Suppose I have 8 blocks of three different colors: red, yellow, and blue. Suppose there are 4 red ones, 3 yellow ones, and 1 blue one. If I randomly line them up, what is the probability that there will be 4 red ones in a row? 3 red ones? 2?
I tried the following:
- 4 red in a row: $\left(4! \cdot {4\choose 4} \cdot 5\right)/8! = 0.30\%$
- 3 red in a row: $\left(3! \cdot {4\choose 3} \cdot 6\right)/8! = 0.36\%$
- 2 red in a row: $\left(2! \cdot {4\choose 2} \cdot 7\right)/8! = 0.21\%$
My reasoning was as follows. In the 4 red in a row case, for example, there are:
- $4!$ ways to order the group of 4 red blocks in a row
- ${4\choose 4}$ ways of choosing 4 red blocks from 4 red blocks (which is trivial in this case but follows the general form)
- $5$ possible positions of 4 consecutive blocks in a group of 8
- $8!$ total possible block orderings
But the probability of 2 red in a row should certainly not be greater than the probability of 3 red in a row, so I must be doing something wrong. Does anyone know how to handle this problem?
The general version would be something like: Suppose there are $n$ objects of $m$ different classes. If these objects are randomly ordered, what is the probability of $k$ objects of the same class being in a row?