## #math trivia solution: Two friends got in the same class …

The chances that the two friends are seated next to each other is 8/95.  This can be calculated as follows:

• The chances that the friends are seated in the same row is 4/19:  For any seat one friend occupies, there are 19 the other friend can be assigned to, all equally likely, and four of them are in the same row.  (The same logic produced the 19/99 figure for the chances that the two friends are in the same class.)
• If the friends are in the same row, the chances that they are seated next to each other is 2/5:  The two friends could be seated in 5*4 = 20 possible ways within that same row, and there are eight ways they could sit next to each other, illustrated as follows (where “A” denotes one friend, “B” the other, and “-“ another student in the row of five):

AB— -AB– –AB- —AB
BA— -BA– –BA- —BA

Because the choice of row is independent of the choice of seat within the row, the two probabilities can be multiplied together, giving (4/19)*(2/5) = 8/95.

Beyond the trivia:  Combining with the 19/99 figure from last time, the overall probability that two friends will be assigned seats next to each other when there are five classes with four rows of five seats is (19/99)*(8/95) = 8/475, which is about 1/60.  This makes sense intuitively, because a typical student will sit next to two other students.  The chances that one of those is a specific friend out of a group of 100 should thus be about one in fifty.  The analysis can be made more precise:  With rows of five seats, the two students at the ends sit next to only one student, and the three in the middle sit next to two.  The average number of neighbors for a student is thus (2/5)*1 + (3/5)*2 = 8/5 = 1.6.  The number of other students in the group is 99. So the chances that one of the neighbors is a specific friend among the other students is 1.6/99, or 8/475.