Probability of the $N$th person to get the seat n passengers board an airplane with exactly $n$ seats. The first passenger has lost the ticket and picks a seat randomly. But after that, the rest of passengers will:
Take their own seat if it is still available,
Pick other seats randomly when they find their seat occupied 
What is the probability that the $n$-th person can get his own seat?
The answer is 
$1$, if $n = 1 $
$0.5$ if $n > 1$ 
Following is my attempt
For $n = 2$ 
let's have passengers $A$ and $B$ 
Sample space = {$AB,BA$} 
Favorable = {$AB$} 
Probability = #{$AB$}/#{$AB,BA$}= $\frac{1}{2} = 0.5$ 
For $n = 3$ 
we have passengers $A,B,C$ 
Sample Space = {$ABC,ACB,BAC,BCA,CBA,CAB$} 
Favorable = {$ABC,BAC$} 
Probability = $\frac{2}{6} = \frac{1}{3}$ 
Can anyone explain me in this way ? 
 A: Let us count the number of possible arrangements under this scheme with $n>1$ people, because for $n=1$, the probability that the $n^{th}=1^{st}$ person gets their own seat is $1$, that being the only possibility.
For $n>1$, we want to count the number of scenarios when the $n^{th}$ person will be sitting in his own seat and divide that by the total number of possible arrangements. So, we have seats $S_1,S_2,\cdots , S_n$ reserved for passengers $P_1,P_2,\cdots , P_n$. In our favourable arrangement $P_n$ is occupied by $S_n$.
So, $P_1,\cdots, P_{n-1}$ must have found seats in $S_1,\cdots,S_{n-1}$ only. 
Suppose after entering the plane, $P_1$ chose $S_5$. 
Then $P_2,P_3,P_4$ will have their seats empty and sit in $S_2,S_3,S_4$ respectively. $(*)$
But $P_5$ sees $S_5$ occupied, so he could have chosen $S_1$ to sit on, or one of the seats from $S_6,\cdots,S_{n-1}$. 

Case $1$: If $P_5$ chooses $S_1$, then everyone from $P_6$ to $P_{n-1}$ will find their respective seats empty and sit there, and the only disturbance in seating arrangement (i.e. not sitting in respective reserved places) would be caused by the people with indices $(1,5)$.


Case $2$: If $P_5$ chooses a seat other than $S_1$, say $S_{17}$, then $P_6,\cdots,P_{16}$ follow the same procedure in $(*)$ and  $P_{17}$ has to make the decision similar to Case 1: or Case 2:, and again if $P_{17}$ chose $S_1$ to sit in, then $P_{18},\cdots,P_{n-1}$ sit happily in $S_{18},\cdots,S_{n-1}$, so that the disturbance mentioned in the above block is caused by people with indices in $(1,5,17)$ or we come to another Case 2: with $P_{17}$ in place of $P_5$.

Thus, we see that specifying a subset $A\subseteq \{2,3,4,\cdots,n-1\}$, for example $A=\{5,17,33\}$, the people in the set $A\cup \{1\}=\{1,5,17,33\}$ correspond to the seating arrangement, where $$P_1\rightarrow S_5, P_5\rightarrow S_{17}, P_{17}\rightarrow S_{33}, P_{33}\rightarrow S_{1}$$ and the rest are seating in their own places due to the logic in $(*)$. The empty subset $A=\phi$ corresponding with $A\cup \{1\}=\{1\}$ indicates the seating arrangement that $$P_1\rightarrow S_1$$ so that everyone else is seated in their reserved places.
Thus we have a bijection, each of the favourable arrangements corresponds to a unique subset of $\{2,3,\cdots,n-1\}$, so there are as many favourable arrangements as there are subsets of $\{2,3,\cdots,n-1\}$ (of size $n-2$), which is $2^{n-2}$.
The number of total possible outcomes, is the exact same analysis as above, with the possible seats being $S_1,\cdots,S_n$ for $P_1,\cdots,P_n$ so that we have to consider all the subsets of $\{2,3,\cdots,n\}$, which are $2^{n-1}$ in number.
So the probability for $n>1$ passengers is $\dfrac{2^{n-2}}{2^{n-1}}=\dfrac12$.
A: Let us slightly modify the problem. If the 1st person is sitting in, say, 5th person's seat, when the 5th person comes, they talk and the 1st person leaves, letting the 5th person getting his seat. This situation is similar to the problem, as there is a person who is going in search of a vacant seat, just that the identities of the persons are not the same in the original problem and is always the 1st person in the modified situation. Now let us calculate the probability that $\ {k}^{th}$ person gets his seat(doesn't need to talk with the 1st person to get his seat back). Essentially, if the 1st person sits in his seat, or $\ {k}^{th}$ seat or $\ {i}^{th}$ seat where $\ i>k$, the fate of the $\ {k}^{th}$ person getting his seat or not is decided. So, the number of actually available choices for the 1st person till the $\ {k}^{th}$ boards the plane is $\ n-k+2$. $\ {k}^{th}$ guy gets his seat if the 1st person chooses one of $\ n-k+1$ seats. Hence, probability is $\ \frac {n-k+1}{n-k+2} $.
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A: As you mentioned for 2 people the probability comes 1/2 because 1) 1st person may seat on his seat so that the other person seats on its seat. OR 2) 1st person seats on other 2nd persons seats so the other person gets the seat of the 1st person. Out of this only 1) is favourable as in that case the nth person i.e. the 2nd person seats on its own place.
The same logic can be applied for the 3 persons also.
Say for A,B,C the possible outcomes are 3! But the favourable outcome i.e. the last person seats on his own sit is 2!.(As A,B could sit on their own places or exchange their places explained in 2 person problem)
So similarly applying this for n person, we get
Favourable outcomes: (n-1)!
Total outcomes: n!
Probability= (n-1)!/n!
Simplifying it we get Probability=1/n.
E
If n=2 P=1/2
If n=3 P=1/3 and so on
