3 or more empty seats in a row of n seats and m occupied seats?

Question

we have a row of $n$ seats and $m$ are already randomly occupied. You select an empty seat. What is the probability that your $2$ friends coming afterwards will sit next to you? i.e. there are either $2$ empty seats on your left or $2$ on your right or you are in the middle of $2$ empty seats.

we know that all possible compilations for are $\binom{n}{m}$. I think the question is what is the possibility i selected the negative event i.e. a seat with no free seat next to it or a seat with only one empty seat next to me. The question is how to calculate these probabilities?

I follow completely your new hint but still I cannot see how to calculate these 2 probabilities!!

Answer 1

Hint:

Compare and contrast the differences between the problems where

• You pick a seat first and then the $m$ other people select their seats

• The $m$ other people select their seats first and then you select your seat

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Extended hint: (since it apparently wasn't enough)

Notice that the problem descriptions where you get your seat first and then the other $m$ people get their seats is identical to the problem where the $m$ people get their seats before you. There is a clear bijection between the outcomes described in each scenario. However, it is a much easier problem to approach when thought of from the perspective where you get your seat first.

So, take your seat first. Now, the question becomes, what is the probability that the other $m$ people will choose their seats in such a way that they leave at least two spaces near you open. Alternatively, as you correctly noted, this could be approached by looking instead at the complementary event, that exactly zero or exactly one neighboring seat is left unoccupied. If we were to do it directly, you might want to use inclusion-exclusion. At a glance, it appears that approaching from the complementary perspective will be easier, so continue that way.

Now, finally before continuing with any actual calculations, recognize that the position you sit will impact things quite a bit, so condition on and break into cases based on whether you:

• Sit on an edge
• Sit next to an edge
• Sit in the middle with at least two seats in either direction

Now, find the probabilities that after you sit and the $m$ other people are choosing their seats that:

• exactly zero seats near you are left open
• exactly one seat near you is left open

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Final hint:

In general, if you have $N$ objects, $K$ of which are of one type and $N-K$ of which are of another type, if we select $n$ of these objects without replacement the probability that we have selected exactly $k$ of the objects of the first type is going to be:

$$\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$

This is the hypergeometric distribution. A more generalized form of the distrubtion with more categories of objects exists as well called the multivariate-hypergeometric distribution.

Now, relate this to our problem with chairs and let "nearby seats" be the objects of the first type and faraway chairs be objects of the second type.

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Solution: (I strongly discourage looking until trying to give my previous hints a chance, but hover over to see)

The following assumes that $n\geq 5$. The cases for $n\leq 4$ can be handled manually very quickly by hand.

In the first case where we sit next to an edge, this will occur with probability $\frac{2}{n}$. Given that this has happened, we ask what the probability is that among the chairs selected by our $m$ people, we either have exactly zero open seats nearby (in which case the adjacent seat to us is taken) or we have exactly one open seat nearby (in which case the adjacent seat to us is not taken but the seat adjacent to that is taken). To find these probabilities, count how many seats are now "restricted" either because we are sitting in it, we are requiring the seat to be unoccupied by the $m$ people, or we are requiring the seat be occupied by one of the $m$ people. Once having forced those conditions to be met, we count how many ways we can seat the remaining of the $m$ people. We divide by $\binom{n-1}{m}$ as there are that many ways to have seated the $m$ people. The first occurs with probability $\dfrac{\binom{n-2}{m-1}}{\binom{n-1}{m}}$. The second occurs with probability $\dfrac{\binom{n-3}{m-1}}{\binom{n-1}{m}}$. This case contributes then $\dfrac{2}{n}\left(\dfrac{\binom{n-2}{m-1}}{\binom{n-1}{m}} + \dfrac{\binom{n-3}{m-1}}{\binom{n-1}{m}}\right)$ to the overall probability of failure.

$~$

Continuing, for the next case where we sit one seat away from the edge, for zero open seats this occurs with probability $\dfrac{2}{n}$ we need both the seat to our left and the seat to our right be occupied. For exactly one seat nearby being unoccupied, this either requires the seat in the direction of the edge to be unoccupied and the seat towards the middle to be occupied, or the seat towards the edge to be occupied, the seat to the middle to be unoccupied and the beyond that to be occupied. This contributes a total of $\dfrac{2}{n}\left(\dfrac{\binom{n-3}{m-2} + \binom{n-3}{m-1}+\binom{n-4}{m-2}}{\binom{n-1}{m}}\right)$ to the chance of failure.

$~$

Finally the case of if it is in the middle, this occurs with probability $\frac{n-4}{n}$. The breakdown of cases is similar to before. This contributes a total of $\dfrac{n-4}{n}\left(\dfrac{\binom{n-3}{m-2}+2\binom{n-4}{m-2}}{\binom{n-1}{m}}\right)$ to the probability of failure.

$~$

For the final answer to the originally asked question, you subtract each of the expressions found in the three cases away from $1$.