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终结了：
The problem is same as asking for the probability of seeing HHT before HTT when the current state is H because you need an H to jumpstart both sequences. Let such probability be P。
Current State: H
There are 2 cases:
1. next flip=H (with 1/2 probability). We have HH. No matter what the next flips are, we will always have a string of ,,,HHHHH(HH). We will always see HHT before HTT as the first T will complete the HHT sequence.
2. next flip=T (with 1/2 probability). We have HT.
If the next flip is T (total probability 1/2*1/2=1/4) then we see HTT before HHT.
If the next flip is H (total probability 1/4) then we get back to the original state H.
Therefore, P=1/2+1/4*0+1/4*P , or P=2/3
The problem is same as asking for the probability of seeing HHT before HTT when the current state is H because you need an H to jumpstart both sequences. Let such probability be P。
Current State: H
There are 2 cases:
1. next flip=H (with 1/2 probability). We have HH. No matter what the next flips are, we will always have a string of ,,,HHHHH(HH). We will always see HHT before HTT as the first T will complete the HHT sequence.
2. next flip=T (with 1/2 probability). We have HT.
If the next flip is T (total probability 1/2*1/2=1/4) then we see HTT before HHT.
If the next flip is H (total probability 1/4) then we get back to the original state H.
Therefore, P=1/2+1/4*0+1/4*P , or P=2/3
The experiment is to toss a coin until we can see a pattern of HTH, for example HTTTHTTHTH or the pattern HTT such as THTHTHTHTT. We do this 100 times for HTH and another 100 times for HTT writing down the results each time so you can create the mean (average) number of toss before each pattern occurs. If you understand the process then pick a following statement based on what you think is true.
A. It will take longer (more tosses) for HTH to occur in random coin tosses than for HTT to occur.
B. They will take the same number of tosses for either HTH or HTT to occur.
C. It will be sooner (less tosses) for HTH to occur in a random coin tosses than for HTT to occur.
We assume that because the coin toss is random that everything that comes from it is also random. This ignores the fact that both HTT and HTH are not random; they are non-random patterns that have been created. Most people think that B is the correct answer, although it is actually A which is correct. The average number of tosses for HTH is 10 before the pattern occurs and 8 tosses before the pattern HTT occurs. To explain why A is correct; it has to do with the third toss in the patterns. When we are looking for the pattern HTH, and we have the HT the next toss could result in H in which case we complete the pattern or T where we have to start over again. When we are looking for the pattern HTT, and we have HT the next toss could result in a T in which case we complete the pattern or a H which would not complete the pattern but it is then possible to use that H in the start of the next HTT. It’s weighted for the benefit of HTT and that is why we need fewer coin tosses to get the pattern HTT.
第八题：
You have a deck of cards, 26 red, 26 black. These are turned over, and at any point you may stop and exclaim "The next card is red.". If the next card is red you win £10. What's the optimal strategy? Prove this is the optimal strategy.
一副扑克，26红，26黑。混合后背面向上，变成一摞。每轮翻开最上面一张。在任意一轮开始之前，你都可以叫停，声明“这轮翻开的是红牌”。如果这轮翻开的真是红牌，你赢钱。 你的最优策略是什么？怎么证明？
You have a deck of cards, 26 red, 26 black. These are turned over, and at any point you may stop and exclaim "The next card is red.". If the next card is red you win £10. What's the optimal strategy? Prove this is the optimal strategy.
一副扑克，26红，26黑。混合后背面向上，变成一摞。每轮翻开最上面一张。在任意一轮开始之前，你都可以叫停，声明“这轮翻开的是红牌”。如果这轮翻开的真是红牌，你赢钱。 你的最优策略是什么？怎么证明？
