crazy guy on the airplane If the last person can't seat in his seat, the only seat he can sit on is the first seat. So the possibility he can't seat in his seat is equal to the possiblity that nobady chooses the first seat. the possibility is 99/100*98/99*97/98...1/2=1/100. So the possibility he will seat in his seat is 1-1/100=99/100
yong
Actually, the answer is that the last guy has a 50% chance of sitting in his assigned seat.
peter
The answer is 1/100. Assumptions are that all seats are equally likely of being chosen, with a single trial being one person choosing or not to sit in the first seat, and success being the that person sitting in that seat, and a failure being that person not sitting in that seat. Then the probability that the seat will be available after 99 other people have chosen seats (after 99 independent trials, each with a failure result, calculated using permutations) is 99/100 * 98/99 ... * 2/3 * 1/2 = 1/100.
Mark Benson
Isn't the solution a 1/2? For instance the last explanation says the number permutations in which the last passenger will be seated in his assigned seat is (n-1)! ,(where n is the number of seats) and the total number of ways that passengers can be seated is n!. This does not take into account the fact that a passenger will sit in his assigned seat if it available. For instance suppose n = 3.
Jeff
There are two ways that I determined that the answer to this problem is 50%: first, I modeled the problem on the computer; second, consider a simpler version of this problem where there are only two seats--which is easy to solve, then apply the same process to higher numbers of seats. You should be able to see that the answer for n seats is recursively related to the answer to the ( n-1 ) seat problem. Done.
peter
Well here is a formal proof:
Jeff
The answer is 1/2.
Aryabhatta
Simpler solution:
LUís Pedro Coelho
Reading everybody's post, I think 1/2 is the correct answer.
VikasRana
Based on how I interpret the question, the answer is definitely not 1/2. The first guy is randomly choosing his seat - so there's a 1/100 chance he chooses his assigned seat and so everyone boards property and passenger 100 gets their seat with a 100% probablity (1/100 * 1). There's also a 1/100 chance the first guy chooses the 100th passenger's assigned seat, so there's no chance of getting their assigned seat (1/100 * 0). The other 98 cases are not equally weighed. For instance, if the first guy selects seat 99, the first 98 passengers all board into their assigned seats and the probability that the last passenger gets their assigned seat is 1/2 (passenger 99 either sits in seat 100 or seat 1) (1/100 * 1/2). This pattern repeats to give us: (1/100 * 1) + (1/100 * 1/2) + (1/100 * 1/3) ...
tom
it's one in two. When the last passenger boards the plane the available seats can only be one of two. Either seat 1 or seat 100. Whatever happens with the other 98 seats is irrelevant; the point is that at any stage there is an equal chance of a person choosing between one and a hundred (of course in real life which end they enter the plane from will skew it) Whenever this happens then all other passengers go to the correct seats.
Stephen Jones
Tom extrapolated a bit too quickly. Let see again.
VikasRana
Thanks VikasRama - I now see the flaw in my logic (I suspected there was one, if the numbers aren't nice...it's probably not a cool question ;-). The comment about "order will be restored within the black box" turned the light on for me. I was looking past the fact that some passagers will still have their assigned seat open even after previous passengers have had to choose a new seat. Thanks :-)
tom
Here's an explanation that doesn't involve complicated mathematics.
Marc Peabody
The answer is 1/4. Here is my logic.
Rogier Wester
You have done all calculations assuming that person #1 doesn't take his seat. What is given is the fact that person #1 is crazy and he MAY or MAY NOT take his own seat. You have completely ignored the fact that he MAY occupy his own seat. That's why you answer is half of the one proposed earlier.
VikasRana
Ok, so it looks like Marc has the right idea (as well as everyone else who said 1/2). Although it seems wierd, it is correct. I had to check it out, so I wrote up a quick C++ program to simulate it using random numbers, and it seems to be normally distributed around .5. Good work everyone!
Tyler
I have learned a lot from you guys. Here I want to consider it from the view of graph. As pointed by Aryabhatta, we can view this problem as a graph problem. Each feasible seating corresponding to a graph with the property that it contains only one cycle that begin form point one, each point on the cycle is smaller then his son, except the last one. We can divide these graphs into two sets: one that includes 100 and the other does not. Then we can see there is a 1-1 correspondence between these two sets: If a cycle does not contain 100, we can insert 100 to get a cycle in the other set and vice versa. In fact for the graph without cycle, that means every one sits on his own seat, we can think the point 1 is a self-cycle, and we can insert 100 to get a cycle, which corresponds to the case that 1 sits on 100 and 100 sits on 1. This shows that there are exactly half cases that 100 sits on 100. So the probability is ½.
shaoji
The answer is definitely 1/2
Paul Bethe
As a real proof it is two step.
Paul Bethe
I think the answer is 1/100.
Shub
The only solution is 50%. All it depends on the seat1 person. Whether he sits in his seat or not? is the only the probability to be considered.
PradeepKumarReddy
Why not try to solve it with boolean logic?
daniel f. martin
I find it amazing that people who claim the solution is a real number other than 1/2 do not take a couple minutes and work out simple cases to see that they have a flaw in there in logic.
Jeff
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