There’s a departure gate at an airport. Outside the gate there is a plane that has 256 passenger seats. At the gate lobby there is one *silly old lady*, one *grumpy old man*, and 254 *energetic businessmen* waiting to board the plane. Every passenger has a ticket that indicates their assigned seat.

*Silly old lady* is a bit absent-minded. When it comes her turn to enter the plane, she randomly chooses an empty seat and sits down, regardless of what her ticket indicates.

*Energetic businessmen* are efficient. They find their assigned seat easily and sit down after they’ve found it. But because they are also polite, they will randomly choose some other seat if they find that their assigned seat is already occupied by someone else.

*Grumpy old man* is… well… grumpy. He sits tight in the lobby while others stand in the line, and only after everyone else have boarded will he go into the plane as the last passenger. And boy, won’t he raise hell if he sees someone occupy his seat when he comes into the plane. I mean, after all, he’s paid full price for the flight and has the right to demand flawless service.

And so the boarding begins.

**What is the probability that the grumpy old man was satisfied with the service after the boarding is over?**

Post your answer and reasoning as a comment. I will post the correct answer by some time on Sunday if not already answered.

Kreeta Palo(08:34:53) :The chance of the old lady finding her own seat is 1/256. If that happens, the bisnesmen go to right seats and the grumpy man gets his empty seat.

But of course if there is some snooty flight attendant, who is not polite enough to the grumpy man… I’d say the propability for him being complitely satisfied is about one out of a million. ;)

rlankine(09:14:00) :>”The chance of the old lady finding her own seat is 1/256″

Hmm… but what if the old lady is NOT the first passenger to enter the plane?

Al Nikolov(19:02:35) :The answer is 1/2, believe it or not. But I’ve used an inductive approach, so the full explanation would be rather long. Let’s see if you have a compact one.

First, let’s say the lady (L) is entering right before the man (M), so there are only 2 seats left (i=2). It’s obviousely 1/2. The i number isn’t actually relevant, since if it is, the answer could be different for different i, but the puzzle supposes just one solution, so we can stop here now. 8)

Anyway, add a businessman (B1) between L and M (i=3). Both “outer” variants are 1/3: L found the right place, or L occupied M’s place. But there also two “inner” variants, when L occupies B1’s place. Both symmetrically resolve to 1/3*1/2, since B1 might take a L’s or M’s place. 1/3+1/3*1/2=1/2.

Add another B2 (i=4). Outer: 1/4. Intermediate (one level): 1/4*1/3. Inner (two levels): 1/4*1/3*1/2. Outcome is always symmetrical.

rlankine(22:53:25) :Spot on, Al, as it could’ve been expected from you. Like you suspect, there is a compact and intuitive solution as well. I’ll post it on Sunday like I planned originally.

rlankine(23:08:55) :… hmmm… on the second thought, with this…

>”The i number isn’t actually relevant, since if it is, the answer could be different for different i, but the puzzle supposes just one solution, so we can stop here now.”

… I don’t completely agree. Because getting 1/2 as a result for i=2 does not guarantee that all i=3..256 share the same result. They happen to, yes, but had they not, then the real answer would have been the weighted average for i=2..256 , no?

Al Nikolov(23:35:09) :I mean, the answer doesn’t depend on the order in which L enters the plane between Bs (doesn’t depend on i). It is proven by the mathematical induction, but the puzzle itself suggests it, this is how I feel. I don’t insist, though.