So, I think that tge question is ambiguous, and a little more direction could clear it up. If you want to know the probability, taking into account the 10 heads results thus far, I would tell the interviewee, "I am going to do this process repeatedly (picking a coin and flipping 10 times), and I'm going to keep track of what percentage do the all one side thing. What will that percentage be approximately?".

That's not ambiguous if the applicant has taken a basic college statistics course, which is what the question is intended to determine. The term "probability" has more precise definitions in mathematics than in everyday language.

Ah, that's the part I was missing. Thanks

Did you miss the bit in the OP where one of the coins has heads on both sides? The coin may or may not be biased, and the results of flipping it give you information about whether or not it is biased.

Explaining the bit you quoted: if you had perfect knowledge of the wind conditions, how hard the person flipped the coin, and so on, you would know precisely which side it would land on. That fact is determined. It's just because you are missing information that there's 50% probability of heads (for an unbiased coin). The probability comes out of your imperfect knowledge, and changes depending on your knowledge: for example, if you have some reason to believe the coin is biased, then you no longer think it's going to land on heads 50% of the time. Since one of the coins is biased, getting a long string of heads is reason to think that the coin they drew and are flipping is the biased coin. This information affects the probability you assign to the coin coming up heads.

It's really interesting to watch this thread. I'm not saying that you should simply trust what the other commenters are saying, but the user crntaylor is the one that originally posted the question, and his solution is that the likelihood of the next flip being heads is 75%

It may be a useful sanity check to simulate the experiment multiple times to see the result:

  def choose_coin
    rand < 0.001 ? 1 : 0.5
  end
   
  def flip_coin(coin)
    rand < coin ? :heads : :tails
  end
   
  def all_10_heads?(coin)
    10.times { return false if flip_coin(coin) == :tails }
    true
  end
   
  next_flip = { :heads => 0, :tails => 0 }
   
  1000000.times do
    coin = choose_coin
    next unless all_10_heads? coin
    next_flip[flip_coin(coin)] += 1
  end
   
  puts next_flip[:heads].to_f / (next_flip[:heads] + next_flip[:tails])

What's really interesting is that there are probabilistic programming languages where you can write a program that does a simulation just like you did, but the execution engine can compute the probabilities exactly and much faster too. It does this by computing along all possible paths in the program, and keeping track of the probability mass of each path, and then summing them all up in the end.

http://en.wikipedia.org/wiki/Probabilistic_programming_langu...

Completely shameless plug for a tiny probabilistic programming language that I wrote as an embedded DSL in Haskell:

https://github.com/chris-taylor/hs-probability

The code that solves this problem is:

  solve = do
    coin   <- choose (999/1000) fair biased
    tosses <- replicateM 10 coin
    condition (tosses == replicate 10 Head)
    nextToss <- coin
    return nextToss
   where
    fair   = choose (1/2) Head Tail
    biased = certainly Head

Likewise, a tire has no knowledge of the past or future. It will transfer a bump from the road to you following the laws of physics, depending on whether it is functioning properly, represented by a probability.