A rumor's loudness scaled by a die each round

Question

A rumor starts at loudness 6. Each round, independently, you roll a fair six-sided die and multiply the current loudness by (that die's face divided by 3). The rumor runs for two rounds. What loudness should you expect after two rounds?

Accepted Answer

Each round multiplies the loudness by a random factor (a die roll over 3), and the rounds are independent. Condition on the loudness before each round with the tower rule. Step 1, the average factor per round. The factor is \frac{	ext{die}}{3}, and a fair die averages 3.5, so E[	ext{factor}] = \frac{E[	ext{die}]}{3} = \frac{3.5}{3} = \frac{7}{6}. Step 2, the recursion. Let Lt be the loudness after t rounds. Given L{t-1}, the next round multiplies it by an independent factor averaging 	frac76, so E[Lt] = 	frac76 \, E[L{t-1}]. Step 3, iterate from L0 = 6 over two rounds: E[L2] = 6 	imes \…

A rumor's loudness scaled by a die each round

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