Comments on: Puzzle : Kissa Kursi Ka ( A matter of a seat )

By: Saurabh Joshi

Saurabh Joshi — Wed, 08 Oct 2008 05:51:43 +0000

@amit,
if you read carefully vikas’ argument is write. It’s upto you to buy the argument or not, but if not then you should produce a counter-example which you did not.

By: Amit Chandel

Amit Chandel — Tue, 07 Oct 2008 14:31:05 +0000

Let us denote seat numbered i as S_i, and person numbered j as P_j

@Vikas
if I consider 3 passangers only then it can happen that P_1 sits at S_2, and P_2 sit at S_3. So P_3 will neither see P_1 on his seat nor an empty seat. So I am sorry I cannot buy your argument.

I have also applied induction to arrive at 1/2 in this way:
Let f(n) denote the probability that the last person will see his seat empty given there are n passangers.
Base Case: n=2 f(2) = 1/2 . Obvious!
Now we have to prove that f(n) = 1/2 assuming f(m) = 1/2 for all 1 < m < n. Here we go.

P_n can see his seat empty in two cases:
Case #1: P_1 sits on S_1. Prob(Case #1) = 1/n
Case #2: P_1 sits neither on S_1 nor on S_n. He chooses S_k where 1 < k < n.
If he sits on S_k, when P_k comes, he will see S_1 empty, P_2 on S_2, P_3 on S_3, …., and P_1 on S_k. Lets ignore seats from 2 to k and passangers P_1 to P_(k-1). So we removed k-1 seats and k-1 passangers. Lets re-number the seats and passangers reducing all by k, except S_1. We get the same problem with k passangers and k seats.
So probability of last person getting his seat in this situation = (prob of case#2 happening) * (probability of last person getting his seat given case#2 i.e. under modified problem )
= ( P_1 neither sit on S_1 not on S_n) * f(k) where 1 < k < n
= ( (n-2)/n ) * (1/2) = (n-2)/(2n)

hence f(n) = 1/n + (n-2)/2n = 1/2.

By: vikas

vikas — Tue, 03 Jun 2008 07:00:21 +0000

A simpler proof would be if you observe that the last person will either find his seat empty or the seat assigned to passenger 1. Any other seat if empty till last would have already been taken by the person to whom it was assigned.
So now among these two possibilities, any person before the last would have no preference for either of the two so the probability that any of them is vacant till the last has to be same which is 1/2

By: Increasing or decreasing subsequence « Me, Myself and Mathematics

Increasing or decreasing subsequence « Me, Myself and Mathematics — Fri, 30 May 2008 09:41:33 +0000

[...] found this problem here. Solution to the second puzzle given in that post is already posted in Kissa Kursi Ka. I tried a couple of approaches to solve the first one but could not succeed. Finally, Deepanjan [...]