Another interesting problem I met in statistic is: suppose we flip a coin which has probability 0.7 to be head again and again and two people choose two different sequences of length 3 (e.g. THH). The people whose sequence appears first wins. If you're allowed to choose first, which sequence will you choose?

I guess most people will choose sequence HHH. However, this sequence is beaten by THH. Suppose we first observe sequence HHH at step \(n(>3)\), then the outcome must be T at step \(n-3\), which means that we have observed THH at step \(n-1\). So for HHH to win against THH, it must appear at step 3. So the probability for HHH to beat THH is 0.7^3=0.343<0.5, i.e., HHH is not as good as THH.

A nature question is that does there exist a best choice in this problem? The answer is no. I have done simulations in MATLAB to find the probability for a sequence to beat another. The simulation result shows that none of the 8 sequence can beat all other choices, which means that there is no benefit to choose a sequence first.

Talking about the probability for one sequence to come out first against another one, there is a much better solution rather than simulation. I will make post about this good solution later.