模式先后：HTH vs HHT

结论：朋友获胜概率为 $\boxed{\frac{2}{3}}$（你获胜概率为 $1/3$）。

直觉上，HHT 比 HTH 更容易“抢先”出现。

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Original Explanation

Both sequences contain the subsequence HT, so imagine writing down an infinite number of toss outcomes and scanning for HT subsequences. In order for HTH to occur first, an H must not precede HT (else HHT occurs) and an H must succeed it (to create HTH). Hence, the probability HTH wins is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ given there is an HT. In order for HHT to occur first, an H must precede HT, which happens with probability $\frac{1}{2}$ given there is an HT. Note that these are the only two outcomes that can occur and thus define our sample space $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$. The probability that your friend wins is $\frac{1}{2} \div \frac{3}{4} = \frac{2}{3}$ and the probability that you win is $\frac{1}{4} \div \frac{3}{4} = \frac{1}{3}$. In other words, HHT is twice as likely to occur before HTH.