That’s an interesting question. I’d like to know the same thing. I’m no math expert; the only idea I could come up with is that perhaps the nineteen different pairs of coins are not completely independent since each coin is a member of two pairs. For example, if you were trying to figure out the probability that #5 and #6 were not both heads, that might be affected by what you know about whether #4 and #5 are both heads and whether #6 and #7 are both heads. If you know that #4 and #5 are both heads as well as that #6 and #7 are both heads, then probability is 1 that #5 and #6 are both heads. For each pair, TT, TH, and HT will result in not having a consecutive pair of heads, but HT cannot be preceded by TH and TH cannot be followed by HT. In that way, each pair doesn’t only have to worry about itself not being HH, but also that any H that does appear is not adjacent to another H. Each pair of TT can be followed by HT, TH or TT, but each pair of HT can only be followed by TH or TT. ]]>

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The playing algorithm is flip one, discard it, then flip another, and keep it if it is greater than the first, otherwise discard it and flip and keep the third.

The reason your odds are 50% can be found by considering which of the 3 in your selection order is the greatest. There is a 1/3 chance it is the first, in which case you lose because you always discard it. There is 1/3 chance it is the second, and you win, and in the 1/3 case it is the last, you win that 1/2 the time, depending on if 1 was greater than 2 or not.

Does it matter if the 3 numbers may not be unique?

I like Scott’s modification, too!

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The other analysis looks something like recursion, or calculating the Fibonacci sequence. It’s really tough to explain only in words (my wife, who very long ago, was a math teacher, didn’t even want to hear it).

It involves imagining the flips sequentially, walking down the possibility tree to the first Tail flip. Then working up a recursion formula for calclulating the probability of two heads following any tail flip, starting from the last flip (0), to the next to last (0), to the next to next to last (.25), etc. That yields 0.983109, or about 16 losses per thousand tries. (A losing bet, the way you describe the problem, losing $600 every thousand games.)

Empirically running the test in Excel yields 12-24 losses per thousand tries, vindicating the more complicated analysis.

Why is the first analysis wrong? Thanks, Mark

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