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If there are 10 seats in a round table and 7 people are already sitting in some random pattern(edit: uniform distribution, thank you for the correction) what is the probability that the next two people entering the room will get to sit in adjacent seats?

I am very challenged in trying to start to think about the problem. I would sincerely appreciate any thoughts and I will definitely indicate my progress thereafter.

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So far you have three completely different answers :). – mjqxxxx Oct 28 '13 at 22:50
This may indicate some ambiguity in how your question is written. – mjqxxxx Oct 28 '13 at 22:50
Two of the answers are not different from each other. – André Nicolas Oct 28 '13 at 22:57
You're right; $21/36=7/12$. – mjqxxxx Oct 28 '13 at 22:59

We assign seats at random, by putting place mats labelled $1,2,3,\dots, A, B, C$ at random in front of the $10$ chairs. The place mats labelled $A,B,C$ indicate where any person apart from $1$ to $7$ may sit. A pair of lovers now enters the room, and they wish to sit together. We find the probability that they can do so without asking anybody to move.

The only thing that matters is the placement of $A, B, C$. And since the table is round, all that matters is the placement of $B$ and $C$ relative to $A$. So we have $9$ empty spaces, and have to choose $2$ of them. There are $\binom{9}{2}$ choices, all equally likely.

We count the bad choices. For a choice to be bad, it must involve choosing $2$ non-adjacent seats from the $7$ not next to $A$. There are $\binom{7}{2}$ ways to choose $2$ seats, of which $6$ give an adjacent pair. So there are $15$ bad choices, out of the $\binom{9}{2}$ choices. Thus the probability there will be an adjacent pair is $\frac{21}{36}$.

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Thank you so much for the detailed response. I edited it to be a uniform distribution. Thank you, I fully comprehend the problem now. – Daniel Synn Oct 28 '13 at 22:58

This is impossible to determine without knowing the distribution function of the patterns in which the $7$ people are already sitting. I will assume a uniform distribution since this seems most natural.

Assuming a linear distribution, the $7$ people are a red herring. We just need to calculate the probability that two random seats are next to each other. After the first seat is chosen, two of the remaining $9$ are adjacent to it, so the probability is $2/9$.

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Yes, I apologize for the ambiguity of the question. Thank you. – Daniel Synn Oct 28 '13 at 22:57

Based on the phrasing of the question, I will assume that the next two people will enter together, and that they'll sit next to each other if they can. If all configurations of the first $7$ people are equally likely, then so are all configurations of the remaining $3$ empty seats; we want to know how likely it is that any pair of those empty seats are adjacent. There are ${{10}\choose{3}}=120$ configurations, of which the number with an adjacent pair is $10\cdot 8 - 10=70$. (Or count the number with no adjacent pairs. After placing the first seat, there are seven spots left. Two of them leave five spots for the third seat; the others leave four. So this number is $10\cdot \left(2\cdot 5 + 5\cdot 4\right)=300$; dividing by $3!$ leaves $50$ blocked configurations.)

So the desired probability is $7/12$.

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Yes, we are to assume uniform distribution. Thank you so much for the detailed response. – Daniel Synn Oct 28 '13 at 22:59

Consider 10 chairs arranged in a circle. The number of ways of leaving three seats empty is C(10,7) = 240.

Of these, the number of ways in which atleast 2 of the empty seats are adjacent is 10.8 = 80. (Think of this situation as having 10 fixed spaces. There are 10 ways to choose a pair of adjacent spaces, considering the first and last also as adjacent-one for each space being the left space of the chosen pair. There are 8 ways to choose one of the remaining 8 spaces)

Hence, the required probability is 80/240 = 1/3.

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There are two mistakes. $\binom{10}7=120\ne240$. And you're overcounting the configurations with three adjacent empty seats. – joriki Jun 22 at 11:21

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