3 ms·

Well if I can count: 00000 00000 0 00000 00000 1 for 999 fair coins = 1998 + 1 for fake coin = 1999 1000 of which has 11. flip 0 So 1000/1999 ~ 50%

• baddox 13y ago
  True. I glossed over the crucial part, which is that you have to enumerate the same number of potential outcomes for the biased coin's ten flips as you do for each fair coin, because each coin is equally as likely to be selected from the jar. Of course, all 2^10 potential outcomes for the biased coin's ten flips are the same: all heads. So we have: 00000 00000 0 00000 00000 1 For 999 fair coins = 1998 And 2^11 instances of 00000 00000 0 for the biased coin = 2048. That's 1998 + 2048 = 4046 equally likely outcomes that begin with ten heads. Of those, only 1998 / 2 = 999 outcomes feature an eleventh tails. So the odds of getting an eleventh heads after seeing ten heads is (4046 - 999) / 4046 =~ 0.753.
  • jonsen 13y ago
    ... because each coin is equally as likely to be selected from the jar That's not the reason. It's because the fake coin also has two sides and therefore also 2^11 different outcomes. It just happens they all look the same. Otherwise I agree with your argument.
    • baddox 13y ago
      True, I could have still used zeros and ones to enumerate all outcomes of the biased coin, but said that both digits represent heads.
  • deleted 13y ago
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