Probability question involving sets of cards I have an infinite deck built out of sets of 10 cards (in other words 10*n cards). The sets are identical so one '2' is identical to another '2'.
A player draws 6 cards. If he draws:


*

*any '1' AND a '2', or

*any '3' AND a '4', or

*any '5' AND a '6', or

*any '7' AND a '8', or

*any '9' AND a '10',


he wins. In other words there are 5 pairs and if the player draws a complete pair he gets a point.
What is the probability he won't win any points at all?
To expand on the problem, if the player gets a point for every pair he completes in a hand, what is the probability he'll get 1, 2, or even 3 points? (3 points being 6 cards of 3 completed pairs)
From what I know of Newton's Binomial, there are : $\binom{10}{6} = 210$ different hand combinations.
To expand even further, how do the probabilities change if the source deck ceases to be infinite? From trial and error I can see that if the deck has only 10 cards then the player has to draw at least 1 complete pair.
Example: For example, a hand of {1,1,3,5,5,9} will get no points. A hand of {1,1,2,3,4,5} will get 2.
Script: I've made a simple js script to roughly calculate the probabilities of the infinite deck to verify if your mathematical answer is on track. I am yet to write a script which simulates a finite number of cards in a deck. http://jsfiddle.net/ch3shirecat/xZ8s5/
After azimut's answer: A slight explanation. If the deck has more than 10 cards (10*n with n>1) then any card can have more than 1 other card as a pair. For example, in a deck of 30 there'll be three '1' cards and three '2' cards with 9 possible pairings between them (with each giving a point). So the hand of {1,2,1,2,1,2} is possible and will give 3 pairings. Does it make sense? Thank you!
 A: I'll answer the question for the infinite deck; the case of a finite deck seems rather complicated.
So I'll assume that we draw $6$ cards with independent uniform probability over $\{1,\ldots,10\}$.
The probability to draw at least one pair can be calculated using inclusion-exclusion. The probability to draw any particular pair is
$$
p_1=\sum_{k=0}^2\binom2k(-1)^k\left(\frac{10-k}{10}\right)^6=\frac{99631}{500000}=0.199262\;.
$$
The probability to draw any two particular pairs is
$$
p_2=\sum_{k=0}^4\binom4k(-1)^k\left(\frac{10-k}{10}\right)^6=\frac{579}{25000}=0.02316\;.
$$
The probability to draw any three particular pairs is
$$
p_3=\frac{6!}{10^6}=\frac9{12500}=0.00072\;.
$$
Thus the probability to draw no pair is
$$
\sum_{k=0}^3\binom5k(-1)^kp_k=\frac{22809}{100000}=0.22809
$$
(with $p_0=1$, the probability to draw zero particular pairs). The probability to draw exactly one pair is
$$
\binom51p_1-2\binom52p_2+3\binom53p_3=\frac{55471}{100000}=0.55471\;.
$$
The probability to draw exactly two pairs is
$$
\binom52p_2-3\binom53p_3=\frac{21}{100}=0.21\;,
$$
where the $3$ counts the number of ways of repesenting a triple of pairs as the union of two pairs of pairs.
The probability to draw three pairs is
$$
\binom53p_3=\frac9{1250}=0.0072\;.
$$
The four probabilities add up to $1$, as they should.
A: I'm not completely sure about some details of the question.
So I'm going to answer a slight modification which hopefully covers your question.
I assume that your deck has $2n$ cards, consisting of $n$ pairs.
You randomly draw $k$ cards. What is the chance to get exactly $s$ pairs?
1) Lets start with the number of possibilities to draw no pair at all.
For the first card, there are $2n$ possibilities. For the second one, there are $2n - 2$ possibilities ($1$ card is already gone, and the second of its kind is forbidden.) For the third card, there $2n - 4$ possibilities etc. Since the order of the cards does not matter, we have to divide the resulting number by the $k!$ possible reorderings of the $k$ drawn cards. So the total number of possibilities is $$\frac{(2n)(2n - 2)(2n - 4)\ldots(2n - 2(k-1))}{k!} = \frac{2^k k!}{(n-k)!\cdot k!} = 2^k \binom{n}{k}.$$
2) Now how many ways are there to draw exactly $s$ pairs? There are $\binom{n}{s}$ ways to select the $s$ pairs. Removing these pairs from the deck, there remain $k-2s$ cards to be drawn from $n-s$ pairs such that there is no pair among them. As already seen, there are $2^{k-2s} \binom{n-s}{k-2s}$ ways for this. So the number of possibilities is $$2^{k-2s}\binom{n}{s}\binom{n-s}{k-2s}.$$
3) For the probability, we have to divide this by the total number $\binom{2n}{k}$ of possible hands. Hence the chance to get exactly $s$ pairs drawing $k$ cards from a shuffled deck consisting of $n$ pairs is
$$2^{k-2s}\frac{\binom{n}{s}\binom{n-s}{k-2s}}{\binom{2n}{k}}.$$
Example: In your initial case, $n = 5$ and $k = 6$.
The chance to get $0$ pairs is $0$ (as you pointed out, there is always a pair if you draw $6$ cards out of $5$ pairs).
The chance for exactly $1$ pair is $\frac{8}{21}\approx 38\%$, the chance for exactly $2$ pairs is $\frac{4}{7}\approx 57\%$, and the chance for exactly $3$ pairs is $\frac{1}{21}\approx 5\%$.
