Puzzle: 100 passengers are waiting to board a flight having 100 seats. Each of them is having a ticket with a particular seat number which is same as their number in the queue. Unfortunately the first person in the queue is little crazy and he can occupy any random seat. Now, all other passengers will see if their seat is occupied by that crazy guy or not? If not, then they will be seated on their own seat numbers otherwise they will also look for a free random seat to sit on. This continues. What's the probability that the 100th person gets his own seat to sit on?
Solution: The crazy guy can occupy any seat randomly, hence the probability of him occupying his own seat = 1/2 (and not 1/100 as there will only be two cases - either he sits on his propely allocated seat or not... rest of the 99 seats can be viewed as only one case here). Similarly, the probability of the second person in the queue getting his own seat is again 1/2 as either the crazy guy would have occupied his seat or not.... and this continues. For any other person in the queue, his eat will either be occupied either by the crazy guy (or some other guy as that guy's seat was already occupied) or free for him to sit on. So, a seat will either be Free to sit on OR Occupied. That's it. Only two cases.
Hence, the probability of the 100th person occupying his own seat is also 1/2 only. Irrespective of which seat the crazy guy sits on, all the other passengers will have only two possible choices - either their own seat or any other random seat. The key here is to visualize all the random choices as only only one case. Confused? You can take an example of 3-4 passengers and probably that can help you understanding the underlying idea.