last 31 positions. Let's count these arrangements.
First, pick a club for position 21 (13 ways).
To put 2 clubs in the first 20 positions, choose two of the remaining clubs (${12}\choose{2}$ ways), choose the two positions for them (${20}\choose{2}$ ways), and choose the order in which to put them ($2$). So there are ${12\choose2}\cdot{20\choose2}\cdot2$ ways.
To put the remaining 10 clubs in the last 31 positions, choose the positions (${31}\choose{10}$ and choose the order in which to put the clubs in those positions ($10!$), to give ${31\choose10}\cdot10!$ ways.
Finally, you have to place the 39 cards of the other suits ($39!$ ways).
So the number of arrangements is $13\cdot{12\choose2}\cdot{20\choose2}\cdot2\cdot{31\choose10}\cdot10!\cdot39!$.
The probability that an arrangement of the deck is one of these is then $$\frac{13\cdot{12\choose2}\cdot{20\choose2}\cdot2\cdot{31\choose10}\cdot10!\cdot39!}{52!}\textrm.$$
