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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.

  1. What is the likelihood you sit next to the individual that you know?
  2. What is the likelihood you are seated opposite to the person that you know?
  3. 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 $1/3$ ($2/6$)?

amWhy
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3 Answers3

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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.

Parcly Taxel
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In total, there are $6! = 720$ ways for all the people to sit on the chairs.

  1. 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 $$\frac {24 \times 2 \times 6}{720}=\frac{2}{5}$$

  2. 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: $$\frac {24 \times 1 \times 6}{720}=\frac{1}{5}$$

  3. 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)

apple
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    hi, for number two, just assume you are already seated, there are 5 seats available, so the chance is indeed 1/5 as you calculated. Same argument can be used at the other questions. – SAJW Oct 17 '18 at 12:00
  • For Q1, the way of counting ways to arrange the seats does make sense. However, if view the individual I know and myself as a bundle, and first arrange as five entries in total, isn't that lead to 5! * 2 = 240? I wonder where's the problem in my logic? – user21 Feb 18 '23 at 05:50
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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 $\frac{72}{120} = \frac{3}{5}$

Satish Ramanathan
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