Was it rigged?
- I don’t get this?
- What don’t you get?
- John and Paul are both on the committee and they sat next to each other!
- Please explain!
- A committee of three was selected at our last meeting. There were 20 people sitting around the round table and three people were selected randomly.
- And your point is?
- Isn’t it incredible that two of the three elected sat next to each other?
- Do you expect the election was rigged?
- I don’t know. I just want to know what the chances are that at least two of the three committee members sat next to each other.
Problem source: Fortnight problem, Department of Computer Science, San Diego State University.
March 5th, 2010 at 1:43 am
There are C(20,3) = 1140 possible committees. There are 20 composed of groups who all sit together, and 16×20 = 320 composed of a singleton plus two who sit together. 340/1140 = 17/57 ~ 0.298 or almost 30% of the time, which is pretty much un-incredible.
March 5th, 2010 at 2:21 am
Paul can sit in any of 19 positions relative to John. The third member of the committee (let’s call him George) can sit in any of the other 18. So that’s 19*18/2 = 19*9 = 171 possibilities for where Paul and George sit relative to John, without regard to which is George’s seat and which is Paul’s.
The three would be sat together iff Paul and George sit relative to John in any of the following ways:
* one sits to John’s left, and the other takes any of the other 18 seats: 18 ways
* one sits to John’s right, and the other takes any other seat except the one on John’s left (because I’ve already accounted for that possibility): 17 ways
* one takes any seat except John’s, the 2 seats next to John’s, and the one 2 places to John’s right (because I’ve already accounted for those possibilities), and the other sits to his left: 16 ways.
This makes 51 ways in total. Thus the probability that the three committee members are sat together is 51/171 ~ .298.
March 9th, 2010 at 4:06 pm
I found the same result but in a third way:
The first one (A) is elected.
The second one to be elected (B) has 15 chances on 19 to seat at least 3 seats away from A and 2 chances on 19 to seat exactly two seats away from A.
In the first case, there’s 14 (20 seats – 2 taken seats – 4 seats nearby) chances on 18 to elect C away from A and B.
In the second case, there’s 15 (20 – 2 taken seats – 3 seats nearby) chances on 18.
Probability no one seats just by another one:
15/19 * 14 / 18 + 2/19 * 15/18
= 210/(19*18) + 30/(19*18) = 240 / (19*18) = 40 / 57.
Probability at least one seats just by another: 17/57.