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There are $n$ balls $b_1, \dots ,b_n$ and $n$ boxes. Each ball is placed in box chosen independently and uniformly at random. We say that $(b_i, b_j)$ is a $\textit{colliding pair}$ if $i<j$, and $b_i$ and $b_j$ are placed in the same box. WHat is the expected number of $\textit{colliding pairs}$?

1. $\frac{n-1}{2}$
2. $0$
3. $1$
4. $\frac{n}{4}$
5. $\begin{pmatrix} n \\ 2 \end{pmatrix}$

### 1 comment

We can reform this problem in another problem like 'n total slots and n total keys and we have to find expected number of colliding pairs in uniform hashing'.

https://gateoverflow.in/159733/uniform-hashing

Correct Option: A

Let $X_{ij}$ be an indicator random variable, such that

$X_{ij}=\left\{ \begin{array}{rcl} 1 & \text{if collide} \\ 0 & \text{otherwise} \end{array}\right.$

$\therefore$ Probability that $i$ and $j$ both hash to the same slot.

$P_r(X_{i,j})=\frac{1}{n}$

$\implies E[X_{i,j}]=\frac{1}{n}$

Now,

$\text{E[No. of Colliding pairs]}= E\left[\displaystyle\sum_{i=1}^n \sum_{j=i+1}^n X_{i,j}\right]$

=$\displaystyle\sum_{i=1}^n \sum_{j=i+1}^n E[X_{i,j}]$

=$\displaystyle\sum_{i=1}^n \sum_{j=i+1}^n\frac{1}{n}$

=$\dfrac{n(n-1)}{2n}$

=$\dfrac{(n-1)}{2}$

by

how did the summation give the result  n(n-1)/2n

can you elaborate that calculation please
in summation, i value is starting from 1 and j=i+1 inorder to consider the condition i<j??
@Aishwarya.R.Menon because number of such pairs are so. And we have to consider all possible colliding pairs. Which are nC2.

The balls b1,b2,....,bn are put in the box uniformly and randomly.

b1 has no chance of collison.

probability of b2 colliding=1/n.

prob of b3 colliding =2/n (j=3; i=1 or 2).

prob of b4 colliding = 3/n.. So on..

.

.

prob of bn colliding = (n-1)/n.

Therefore, expected no of colliding pairs

= 1/n+2/n+3/n+....+(n-1)/n = 1/n*[1+2+3+..+(n-1)] =1/nX[n*(n-1)/2]=(n-1)/2 (A)

by
Actually I have a  different  understanding of the problem.
I felt the other approaches discussed here misses an important point though they arrive at the correct answer.
Here is how I understood the problem and have arrived at the solution.

Lets model the problem  as follows:
"ABCD" means b1 is given to box A b2 is given to box B b3 is given to box C and b4 to box D.In this case there is no collision pair.

"AABB" means b1 is given to box A b2 is given to box A b3 is given to box B and b4 to box B.In this case there are two collision pairs.One from Box A and other from Box B.

"AAAA" has 4*3/2 collision pairs
and so on.

So different strings lead to different collision pairs.The point other approaches are missing is that for a string like "AABB" collision pairs can be coming from two different boxes too.

The question asked here is not about the "expected number of collision pairs for a single box" but for the
entire distribuion of n balls across n boxes.In other words if I take a random string what is the expected number of collision pairs that could come from all boxes.

Also if there are r balls in a box then number of collision pairs that could come is r*(r-1)/2=rC2

So E(CP)=Total of Cp from each random distibution/total number of possible distribution

Total number of possible distribution=n^n

Total of Cp from each random distibution is done by choosing 1 alphabet out of n alphabets after which selcting a r positions out of n positions and placing that alphabet.After which the remaining n-1 alpahbets are filled.
Since there are r repetitions of the chosen alphabet so there are r*(r-1)/2 CP with respect to chosen alphabet.
(Observe that the permutation like AABB will be counted twice one wrt A and other wrt to B and in each case 1 will be added making total Cp as 2)

Total of Cp from each random distibution=

$$(\sum_{i=2}^n(nCr*(n-1)^{(n-r)}*rC2))*n$$

$$(\sum_{i=2}^n(n*(n-1)*n-2Cr-2*(n-1)^{(n-r)}))*n*1/2$$

$$(\sum_{i=2}^n(n-2Cr-2*(n-1)^{(n-r)}))*n*n*(n-1)*1/2$$

$$((n-1+1)^{n-2})*n*n*(n-1)*1/2) [Bionomial theorem]$$

$$So E(CP)=(n^{(n-2)})/n^n *n*n*(n-1)*1/2$$

$$E(CP)=(n-1)/2$$

@arjun sir?