How would you compute this? I am sure there is some (relatively) simple trick that I am missing.
I wrote a computer program to simulate the situation and came up with a probability of just under 9%, so I am assuming the 1 in 10 answer is the one they are looking for. I don’t know how you would actually calculate it though.
A: 1 out of 2.
B: 1 out of 10.
C: 2 out of 50.
D: 1 out of 100
E: Zero.
E is out. Obviously it is possible that you get the right seat.
Let’s say passengers and seats are both numbered 1 to 100, and passengers get on starting with number 1, then number 2, etc… You’re passenger one hundred.
Probability that passenger 1 sits in any given seat: .01 .
Probability that passenger 2 sits in any given seat other than their own: .01 * .99 = .0099
Probability that passenger 3 sits in any given seat other than their own: (.01 + .0099) * .98 = … now it gets hard.
Let me think on it for a while.
Well, it’s not very interesting if the first passenger sits in your seat, is it?
The probability that the first passenger doesn’t sit in your seat is 99/100. The probability that the first two passengers don’t sit in your seat is 99/100 * 98/99. The probability that the first three passengers don’t sit in your seat is 99/100 * 98/99 * 97/98.
See the pattern?
I suspect the answer is:
1 out of 2
Because:
There are two important seats, the person who dropped his ticket’s seat, and your seat. If the dropped seat, or your seat is ever taken, then each of the subsequent boarders will take his own seat. Naturally that means that there are only two seats that might be available at the end – the dropped seat, or your seat.
Hmm. By buying a ticket, you’re randomly selecting one seat.
There are 99 full seats and one empty seat. If you randomly select a seat there is a 1/100 chance that you will select the empty seat.
D: 1/100. QED.
Whoops… :smack:
Teach me to read the problem carefully. That’s where I lose the majority of my lost points in math class.
For the answer Tom wanted, he should’ve specified that every passenger takes a seat at random. With the stipulation that a passenger sits at random only if their seat is taken, the problem may be more complicated.
Wait.
At any given time, there is a line of people outside the gate. Of those people, at most one has their correct seat filled by someone else at any given time. But what we need to know is what is the probability that one person has their given seat filled, and what is the probability that the number of such people is one, and what is the probability that it is zero?
Think of it this way. Suppose we had the same problem, but only two people were on the plane. Then the answer is .5, because the probability that the dropped ticket guy takes your seat is .5 .
Now suppose the plane has three people. The dropped ticket guy has a 1/3 chance of taking your seat. The second passenger has a (1/3)(1/2) chance of taking your seat. So the answer, is that there’s a 1/3 + (1/3)(1/2) chance that your seat gets taken before you show up. That adds up to .5 again.
Now suppose four people. There’s a 1/4 chance that the dropped ticket takes your seat, a (1/4)(1/3) chance that the second passenger takes your seat, and a (1/4)(1/2) + (1/4)(1/3)(1/2) probability that the third passenger takes your seat. And that all adds up to .5 .
So given those results, it seems likely to me that the answer for any number of seats (other than one seat) is always .5 . But I might be wrong.
Nope, it is simpler.
Not necessarily. If the person who dropped their ticket takes the seat of the next-to-last passenger to get on, that person won’t be sitting in their seat.
Probability first person DOES NOT sit in your seat = 99/100, second person 98/99, third person 97/98, and so on, last person 1/2.
So that’s (n-1)!/n*(n-1)* … * 2 = 1/100.
I don’t have a nice closed form solution, but I can sketch out a recursive one (I’ll come back later and fill it in if no one else has)
Basically, in the general case, what you want is the probability that at some point in the n people boarding the plane, they will all be sitting in each other’s seats (or all their own). Once the plane reaches that state, every other person will get his own seat, including you.
(Aplologies for the lack of coding)
Let P(i, x) be the probability that the ith person leaves the plane in a state such that there are x people in seats claimed by people not currently on the airplane.
P(i, x) = P(i-1, x+1)x/(n-i) + P(i-1, x)(n-x-i)/(n-i)
(ie, the probability that any given person leaves the plane in such a state is the sum of 1) the probability that the previous person left it with one more outside-claimed seat and this person sat in an outside-claimed seat, thus reducing it and 2) the probability that the previous person sat in any other seat (either his own, or one of somebody currently on the plane), thus leaving the number of outside-claimed seats the same).
We know that
P(1, 0) = 1/n and
P(1, 1) = (n-1)/n,
and we want the sum of P(i,0) from i = 1 to n.
This was totally back-of-the-envelope, but I think that that works.
The chance is 1 in 2. My reasoning is that every time this first man sits in someone else’s seat he’s likely to be asked to move. This just confuses the issue until the 99th passenger arrives. At that point the first passenger can only be in one of two seats, his own or yours.
I’m still thinking about this. My first answer is definitely right if everyone sits at random, but maybe not if some people might not.
However, in that scenario, everyone except for the first person to get on, and the next-to-last person to get on will be sitting in his own seat. Which means the next-to-last person will still have a 50% chance of taking the dropped seat or your seat.
I am glad to see this problem is not as simple as I thought it might be. My computer program may not be correct but it is seating the last person in their assigned seat just under 9% of the time.
Just to clarify, the first person chooses a seat at random and each person thereafter takes their seat if it is not occupied, otherwise they choose a seat at random.
OK, for a 3 and 4 seat airplane, the probability is 1/2. I’ve discovered that by perfect induction (i.e., brute force).
It’s worth noting that there are 4 possible outcomes for a 3-seater, and 8 for a 4-seater. I’m guessing there’ll be 16 for a 5-seater.
My computer program says 1 out of 2. Haven’t derived it analytically yet.
Huh. You’d think the question would be equivalent to how what proportion of the elements of S[sub]n[/sub] fix n, but that’s 1/n, not 1/2.
What’s the chance EVERYONE sits in their assigned seats? The same chance as the first guy is sitting in his, 1/100.
Once the first guy doesn’t sit in his assigned seat, what’s the chances he sat in yours? 1/100.
If he didn’t sit in yours, what’s the chances NONE of the others will sit in your seat?
99/100 * 98/99 * 97/98 * … * 1/2 = (n-1)!/n*(n-1)(n-2) … * 2 = 1/100.