Expected wait for the next bad day in a two-state chain

Question

Each day is either good (G) or bad (B). After a good day, the next day is good with probability $\tfrac35$ ; after a bad day, the next day is bad with probability $\tfrac{7}{10}$ . Today is a bad day. On average, how many days will pass until the next bad day?

Short answer

$7/4$ . With probability $0.7$ tomorrow is bad (1 day). Otherwise tomorrow is good, and from a good day the wait for a bad day is geometric with success chance $\tfrac25$ , mean $\tfrac52$ . So the answer is $1+0.3\cdot\tfrac52=\tfrac74$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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