Consider a group of four people. Everybody writes down the name of one other (random) member of the group. What is the probability that there is at least one pair of people who wrote down each others name?

Answer is 17/27. I think it should be 19/27. how to calculate it ?

  • $\begingroup$ Is this homework? What have you tried? $\endgroup$ – JavaMan Jan 23 '12 at 22:21
  • $\begingroup$ no not exactly homework. I am trying to teach myself some maths. I am stuck at some problems. my solution was to find if there was no pair so to me it seemed the solution was (3*2*2*2)/3*3*3*3 $\endgroup$ – user669083 Jan 23 '12 at 22:25
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    $\begingroup$ Maybe you can tell us what makes you think the answer is $19/27$? Writing the argument out explicitly might help you figure out where you are making a mistake, or persuade you even more firmly that the answer in the book is incorrect. Answers in books are incorrect in some cases. $\endgroup$ – Dilip Sarwate Jan 23 '12 at 22:28

The numerical answer is $17/27$.

Divide our set of $4$ people into groups of two.

One grouping is $\{A, B\}, \{C,D\}$. There are $2$ other groupings, $\{A, C\}, \{B,D\}$ and $\{A, D\}, \{B,C\}$.

The probability that $A$ and $B$ write each other's names is $\dfrac{1}{9}$. The same applies to $C$ and $D$. Let us compute the probability that both these things happen. It is $\dfrac{1}{81}$.

So the probability that $A$ and $B$ write each other's name, or that $C$ and $D$ do (or both), is $$\frac{1}{9}+\frac{1}{9}-\frac{1}{81}.$$

We subtract the $1/81$ to avoid "double-counting" the situations where $A$ and $B$ pick each other, and $C$ and $D$ also do. Or else we can think of it as following from the formula $$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y).$$

The same calculation applies to the other two pairings. So we multiply $\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{81}$ by $3$. The result is $2/3-1/27$, which is $17/27$.

  • $\begingroup$ +1! That's a much better way to think of the problem than what I was writing :-) $\endgroup$ – JavaMan Jan 23 '12 at 22:36

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