Taking Seats probability question tweeked I was just thinking about a possible change to the taking the right seat probability question here. I can see how this works when there is 1 passenger with the wrong pass through symmetry. However, what if there were two passengers with the wrong passes, namely the first and the second? Would the probability of the passenger sitting in the 100th seat getting his seat be 0.25: As it would be the probability that both of the two passengers displaced by these two don't sit in seat 100, which is $0.5^{2}$? 
 A: If two passengers board the plane and have lost their passes and don't know their seat number, then by the time the last passenger boards we could be in the following situation:


*

*The first and second passenger are in their correct seats and so the 100th passenger gets the right seat

*The first and second passenger are in each others seats and so the 100th passenger gets the right seat

*Passenger two is in the 100th passenger's seat and the first passenger is in his seat

*Passenger one is in the 100th passenger's seat and the second passenger is in his seat

*Passenger two is in the 100th passenger's seat and the first passenger is in the second passenger's seat

*Passenger one is in the 100th passenger's seat and the second passenger is in the first passenger's seat


Only two of these scenarios will result in the last passenger getting his correct seat. Therefore the probability is $\frac{1}{3}$.
A: There are only 3 seats that matter, 1,2, and 100 and passenger 100 wiil get her seat only if it is the last of the 3 seats to be occupied. 
Since each seat has equal probability of being occupied last, Pr = $\dfrac13$
Note that who occupies seats 1 & 2 is irrelevant. It may be passenger 54 occupying seat 1 and passenger 35 occupying seat 2, as long as seats 1 and 2 are filled before seat 100, the 100th passenger will get her seat.
