There's a hundred seats on a plane. Every passenger has an allocated seat on the plane, but the first man on is French, so he has jumped to the front of the queue and sits in a seat at random. Everyone else sits in their own seat. However, being British and not wanting to cause a fuss, if there is somebody sitting in their seat, they sit in another seat at random. What is the chance that the last person on the plane will sit in their correct seat?
A MathS test I'm positive only 1% of you will pass
- Thread starter Tastycles
- Start date
Viola Moonlight
I'm Literally Just Here for WZCW
If a French man takes a British man's seat on an airplane, then the British guy isn't going to let that tobacco-smelling wine-loving bastard get away with it. I've got enough real life experience to realise that British people don't appreciate me when I inform them of my French heritage.
Close to 100% assuming a nasty little trick question involving stewards. 50% otherwise.
Hulk Hogan's Brother
Stop asking me what I'm gonna do!!!
50%. There are only two outcomes possible after all.
Is that the WWE dynamic? There's 4 guys in a title match so logically they all have a 25% at winning?
99% chance of sitting in the right seat. There's only one french guy and all the non-brits sit in their correct seats. Meaning only one brit would sit in the wrong seat. That one person would be taking the seat belonging to the frenchman. The thing to think about is if the frenchman's seat is the last one left.
Constipated Badger Bottom
Pre-Show Stalwart
It's a trick question because the last person on the plane is Scott Stiener so by applying the laws of Stienermatics there is a 137.6% chance that Steiner will sit in the correct seat
If you listen to that promo, Steiner actually did those calculations correctly.
Captain Morgan Freeman
Taught Elvis how to play karate
If the last "person" on the plane is a steward and therefore not a "passenger" then it is 100%.
Plane crew is always in first.
Captain Morgan Freeman
Taught Elvis how to play karate
I'm sticking by my answer unless Tasty clarifies that the last "person" on the plane was in fact a "passenger".
Constipated Badger Bottom
Pre-Show Stalwart
Depends on the Airline, on some KLM flights I've been on there has sometimes been one of the stewards standing just outside the plane greeting the passengers so yeah we need some more details from Tasty
Hulk Hogan's Brother
Stop asking me what I'm gonna do!!!
There is only one seat left by the time the last person enters. Either it is the correct seat or it is the wrong one. Two possible outcomes, one correct result.
It cannot be 99% in any case. It can be 100 percent if you can prove that the seat left will be the right one, it will be 0 percent if you can prove that the seat left will be the wrong one. If you cannot do either, it is 50%.
Awesome_Miz
Proud LSN mentee
I will have to go with 33.33%.
Obviously there’s a 1% probability that the first passenger chooses his own seat, in which case the last one automatically gets his. Then there’a a 1% chance that the first chooses the second passenger’s seat, who then has a 1/99 chance of choosing her own, or a 100% chance of this all getting way too complicated.
So let’s look for something not quite so obvious. There are various ways of solving the problem, but the one I’m interested in uses the fact that what we’re really looking at is a permutation!
This also has the property that the sequence of
only increases. We need the probability that
is not 100.
Let’s say that it is. Then we have the cycle
. We can stick the transposition
after it to give
. This is a valid cycle that doesn’t end with 100.
What if we have a cycle that doesn’t end with 100? If we write it as
we can again stick the transposition
after it. Now this gives
, which does end with 100.
Even better, if we start with any cycle, ending with 100 or not, and do both transformations we get back the exact same cycle we started with. We say that the two transformations are inverses of each other, which just means that each undoes what the other does. A fancier word for an invertible function between two sets is “bijection”.
So we’ve split all valid sequences into ones where 100 gets his own seat and ones where he doesn’t, and shown that there’s a bijection between them. That means that the two sets have the same size! So picking a sequence at random we find that in exactly half the cases 100 will get his own seat.
PIPEBOMB!
So let’s look for something not quite so obvious. There are various ways of solving the problem, but the one I’m interested in uses the fact that what we’re really looking at is a permutation!
This also has the property that the sequence of
Let’s say that it is. Then we have the cycle
What if we have a cycle that doesn’t end with 100? If we write it as
Even better, if we start with any cycle, ending with 100 or not, and do both transformations we get back the exact same cycle we started with. We say that the two transformations are inverses of each other, which just means that each undoes what the other does. A fancier word for an invertible function between two sets is “bijection”.
So we’ve split all valid sequences into ones where 100 gets his own seat and ones where he doesn’t, and shown that there’s a bijection between them. That means that the two sets have the same size! So picking a sequence at random we find that in exactly half the cases 100 will get his own seat.
PIPEBOMB!
shattered dreams
Hexagonal Hedonist
1/n + 1/n * .5 * (n-2) gives the answer where n is the number of seats on the plane (when n is 2 or greater). It will = .5 or 50% for any n. Although I'll assume that since the guy is French he isn't crazy but simply a douche and is actually choosing any of the 99 seats that isn't his, in which case the answer is approximately 49.5%.
How the hell can it be 50% if there's 100 seats and one wrong one?
shattered dreams
Hexagonal Hedonist
Do it for a plane with two seats and then one with three seats killjoy and see if you start to see what is happening.
Guess I saw the question a little differently.
Order of Events in question:
1st: French guy was 1st, he took the wrong seat.
2nd: Everyone else takes the right seat.
3rd: British guy takes the wrong seat.
From the way the question is worded, there is a 0 percent chance of the last guy (british guy) is getting the correct seat.
Otherwise, if the question was simply worded incorrectly, and looking for a mathematical answer, the answer is: 98.9898...% chance that someone other than the British guy takes the last seat.
While there is only a 50/50 outcome regarding the last seat, the fact that there is 98 times the likelihood that the last seat is the correct seat increases the outcome over 50/50.
Another way to view the question: If I have a deck of card consisting of 98 regular cards and 2 jokers, if the first card is a joker, what is the chances of the last card being a joker as well.
Considering that this question comes from a job interview, I would guess it was a logic question not an actual math equation.
Order of Events in question:
1st: French guy was 1st, he took the wrong seat.
2nd: Everyone else takes the right seat.
3rd: British guy takes the wrong seat.
From the way the question is worded, there is a 0 percent chance of the last guy (british guy) is getting the correct seat.
Otherwise, if the question was simply worded incorrectly, and looking for a mathematical answer, the answer is: 98.9898...% chance that someone other than the British guy takes the last seat.
While there is only a 50/50 outcome regarding the last seat, the fact that there is 98 times the likelihood that the last seat is the correct seat increases the outcome over 50/50.
Another way to view the question: If I have a deck of card consisting of 98 regular cards and 2 jokers, if the first card is a joker, what is the chances of the last card being a joker as well.
Considering that this question comes from a job interview, I would guess it was a logic question not an actual math equation.
