A combinatorial problem on coincidences

In response to a recent editorial by Dick Cavett on coincidences,  Jack (4th commenter) responded with the following story:

A relative was married and there was a sit-down dinner for all guests at the reception, including my father, mother, sister and me. I suppose there were 150 people. The hostess wanted totally arbitrary seating, so she placed 150 little cards in a basket, numbered 1 -150, and each guest was to reach in a pick a number as they entered the room. Then she randomly numbered the 150 seats at the tables. It made for some confusion, as each person had to search for his or her seat. In case you haven’t guessed, the four of us where seated in a row. I mean, what are the chances!

You can try solving it yourself before reading on.  Assume that there were 15 circular tables, each with 10 persons each.

Solution. The assignment of people to numbers can be ignored since the probabilities are exactly the same for every possible assignment of people to numbers.  The analysis assumes that the assignment of people to numbers has already been carried out.

There are 150! (that is, 150 factorial) ways of placing the 150 numbers at the tables.  There are 10 ways of selecting four consecutive seats at any specified table, and thus 150 different ways of selecting four consecutive seats at one of the 15 tables.  If the family sits together, this leaves 146 seats for everyone else, and these numbers can be assigned to seats in 146! ways.   Thus, the number of configurations in which the four family members are seated in a row are

146! (150)(4!),

where the 4! is the number of different ways the four family members can be arranged within the four seats.  Therefore, the probability of the four family members being together is

146!(150)(4!)/150! = 24/(149*148*147),

which is approximately 1 in 135,000.  Incidentally, the probability is independent of the number of tables or the size of tables, so long as there are at least 5 persons at each table.   For example, the probability is the same if there was a single circular table with 150 seats.

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2 Responses to “A combinatorial problem on coincidences”

  1. Geoffrey De Smet Says:

    The Miss Manners problem is similar: it’s about seating guests at tables under several constraints, such as 1 democrat and 1 republican politician per table.

    I’ve solved it with drools-solver:
    http://blog.athico.com/2009/05/miss-manners-2009-yet-another-drools.html
    It’s included in the examples in the download zip.

  2. strawberrymargaritas Says:

    Maybe all four of you successively picked cards from the basket and your hostess forgot to shuffle them well? Even so, the random labels on seating encourages more possibilities.

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