How many same-color neighbors in a row of balls?

Question

Four red balls and four blue balls (8 balls total) are lined up in a single row in a completely random order. On average, how many of the 7 neighboring pairs consist of two balls of the same color?

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A row of 8 has 7 neighboring pairs (positions 1-2, 2-3, up to 7-8). Put one indicator on each of these 7 gaps and add. Fix one neighboring pair of positions. The two balls there are a random pair drawn from the 8 without replacement. The chance they are both red is P(	ext{both red}) = \frac48 \cdot \frac37 = \frac{12}{56} = \frac{3}{14}, and by symmetry the chance both are blue is also 	frac{3}{14}. So P(	ext{same color}) = \frac{3}{14} + \frac{3}{14} = \frac{6}{14} = \frac37. Let Xi = 1 if gap i holds a same-color pair; E[Xi] = 	frac37. With 7 gaps, E[X] = 7 	imes \frac37 = 3. Takeaway.…

How many same-color neighbors in a row of balls?

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