What is the probablity of sitting next to my friend? Let's say you are at a table with $5$ others, everyone is seated randomly around a $6$ person table, and you only know $1$ person at this party. 


*

*What is the likelihood you sit next to the individual that you know?

*What is the likelihood you are seated opposite to the person that you know?

*What is the likelihood that you sit next to two strangers?


The table has $6$ seats so if you sit in any one seat then there are $5$ chairs left over. Since your friend can be seated on either side of you that leaves 3 chairs.  With that reasoning would it be $1/3$ ($2/6$)?
 A: In total, there are $6! = 720$ ways for all the people to sit on the chairs.


*

*First there are $6$ ways for you to take a seat, $2$ ways for your friend to sit next to you. Now with $4$ people left with $4$ chairs, there are $4!=24$ ways for them to sit. So the probability would be $$\frac {24 \times 2 \times 6}{720}=\frac{2}{5}$$

*Similar to 1. , there are $6$ ways for you to have the first seat, $1$ way for your friend to sit opposite you and $24$ ways for the rest to sit. The probability would be: $$\frac {24 \times 1 \times 6}{720}=\frac{1}{5}$$

*In fact this is the complement of 1. so the probability would be $$1- \frac{2}{5}=\frac{3}{5}$$
(Sorry, I was late and English is my second language)
A: Total number of ways six people can sit around the table is $(6-1)! = 5!$ . The first problem.  Fix yourself, there are 2 ways your known friend can sit either side. The rest  of you can sit in 4! ways to a total of 48.  Thus the probability for the first question is $\frac{48}{120} = \frac{2}{5}$.  The second question is you occupy one chair and your known friend occupies on opposite to it and the remainder of 4 will occupy the chairs in $4!$ ways to get a probability of$\frac{24}{120} = \frac{1}{5}$.  The third question is  the two strangers can be picked in ${4\choose2}$ ways and they can be permuted in 2! ways and the remainder 3 can be permuted in 3! ways to a total of $(2\times6\times6) = 72$.  Thus the probability is $\frac{72}{120} = \frac{3}{5}$
A: The answer to #1 is not $\frac26$ but $\frac25$, since you can consider your own seat fixed without loss of generality, leaving 5 chairs, two of which are adjacent to you. Similarly, the answers for #2 and #3 are $\frac15$ and $\frac35$ respectively, both obtained by counting the number of chairs where the desired result is obtained by your friend sitting there by the number of empty chairs.
