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'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 $\frac{1}{21}\approx 5\%$.