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I'm wondering why my answer here is incorrect.

The question is: John and Steve are members of a class of $20$ students. What is the probability that they are paired together?

My thought was simply $1$ way to pair John and Steve and $20 \choose 2$ total pairs so $\frac{1}{190}$. But the correct answer is $\frac{1}{19}$. How is it that my method doesn't work?

ClownInTheMoon
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4 Answers4

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You calculated the number of combinations of two people selected from a class of 20.

Consider how many people John could be paired with: There are 19 other students in the class. There is one chance in 19 that John's partner will be Steve.

Dean
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In your approach, you are looking at all possible pairs (two element subsets) that can be made out of $20$ objects. That means in your method you are looking at both pairs (John, Bill) and (John, Steve) at the same instance, whereas if you divide the class in pairs then for a given pairing John can appear only in one pair, I.e. Both pairs (John, Bill) and (John, Steve) cannot coexist in a given pairing. That is why your answer is incorrect.

Anurag A
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  • I see... so then my approach would require me to realize that the numerator must be the number of ways to choose a pair to be John and Steve from a particular pairing of the entire class. That is, $\frac{10 \choose 1}{20 \choose 2}$. – ClownInTheMoon Dec 10 '16 at 17:16
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    While you are correct that the (ordered) pairs (John, Bill) and (Bill, John) cannot both be part of the same pairing (assuming that there's only one John and only one Bill in the class, anyway), that is not actually the reason why the OP is getting the wrong result. – Ilmari Karonen Dec 10 '16 at 23:12
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There are indeed ${20 \choose 2} = 190$ possible pairs among the $20$ students. Thus, if only one pair was chosen randomly from among the students, the probability that this pair happened to consist of John and Steve would be $\frac1{190}$.

However, the question strongly implies (even if it does not quite say so outright) that all the students are paired up, i.e. that $10$ pairs are formed. The probability that one of these pairs contains John and Steve turns out to be $10 \cdot \frac1{190} = \frac1{19}$.

This is a somewhat surprising result, if you think about it, since the pairs are chosen without replacement, and are therefore not independent.* Thus, it may not be at all obvious that the correct result should be obtainable by simple multiplication. One way to prove that this does, in fact, give the correct result is by induction on the number of students.

However, in this case, there's a much simpler alternative way to arrive at the same solution: simply observe that John will always end up paired with some other student, and consider the probability that, out of the $19$ possible choices, that random other student happens to be Steve.

*) In fact, if the $10$ pairs were chosen independently with replacement, the probability that John and Steve would be paired together at least once would not be $10 \cdot \frac1{190} = \frac1{19}$, but rather $1 - \left(1 - \frac1{190}\right)^{10} \approx 0.05140 \approx 1/19.45 $.

Ilmari Karonen
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3

There are ten total pairs that they could be in.

Another way of looking at it is that John is definitely going to be paired with someone, and there are $19$ possible people for him to be paired to.

AlgorithmsX
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