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100 Seats and 100 passengers
Posted: January 6th, 2007, 11:54 pm
by liuliudada
100 passengers are boarding on a plane. They all got their own unique seats. The first passenger to enter the plane is blind. He will randomly pick a seat. For all other passengers, he will sit on their own seat, but if his seat is taken he will randomly pick another seat. What is the probability of the last passenger to be seated in his seat?
100 Seats and 100 passengers
Posted: January 7th, 2007, 5:34 am
by sevvost
1/2
100 Seats and 100 passengers
Posted: January 7th, 2007, 5:39 am
by mj
i make it 0.5
100 Seats and 100 passengers
Posted: January 7th, 2007, 11:05 am
by Msccube
This problem can be solved by recursion.
100 Seats and 100 passengers
Posted: January 7th, 2007, 8:29 pm
by mj
this problem can be done in about 3 lines without recursion
100 Seats and 100 passengers
Posted: January 8th, 2007, 11:03 am
by Msccube
QuoteOriginally posted by: mjthis problem can be done in about 3 lines without recursionYes, I agree.The xth seat cannot be empty, where x is not equal to 1 or 100, otherwise the xth guy will sit on it.He can only sit at the 100th or the 1st one. The prob of 1st seat is empty = the prob of the 100th seat is empty by symmetry,So P =0.5.
100 Seats and 100 passengers
Posted: February 23rd, 2007, 1:22 pm
by alexfore
That is not logically correct. Let us just consider the guys who randomly pick a seat:- if he sits in 1st seat, then last guy gets his seat.- if he sits in last seat, then last guy doesn't get his seat.- if he sits anywhere else, then we have the same argument for the next guy who picks at random.now the prob of guy choosing the last seat is the same as the prob of him choosing 1st seat. and eventually someone wil choose one of the two. thus is it 1/2. you cannot invoke a symmtery argument because the way the plane is board breaks that symmetry. i.e. the prob that 1st guy is in his seat is 1/100 but the prob that the last guy is in his seat is 1/2.
100 Seats and 100 passengers
Posted: February 23rd, 2007, 10:27 pm
by Msccube
QuoteOriginally posted by: alexforeThat is not logically correct. Let us just consider the guys who randomly pick a seat:- if he sits in 1st seat, then last guy gets his seat.- if he sits in last seat, then last guy doesn't get his seat.- if he sits anywhere else, then we have the same argument for the next guy who picks at random.now the prob of guy choosing the last seat is the same as the prob of him choosing 1st seat. and eventually someone wil choose one of the two. thus is it 1/2. you cannot invoke a symmtery argument because the way the plane is board breaks that symmetry. i.e. the prob that 1st guy is in his seat is 1/100 but the prob that the last guy is in his seat is 1/2.The prob that the 1st guy/lady choose the 1st seat is equal to the prob that he/she chooses the 100th seat. This is how the symmetry argument comes in.
100 Seats and 100 passengers
Posted: February 27th, 2007, 2:25 pm
by Advaita
Here's how in some more visual detail:suppose we have 10 seats._/_/_/_/_/_/_/_/_/_/now the prob(1 can sit in his seat) = prob(he sits in the last seat)1/_/_/_/_/_/_/_/_/_/or _/_/_/_/_/_/_/_/_/1/the REMAINING times, he sits in X's place. Suppose X = 7,Now 2,3,4,5,6 all sit in their respective seats. Mr. X takes any of the remaining ---_/2/3/4/5/6/1/_/_/_/We now have a 4-SEAT PROBLEM *with Mr X taking Mr 1's randomness responsibilities*, and we apply the symmetry again to seat 1 and seat 10, as is there were FOUR remaining people and seat, and FORGETTING altogether about Mr. X, 2, 3, 4, 5, 6.
100 Seats and 100 passengers
Posted: March 10th, 2007, 1:54 pm
by Guest
My approach is tedious one. Consider this as a two state Markov Chain, good or bad. Effectively we need to find out the prob of bad (or good) at the end of chain. The transition prob from bad state at last nth passenger to bad state at last n-1-th passenger is P(n-1)=P(n)*(1-1/n^2), Note P(N)=1-1/N, so P(1)=P(N)*(1-1/(N-1)^2)*...*(1-1/2^2)=1/2. The anwser does not depend on N.