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!

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.
