Answer is 1/2.

Let F(N) be the probability of last person getting the right seat if there are N seats left, and the last person’s seat is still unoccupied. Note that this is the initial condition, before first person has sat down.

If the first person sits in his right seat then last person gets their right seat. The probability that first person sits in his right seat is

1/N  –>(1)

If the first person sits in the last person’s seat then probability of last person sitting on their seat is zero.

1/N x 0 = 0 –>(3)

If the first person sits in one of the (N-2) remaining seats, then the probability of last person sitting on their right seat is dependent on which of the N-2 seats did the first person sit on. Did he sit on seat belonging to persons just behind him, or two person behind him, three person or N-2 persons behind. Each of which will produce a different result. Mathematically reducing it, it comes out to be

(N-2/N ) x (SUM of F(i) where i=2, N-1) x 1/(N-2)

1/N x (SUM of F(i) where i=2, N-1) –>(2)

Bringing them all together,

F(N) = (1) + (2) + (3)

F(N) = 1/N + 1/N x (SUM of F(i) where i=2, N-1)+ 1/N x 0

Reducing it we get,

F(N) = 1/N + 1/N x (SUM of F(i) where i=2, N-1) –> (4)

IF F(k)=1/2 where k=2, N, using induction we can prove that F(k+1) is also 1/2. –>(5) (induction proof given below)

We know F(2) = 1/2.  That is, if there were only two passengers, the probability that the last person will get the right seat is 1/2.

Using (5) F(3)=1/2, F(4)=1/2.. and F(100)=1/2.

Proof by induction of (5).

Assuming F(j)=1/2 where j=2, k , we will prove that F(k+1) is also 1/2. 

F(k+1) using (4) is

F(k+1) = 1/(k+1) + 1/(k+1) x (SUM Of F(j) where j=2, k)) –> (7)

We know,

(SUM Of F(j) where j=2, k)) = 1/2 x (k-1)  

as F(j) for j=2, k is 1/2

Replacing that in above equation (7) we get,

F(k+1) = 1/(k+1) x (1 + (k-1)/2)
                = 1/(k+1) x (k+1)/2
                = 1/2