A difficult probability question about choosing seats $100$ women board a plane with $100$ seats. Each of them has a seat assigned in advance. For some reason the first woman who gets in takes a seat at random. Then the second passenger takes he allocated seat if it is not occupied (by the first), and picks a seat at random if her own seat is occupied. The third passenger takes her own if not occupied by one of the first two, and a random seat if it is. And so on. What is the probability that the last passenger will sit in her own seat?
I am bad at probability questions, help please.
 A: I'm not sure about the proper mathematics behind it, but I can give an intuitive answer to this: 
This problem is equivalent to the first passenger choosing a random seat, and when the passenger meant to be sitting in that seat arrives, she replaces the first passenger to retake her seat and the first passenger then chooses another random seat.
At any time, she will choose her own seat and the last passenger's seat always with the same probability (ignoring other choices that will be "cancelled" later). Whenever she chooses her own or the last passenger's seat, it's final.  In other cases, the choice will be repeated. Therefore, at the end, the probabilities she will be in her own seat and the last passenger's seat are equal. Therefore the answer is $\frac{1}{2}$.
(help from @PeterFranek)
A: Replacing 100 by $n$, we get a recursive equation $P(n)=\frac{1}{n}+\frac{1}{n}(P(2)+\ldots, P(n-1))$. 
This is because the first one chooses the position of the $k$'th with probability $1/n$ (for any fixed $k$). If he has chosen his own seat, then the last one will sit in his own seat as well: that's the first $\frac{1}{n}$. Otherwise, people with indices $2,3,\ldots, k-1$ will sit in their seats and the $k$th will either choose seat Nr. 1---then all remaining passengers will sit on their places---or some other from the $n-k+1$ remaining seats. This is completely equivalent to the original problem for $n:=n-k+1$. The only exception is $k=n$ in which case the probability is $0$: therefore, $P(n-n+1)=P(1)$ is not included in the sum.
You can prove by induction that $P(1)=1$ and $P(2)=P(3)=\ldots =P(100)=1/2$.
