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Comment by knughit | Hacker News Reader

Explain the difference?

Are you saying that HH causes an early failure for HT , instead of a potentially longer success HHHT? If so , it is poorly worded, to be ambiguous about how to count failures. (In the 4-6 variation, there are no failures)

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  flip coins
    if HT or HH
      stop

  flip coins
    if HT
      stop

  flip coins
    if HH
      stop

two different scenarios

I'm saying that if you just keep flipping coins (So, no Bob/Alice situation), and stop whenever you hit HH or HT, that the average number of flips to get to HH is 3, and the average number of flips to get to HT is 3.

But, if you are focussing on a particular scenario that you will flip coins until you get to HT, the average number of flips will be 4, and if you flip coins until you get to HH, the average number of flips will be 6.

I just find that really hard to grasp intuitively.

I don't think that's true. I think that the average number of flips before _stopping_ might be 3, but it doesn't become more likely to get HT simply because you're also looking for HH (or vice versa).

It turns out that the average number of flips to get HH and HT is 3 - someone else on the thread described it well - on average, the number of flips to get to a "H" will be 2, and then you will always stop on the next flip - 50% at flip 3 with a HT, and 50% at flip 3 with a HH. Explained that way, it makes sense that the average number of flips is 3.

But, I still find it strange that if you are flipping with one particular scenario in mind, HT or HH, that the average number of flips goes from 3 to 4 or 6, even if I can reason it out with a bit of thinking.

Do I understand your first scenario correctly, that if for example you flip TTHH, you stop there, count that as "took four flips to get to HH," count nothing for HT, and then start over?

Seems like you're just taking a biased sample, which cancels out the differences. To take an extreme example, imagine one candidate is HHHHHHHHHH and the other candidate is any other sequence of ten flips. In the "try until you get either one" scenario, the average number of flips for either one will be 10. Testing them independently, the average number of flips for the second one will be slightly over 10, and for HHHHHHHHHH it'll be huge.

"and stop whenever you hit HH or HT" is equivalent to say "whenever you hit H, flip one more then stop". Average to hit H is obviously 2, plus one is 3, so you are right, but IMO it's not that counter-intuitive.

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