Taking seats on a plane - with biased probability This question is an extension of Taking Seats on a Plane.
The extension is that a displaced passenger has a 0.9 probability of sitting in seat 1. Now what is the probability that the last passenger sits in seat 100?
I tried to apply the same logic as in the original problem, which in the original problem, was that at every step, seat 1 and seat 100 have equal probability of being chosen, so the answer is 0.5. But I think that logic fails here because the ratio between the two probabilities changes at each step. e.g., for the first person, they have a 0.9 probability of choosing the first seat, and a 0.1/99 probability of choosing the 100th seat. If they chose, e.g., the 5th seat, then persons 2-4 will sit in the correct seat, and now the 5th person has a 0.9 probability of choosing the first seat and a 0.1 / 95 probability of choosing the 100th seat.
So I am confused on how to solve this problem in a simple way, and was hoping someone can suggest an analogous approach.
I ran a simulation, and it appears the answer is 0.999+.
 A: Assume that the probability of picking the seat belonging to passenger #1 for every displaced passenger is $p$, and that they pick one of the remaining seats uniformly.  Inspecting the first few cases carefully, a pattern emerges concerning the quantity $P(n)$ denoting the probability that the last passenger obtains her assigned seat, given that there are $n$ total seats in the plane. This quantity satisfies the recursion relation
$$P(n)=p+\frac{1-p}{n-1}\sum_{k=2}^{n-1}P(k)$$
For comparison, remember that in the traditional puzzle it satisfies a similar recursion
$$P(n)=\frac{1}{n}+\frac{1}{n}\sum_{k=2}^{n-1}P(k)$$
which for $P(2)=1/2$ indeed forces $P(n)=1/2, n>2$.
The first recursion relation can be recast in the form of a first order recursion relation:
$$(n-1)P(n)-(n-1-p)P(n-1)=p$$
which is in principle completely solvable in terms of sums. An analytical expression can be given in terms of gamma functions (mostly for notational simplicity)
$$P(n)=p\frac{\Gamma(n-p)}{\Gamma(n)}\sum_{k=2}^n\frac{\Gamma(k-1)}{\Gamma(k-p)}$$
A fairly short calcualation in Mathematica shows that for $p=0.9$, $P(100)\approx 0.998304$.
