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Supposed we are using a round table with 10 seats, there are 7 people who pick their seats randomly. you and your friend are late and you two want to have 2 seats next to each other. Another person is also late and he will arrive before you.

  1. One friend call, he assures that there are 2 seats next to each other at that time. What is your probability of having 2 seats next to each other?

  2. No one calls, and you are not assured there are 2 seats next to each other at that time (you don't know). What is your probability of having 2 seats next to each other?

My approach for 1 is to calculate the number of times that there are 3 seats, then divided by the total number of time that the seats are taken by 7 people. How do we solve /2/ then?

Thank you

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Question 1 seems unclear: have the other 7 people already picked their seats? If so, then obviously you and your friend will have 2 adjacent seats, and your calling friend can take the third. – Newb Oct 27 '13 at 22:42
yes the 7 people have already picked up their seat and sat. – Kiddo Oct 27 '13 at 22:57
up vote 0 down vote accepted

I'm assuming the friend who calls is not the same as the friend who is going to arrive late, while the "other person" is not your friend, will take a seat randomly just like the first $7$, and comes after your friend called.

I'm also assuming that when the friend says there are two seats next to each other, that doesn't exclude the possibility that all three vacant seats are adjacent. If all three are adjacent, the eighth person prevents you from getting two together only by taking the middle seat of the three (thus with probability $1/3$), while if there are just two together and a third separate the eighth person has probability $2/3$ of preventing you from getting two together.

Now there are just $10$ ways to have three adjacent vacant seats around a $10$-seat table, but $10 \times 6 = 60$ ways to have two adjacent and one separate (i.e. given the positions of the two adjacent vacant seats, there are $6$ possibilities for the third vacant seat). Since all ways of having three vacant seats are equally likely, the probability in (1) is $$ \dfrac{10}{70} \times \dfrac{2}{3} + \dfrac{60}{70} \times \dfrac{1}{3} = \dfrac{8}{21}$$

In (2), after 8 people have chosen seats randomly there are just $10$ cases where the remaining two seats are adjacent, out of a total of ${10}\choose 8$.

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