July 25th, 2007
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August 9th, 2007 at 11:34 pm
the answer is 1/2 (I think)
let A(n) be the corresponding event for n people (in the text n = 100)
A recurrence equation for the probabilites : P(A(n)) = 1/n* (p(A(2)+.. P(A(n-1)) + 1).
since P(A(2)) = 1/2 (…) , the result follows by induction.
now I’d love to hear a combinatorical explanation!
August 11th, 2007 at 11:36 pm
Ronnie,
Your solution is absolutely right.
A slightly simpler way of putting it: Let A be the event of the girl sitting in the seat of the last man, or in her seat. Let B be the event that the girl sits in any of the other seats. In case of A the probability of the last man sitting in his own sit is obviously 1/2. In case of B, the probability is 1/2 as well, by induction. Thus the unconditional probability is 1/2.
August 30th, 2007 at 4:20 pm
Another solution: After man #i sits down, his seat is definitely taken (because if it wasn’t, he will sit in it). Therefore after the first 98 men sit down, all of their 98 seats are taken, so the seat left free is either #99′s or the girl’s. As none of the other passengers know the difference between those two seats when they decide where to sit, it is obvious that either one will be left vacant with equal probability.
August 30th, 2007 at 5:36 pm
Dan,
Very nice solution!
October 5th, 2007 at 7:22 pm
you can also think of it this way: the solution remains the same if i say that the woman first sits, and when the owner of that seat comes, he makes the woman stand up and go somewhere else (kind of like the ant riddle).
therefore, we could say that the woman keeps drawing random seats, the question is whether she’ll hit the vacant seat or the last man’s seat first. because of the symmetry of the situation, the probability is exactly 1/2.