Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$n$ men are getting on a plane which contains $n+k$ seats. Each one has a seat number but among them, $m$ men forgot his seat number. They get on the plane one by one. For person $X$ if he knows his seat number and if the seat is empty then he take it but if the seat if occupied then he chooses randomly a chair and sits. On the other hand if $X$ does not know his seat number them he chooses randomly a chair and sits. What is the probability that the last $i$ persons sit on their proper seat!? (Those who forgot their seat number are not necessary the first $m$ persons who get on the plane)

COMMENT: Some versions of this problem are posted to SE. You can see the following ones: Number 1, Number 2, Number 3

share|improve this question

1 Answer 1

Let us assume that the $i$ first-class passengers wait in their luxury lounge and will sue the company if they do not find their seat empty.

Let us number the people and their places from 1 to $n$. (The numbers of the $k$ extra seats do not matter.) I will also assume that the $i$ last passengers are disjoint from the $m$ forgetful people.

In that case, each of the $m$ people starts a rising chain of displaced persons among the $n-i$ ordinary passengers. For each pattern of $m$ chains among them, the only thing that interests us what the last persons in the chains do.

There are $\binom{m+i+k}m$ choices for their seats (this includes one forgetful person choosing immediately their own seat), but only $\binom{m+k}m$ choices will make the airline company happy.

So, I get as success probability $$\dfrac{\binom{m+k}m}{\binom{m+i+k}m} = \dfrac{(m+k)(m+k-1)\dots(k+1)}{(m+i+k)(m+i+k-1)\dots(i+k+1)}.$$

share|improve this answer
    
Thanks. But you almost chained all my assumptions!! –  user117432 Jan 8 at 9:04
    
@user117432 I do not understand your comment. –  Phira Jan 8 at 9:22
    
I mean you added some extra (and strong) assumptions to the question that makes it far from what I had in mind. –  user117432 Jan 18 at 5:30
    
@user117432 I still do not know what you are talking about. The only assumption is that the $i$ last passengers are disjoint from the $m$ forgetful people. I did NOT assume that the forgetful people board the plane first. –  Phira Jan 20 at 9:43
    
@user117432 This one assumption is not particularly strong, it is more a question of being a case where it is not clear what outcome you really want to measure as this puzzle is often worded "find their seats empty" which is not the same thing to "sit on their seat". –  Phira Jan 20 at 9:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.