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Solution to puzzle 38: Twelve marbles

A boy has four red marbles and eight blue marbles.  He arranges his twelve marbles randomly, in a ring.  What is the probability that no two red marbles are adjacent?


This problem is a counting exercise.  We will count the number of distinct marble arrangements such that no two red marbles are adjacent, and then divide this by the total number of distinct marble arrangements to obtain the required probability.

To simplify the counting process, select any blue marble, and consider the remaining 11 marbles, arranged in a line.  The proportion of such arrangements for which no two red marbles are adjacent will be the same as for the original 12 marbles, arranged in a ring.

The number of ways of choosing k outcomes out of n possibilities, ignoring order, nCk, is equal to n! / [k! (n − k)!].
So the total number of ways of arranging 4 red marbles out of 11 is 11C4 = 330.

To count the number of arrangements such that no two red marbles are adjacent, observe that there must be at least one blue marble between each two would-be adjacent red marbles:

[red][blue][red][blue][red][blue][red]

Having fixed the positions of three blue marbles, we have four blue marbles left to play with.  We can think of the four red marbles as dividing lines, around which the remaining four blue marbles must be slotted.  Given four dividing lines and four marbles, we have 8C4 = 70 distinct combinations.

Therefore the probability that no two red marbles are adjacent is 70/330 = 7/33.


Generalization

Clearly, we can generalize to r red marbles and b greater than or equal to r − 1 blue marbles, arranged in a line.  For marbles arranged in a ring, where we require b greater than or equal to r, replace b by b − 1 in the working below.

We have r − 1 fixed blue marbles, and r red marbles (dividing lines), around which the remaining b − r + 1 blue marbles must be distributed.

Therefore, for marbles arranged in a line, the probability that there are no two adjacent red marbles is:

b+1Cr / b+rCr  =  [b! (b + 1)!] / [(b + r)! (b + 1 − r)!]

Hence, for example, in a lottery where 6 balls are drawn (without replacement) from balls numbered 1 through 49, the probability that no two winning numbers are consecutive is:

(43! 44!) / (49! 38!)  is approximately equal to  0.5048.

Source: Original

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