How many streaks with a biased coin?

Question

A biased coin lands heads with probability $0.6$ and tails with probability $0.4$ . You flip it 10 times in a row. A "streak" is a maximal block of the same result.

On average, how many streaks appear in the 10 flips?

Accepted Answer

As always, the number of streaks equals 1 plus the number of switches between neighboring flips. There are 9 neighboring pairs among 10 flips. For each pair, the two flips differ when one is heads and the other tails, in either order: P(	ext{a given pair differs}) = 2 	imes 0.6 	imes 0.4 = 0.48. By linearity the expected number of switches is 9 	imes 0.48 = 4.32, so E[	ext{streaks}] = 1 + 4.32 = 5.32 = \frac{133}{25}. Note this is a bit lower than the fair-coin value of 5.5, because a biased coin repeats itself slightly more often, producing fewer switches. Takeaway. For a coin with head…

How many streaks with a biased coin?

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