Comments on: You are in my Seat!

By: yaniv

yaniv — Sun, 20 Mar 2011 19:56:32 +0000

Hi Jim,
Nice guess, but it is not right.
If you want to see some solutions, switch to the second page.

By: Jim

Jim — Sun, 20 Mar 2011 19:13:22 +0000

1 in 100

By: Jim

Jim — Sun, 20 Mar 2011 19:12:34 +0000

1 in 100 ?

By: srulix

srulix — Fri, 05 Oct 2007 16:22:03 +0000

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.

By: yaniv

yaniv — Thu, 30 Aug 2007 14:36:33 +0000

Dan,

Very nice solution!

By: Dan

Dan — Thu, 30 Aug 2007 13:20:36 +0000

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.

By: yaniv

yaniv — Sat, 11 Aug 2007 20:36:34 +0000

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.

By: Ronnie

Ronnie — Thu, 09 Aug 2007 20:34:12 +0000

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!