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A random person (let's call him Bob) is given 3 cards, containing the names of 3 different countries. Bob is also given the names of the capital cities of these countries (in random order), and his task is to place the country cards with the correct capitals, i.e. to form correct pairs. It just so happens that Bob has absolutely no clue, so he just makes pairs randomly. We call the random variable that counts the number of correct pairs $X$.

So I have to find the mean and the standard deviation of $X$, but I'm not exactly sure what the correct probability distribution of this problem is. From all the probability distributions I have encountered as of yet, this seems most like a hypergeometric probability distribution, because there is no replacement in this problem. But I don't intuitively understand why. Wikipedia says:

The hypergeometric distribution is a discrete probability distribution that describes the probability of $k$ successes in $n$ draws, without replacement, from a finite population of size $N$ that contains exactly $K$ successes, wherein each draw is either a success or a failure.

I have trouble relating this to our problem. We also want to find the probability of $k$ successes in $n (\stackrel{?}{=}3$) draws, without replacement, from a finite population size $N(\stackrel{?}{=} 3)$ that contains exactly $K(\stackrel{?}{=} 1)$ successes, wherein each draw is either a success or failure.

So that would give us a mean of $n \times \dfrac{K}{N} = 3 \dfrac{1}{3} = 1$ and a standard deviation of $\sqrt{ n\dfrac{K}{N} \dfrac{(N-K)}{N} \dfrac{N-n}{N-1}} = \sqrt{\dfrac{1}{3} \times \dfrac{2}{3} \times 0} = 0$, and herein lies the problem, because this is obviously wrong.

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  • $\begingroup$ As you point out, the "matching distribution" is not hypergeometric. The standard hypergeometric problem is "red and blue balls in an urn, drawn without replacement. Red=success. Variable measures number of successes." Note that a lot of successes decrease the probability that the next is a success...in your situation, a lot of successes increase the probability that the next is a success (it's impossible to have only one failure, for instance). Not sure there's a name for the matching distribution...you can read about it e.g. here $\endgroup$ – lulu Sep 21 '16 at 10:24
  • $\begingroup$ Just to say: with only three samples, there's no difficulty in simply enumerating the options (there are only $6$ permutations, after all). The link I included in the earlier comment does the computation in general. $\endgroup$ – lulu Sep 21 '16 at 10:32

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