A group of $n ≥ 4$ people, including Alice and Bob, are seated at a round table, at random (where all the possible ways are equally likely).
(a) What is the probability that Alice and Bob sit next to each other?
*(b) If Cindy also sits next to the table, what is the probability that Bob sits next to either Alice or Cindy (or both)?

For part a, I am pretty confident: $$2\frac{(n-2)!}{(n-1)!} = \frac{2}{n-1}$$

But I am not sure am I right for part b, my solution: $$4\frac{(n-3)!}{(n-1)!}$$

Thank you!

  • 1
    $\begingroup$ please use latex $\endgroup$ – Alex Jan 18 '16 at 11:31
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Jan 18 '16 at 11:35

Let Bob sit down as first.


The probability that Alice will choose for one of the chairs next to Bob is:$$\frac2{n-1}$$

This simply because $2$ of the open $n-1$ chairs are next to Bob. Likewise thinking we can solve:


The probability that Alice will not choose for one of the $2$ chairs next to Bob is $\frac{n-3}{n-1}$.

If - after that - Cindy takes her seat then the probability that Cindy will not choose for one of the $2$ chairs next to Bob is $\frac{n-4}{n-2}$.

That gives chance: $$1-\frac{n-3}{n-1}\frac{n-4}{n-2}$$ that the described event will not occur, i.e. that Alice or Cindy (or both) will sit next to Bob.


For the first you're correct, but your approach is not the easiest one and one that would maybe not lead easily to the answer in the second case. The trick is to disregard the rest of the company and only consider the seating of Alice, Bob and Cindy.

Instead you consider the number of ways Bob could be seated in relation to Alice which is $n-1$, two of these seats are next to Alice giving the probability $2/(n-1)$.

In a similar way the number of seatings of Bob and Cindy is $(n-1)(n-2)$ and the number of ways that would mean Bob is sitting next to Alice is $2(n-2)$ and the same for Cindy, but you would then count the cases where both are sitting next to Alice which is in two ways. So there's $4(n-2)-2$ ways out of $(n-1)(n-2)$. So the probability would be

$$P = {4(n-2)-2\over(n-1)(n-2)} = {4n-10\over(n-1)(n-2)}$$


a) Looks right.

b) Let $A$ be the event "Bob sits next to Alice" and $C$ "Bob sits next to Cindy", thus

$$ P(A\cup C) = P(A) +P(C) - P(A\cap C) = \frac{(n-2)!2!}{(n-1)!}+\frac{(n-2)!2!}{(n-1)!} - \frac{(n-3)!2!}{(n-1)!}, $$ where the last term $\frac{(n-3)!2!}{(n-1)!}$ calculated by treating Bob, Alice and Cindy as one body, and Alice can switch places with Cindy ($2!$).


Your answer to part (a) is correct. One of the $n-1$ others will be seated on Bob's right; the probability that Alice is chosen for that honor is $\frac1{n-1}.$ The probability that she is seated on Bob's left is also $\frac1{n-1},$ and the sum of those two probabilities is $\frac2{n-1}.$

Part (b): Two of the other $n-1$ people will be chosen to sit next to Bob. There are $\binom{n-1}2$ ways to make the selection. There are $\binom{n-3}2$ ways to do this without picking Alice or Cindy, so the probability that neither Alice nor Cindy sits next to Bob is $$\frac{\binom{n-3}2}{\binom{n-1}2}=\frac{(n-3)(n-4)}{(n-1)(n-2)}$$ and the probably that at least one of them sits next to him is $$1-\frac{(n-3)(n-4)}{(n-1)(n-2)}=\frac{4n-10}{(n-1)(n-2)}.$$

It's the same as the probability of drawing at least one red ball if you draw two balls without replacement from a bag containing two red balls and $n-3$ black balls.


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