Most of you would be familiar with the plane seating problem. See this if not
I was trying to solve a similar problem. What is the probability that the $i^{th}$ chair is not occupied, after $k$ people are seated? ($k<i)$
I will use the notation $P(i,k)$ for denoting the probability.
I tried to write a recursive relation and came up with this: $$ P(i,k) = P(i,k|i,k-1)*P(i,k-1)$$
Now for writing P(i,k|i,k-1) which encapsulates one trial where the $k^{th}$ person comes and picks a seat, I considered two cases naturally: If chair $k$ was picked earlier and if it was not.
If chair $k$ was picked earlier, then probability that chair $i$ is not picked is $\frac{n-k}{n-k+1}$. This is because there are $n-k+1$ remaining chairs after $k-1$ people are seated.
If chair $k$ was not picked earlier, then probability that chair $i$ is not picked is 1, as the person will pick the assigned chair.
So summing it up $$ P(i,k | i,k-1) = P(k,k-1)*1 + (1-P(k,k-1))\frac{n-k}{n-k+1} $$
So the final recursion becomes $$ P(i,k) = P(i,k-1)(P(k,k-1)*1 + (1-P(k,k-1))\frac{n-k}{n-k+1})$$
Now I use some intuitive reasoning to claim that $P(m,k) = P(n,k)$ for all $m,n$ where $m,n > k$. The logic is that for all the $k$ people coming, seats $m$ and $n$ are symmetric. They are part of a random pool of other seats from which they choose from if their seat is occupied. They only have a preference to their seat. So the probability that 3rd seat is unoccupied after 2 people are seated is the same as the probability for the 4th seat. So using this fact the recursion becomes linear in space to be : $$ P(i,k) = P(i,k-1)*(P(i,k-1)+(1-P(i,k-1))\frac{n-k}{n-k+1}) $$
However the answer comes out to be wrong. I checked for the case (100,99) (Probability that 100 is unoccupied after 99 are seated) which should be equal to 0.5. Can someone find the flaw (or flaws) in the argument above?
