grad_mba wrote:

Well .. here is the logic I used,

we just need the probability that the last guy sits in the proper seat - which means the 99 guys before him DO NOT sit in the last guy's allocated seat

now,

1st guy entering - probability that he will not sit in the last guys seat - 99/100

i.e he can sit nywhere in the 99 seats but for the last guys seat.

similarly 2nd - 98/99 ; 3rd - 97/98 and so on

Finally it will be 99!/100! = 1/100

how do we get 1/2 - what is it I am missing here ?

The problem starts with the first guy. According to the problem everyone else will sit in his seat if it is available.

So the probabilty of the first guy sitting on his on seat = Probability of the last guy getting his allocated seat = 1/2.

probabilty of the first guy sitting on his on seat = 1/2 (he either sits or does not sit. It is not necessary to consider where he sits if he picks the wrong seat.)

Lets not consider 100 seats for the time being. Let there be just 2 seats. Then, the probability of the last guy (2nd here) getting the seat is 1/2. Agree?

Now, consider 3 seats. If the first guy sits in the second guy's seat then only there will be a need to calculate the probabilty (other cases would represent absolute events - either the last guy gets his seat or he doesn't). And this probabilty will be 1/2.

By induction we can extend this for any number of seats and the probabilty will remain 1/2.