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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.

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!

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Hint: what combination can he draw to get no points? – gt6989b Mar 4 '13 at 17:20
Edited. I tried different things all day and I am not any closer to the answer... – antonAdelson Mar 4 '13 at 17:25
With an infinite deck, I don't see how to assign probabilities to the 6 card hands drawn. – coffeemath Mar 4 '13 at 17:41
@coffeemath At each step, drawing any value with uniform probability 1/10, independent of what has been dreawn already. If you like, sample from $\{1..10\}$ with replacement. – gt6989b Mar 4 '13 at 17:43
OK now it makes sense, and seems a nice question. +1 – coffeemath Mar 4 '13 at 18:49

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

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In reference to my question, there's a problem with that solution. In my question, there can be a deck of 20 cards and only 5 pairs with some pairs being interchangeable. For example, in your example a deck of 20 cards would look like this: {1,1pair,2,2pair,...,9,9pair,10,10pair) with '1' having only ONE pair in the whole deck. In my question, the 20 card deck would look like this {1,1pair,1,1pair,2,2pair,2,2pair,...,5,5pair,5,5pair} (20 cards) with '1' having TWO other cards to go as a pair with it. Does it make sense? – antonAdelson Mar 4 '13 at 20:58
Could I contact you in private for you're the only person to attempt to answer the question, please? Thank you. – antonAdelson Mar 4 '13 at 21:00

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.

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