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!