1

There are 3 friends each with $n$ books, each friend has a set of their own $n$ same books, each one takes a random permutation of the $n$ books. They all read one book every week (n consecutive weeks). Now, we let $X$= no. of weeks they all read the same book. The problem is asking for the variance, but I figured the first thing to do was to figure out the expected value or more importantly the probability of reading the same book for no. of weeks.

I did some math and I was able to figure out that there is $3 (n!)^3$ outcomes in the sample space.

I did the same problem with 3 friends and two books (n=2) and I figured out the probability of all the books matching for $X$= 2 weeks is $\frac{1}{4}$ or maybe $\frac{1}{n!}$ and for $X$ = 0 weeks is $\frac{1}{2}$ or perhaps $\frac{1}{n}$. I am not sure if the problem applies to the other, because the middle ranges are not accounted for i.e $x=1,3,4,...,n-1$ so I am not sure if this scales. If you can point me in the right direction, I would greatly appreciate it!

user168915
  • 155

1 Answers1

2

For the mean, for $i=1$ to $n$, let $X_i=1$ if they are all reading the same book, and $X_i=0$ otherwise. The number $Y$ of weeks they are reading the same book is given by $Y=X_1+\cdots+X_n$.

By the linearity of expectation, we have $E(Y)=E(X_1)+\cdots+E(X_n)$.

For any $i$, the probability they are reading the same book in Week $i$ is $\frac{1}{n^2}$. So $E(X_i)=\frac{1}{n^2}$ and therefore $E(Y)=\frac{1}{n}$.

Remark: The variance is found by a similar but but more complicated calculation. It is enough to find $E(Y^2)$.

Expand $Y^2$. We get $$\sum_1^n X_i^2 +2\sum_{1\le i\lt j\le n} X_iX_j.$$ We will need $E(X_iX_j)$, so we will need the probability that $X_i=1$ and $X_j=1$.

André Nicolas
  • 507,029
  • 1
    Note that the argument has a structure that should be becoming familiar. – André Nicolas Aug 10 '14 at 21:06
  • Thank you very much! I was wondering how you got the probability that they are reading the same book in any week $i$ as $\frac{1}{n^2}$. I mistakenly thought it would be $(\frac{1}{n})^3$. Also $E(X_i^2)=E(X_i)$ and $E(X_iX_j)=Pr(X_i=1|X_j=1)=\frac{1}{n^2}\frac{1}{(n-1)^2}$ so for $E(Y^2)=\frac{1}{n}+n(n-1)\frac{1}{n^2}\frac{1}{(n-1)^2}=\frac{1}{n}+\frac{1}{n}\frac{1}{(n-1)}$ does this seem correct to you? – user168915 Aug 10 '14 at 21:21
  • 1
    For same book, the probability they are all reading Book $1$ is, by a calculation you made, equal to $\frac{1}{n^3}$. Same for Book $2$, and so on, so multiply by $n$. Or else whatever friend $F_1$ is reading, the probability Friend 2 matches it is $\frac{1}{n}$, and the probability Friend 3 matches is $\frac{1}{n}$. Your calculation looks right, but I am trying to read a low quality display in bright sunshine (I have no choice, want to smoke, and a pipe delivers its nicotine slowly). – André Nicolas Aug 10 '14 at 22:27
  • I see the way you have calculated it now. Thanks for dissolving my doubt André. Enjoy this beautiful day, as you have made mine much better! Bless you sir! – user168915 Aug 10 '14 at 23:33