[baddox]

I think you're incorrect because your conceptual model of what constitutes "probability" is incorrect for this type of problem.

Try thinking about it in a more brute force way: imagine literally all possible outcomes of performing this experiment. In other words, create a list like this (each coin in the jar is numbered from 000 to 999 with 999 being the only biased coin, and coin flips are represented by 0 being heads and 1 being tails):

    Picked fair coin #000, flipped 00000000000 (eleven flips)
    Picked fair coin #000, flipped 00000000001
    Picked fair coin #000, flipped 00000000010 ...
    Picked fair coin #000, flipped 11111111111
    ....
    Picked biased coin #999, flipped 11111111111

Now select all of the lines above where the first ten flips are heads. Of these outcomes, how many have an eleventh flip of heads and how many have an eleventh flip of tails? Unless my idea of probability is flawed, this should be the same answer that the mathematicians in this thread are providing, so something right around 75% heads.

[jules]

Where is the unfair coin in your list? Without that you just get 50% heads. This method can definitely work to get the correct answer, but you have to account for all possibilities and weigh them by their probability.

[baddox]

Sorry, I edited coin #999 to be the biased coin.

[jonsen]

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]

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]

... 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]

True, I could have still used zeros and ones to enumerate all outcomes of the biased coin, but said that both digits represent heads.