抛硬币序列：HTH 与 HTT

结论：HTH 的期望抛掷次数更大。

直觉：两者的“重叠结构”不同。HTT 在失败时更容易保留部分前缀（例如出现 H 后仍可能作为下一轮的开头），而 HTH 更容易被打断而需要从头开始。

常见结果：

• $E[\text{HTH}]=20$，
• $E[\text{HTT}]=18$，

所以 HTH 更慢。

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Original Explanation

We are comparing the expected number of flips needed to see either the sequence HTH or HTT for the first time in an infinite sequence of fair coin tosses.

To analyze this, we look at the structure and overlapping potential of each sequence.

For HTH:

• Once you get HT, to complete HTH you need a head next.
• If you get a tail instead, the entire pattern is broken and you must start from the beginning.
• That is, HTT forces you to reset because T does not start the HTH sequence again.

For HTT:

• After seeing HT, if the next flip is a tail, you have successfully completed HTT.
• If the next flip is a head, you now have H — which is the beginning of HTT again.
• So even if you fail to complete HTT, you're partially progressed and already set up for the next sequence.

This difference in structure means HTT allows for a "head start" when you fail to complete the sequence, while HTH does not — it forces a full reset.

Expected Number of Flips:

• The expected number of flips to get HTH is 20.
• The expected number of flips to get HTT is 18.

Conclusion:

On average, it takes more flips to obtain HTH than HTT due to the lack of partial overlap. Therefore:

HTH takes more flips on average than HTT.