4 days, no anser. I can only conclude that no one's still trying to break it.
let a=highjacker, b=second person, c=third person
For 3 seats:
There are two different scenarios that could take place for the plane not getting cancelled.
p1 => a -> a 1/3
p2 => a -> b, b -> a 1/3*1/2
p1+p2 = 1/3+1/3*1/2=1/3+1/6=2/6+1/6=3/6=1/2
Hence there's 50% risk of it not being cancelled, it's 50% chance for it to be cancelled.
For 4 seats
Scenarios:
p1 => a -> a: 1/4
p2 => a -> b, b -> a: 1/4*1/3
p3 => a -> b, b -> c, c -> a: 1/4*1/3*1/2
p4 => a -> c, c -> a: 1/4*1/2
p1+p2+p3+p4=1/4+1/4*1/3+1/4*1/3*1/2+1/4*1/2=6/24+2/24+1/24+3/24=12/24=1/2
And again; Hence there's 50% risk of it not being cancelled, it's 50% chance for it to be cancelled.
(3!/4!+2!/4!+1!/4!+3!/2*4!=(2*3!+2*2+2+3!)/2*4!=(3*3!+2*3)/2*4!=(3*3!+3!)/2*4!=4!/2*4!=1/2)*
Seems to function for both 3 and 4 seats, and you will get the same answer if you try with 5, 6, 7... but I'm not going to do them, since the calculation is too long. Though, the induction would be something like p!/(2*p!)=1/2, where p!=1*2*3*...*(p-1)*p. Look at the solution marked with*
Though, this is not an evidence, since I have not proven it for every number.
Let´s look at it logicly.
The propability of the highjacker getting seated on his own seat is just as big as the propability of him getting seated on the CSI-agent´s seat.
The same goes for the next person; if his seat's taken, the propability is just as big as him getting seated on the highjackers seat as the CSI, and the next the same, and this will all continue until the last person before the CSI-agent gets seated and he has the same condition as the other people; the propability of him getting seated on the highjackers seat is just as big as getting seated as the CSI-agent´s seat which must then be 1/2.
The answer's ultimately 1/2
