Dice sequence: how many neighboring pairs increase?

Question

You roll a fair six-sided die 10 times and write the results in a row. On average, how many of the 9 neighboring pairs show a strictly larger value on the second roll than on the first?

Accepted Answer

Ten rolls make 9 neighboring pairs. Put one indicator on each pair and add. This time the two values can tie, so the rising probability is not 	frac12. Fix a neighboring pair of rolls, calling them the first and second die. Out of the 36 equally likely outcomes, the second beats the first in the \binom{6}{2} = 15 outcomes where the two faces differ and the larger one lands second. So P(	ext{second roll strictly larger}) = \frac{15}{36} = \frac{5}{12}. Let Xi = 1 if neighboring pair i strictly increases; E[Xi] = 	frac{5}{12}. With 9 pairs, E[	ext{increasing pairs}] = 9 	imes \frac{5}{12} …

Dice sequence: how many neighboring pairs increase?

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