# 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+.

• ok, but you need to specify what happens in the case when seat 1 is actually taken... What is the probability that a subsequent passenger will choose a seat after seat 1 is taken? Commented Mar 31, 2021 at 17:21
• @DinosaurEgg If seat 1 is taken, then the remaining passengers will sit in their correct seats. Commented Mar 31, 2021 at 17:26
• this involves assuming that the original seat assignment for the first passenger is seat #1. Usually plane seating problems do not assume that. Commented Mar 31, 2021 at 17:29
• @DinosaurEgg I use seat #1 to denote the seat the first passenger is assigned... Commented Mar 31, 2021 at 17:30

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 $$p=0.9$$, $$P(100)\approx 0.998304$$.

• For $p = 0.5$, I am getting $P(100) = 0.943$ with your expression. Shouldn't it be $P(100) = 0.5$? Commented Apr 1, 2021 at 1:34
• No, the last formula is not reducible to the classical airplane seat puzzle. You would have to use formula number 2 above (which reduces to $P(n)=P(n-1)$ after some manipulation). When setting $p=0.5$ you basically assume that seat 1 has probability $1/2$ to be chosen and all the other seats $1/2(n-1)$. The OG version of the puzzle has uniform distribution of probability $1/n$. Commented Apr 1, 2021 at 2:35
• Oops you're right, I brain farted earlier. I should set it to $p = 0.01$, in which case I obtain $P(100) = 0.05$ Commented Apr 1, 2021 at 2:49
• That doesn't reduce to the problem either. Indeed, the first passenger will have the correct probability distribution but any subsequent passengers that get displaced will NOT. The two formulas are not connected, it's futile to try to check one through the other Commented Apr 1, 2021 at 2:54