[jules]

You didn't read it wrong, but you probably did fail the test ;-) There is no gotcha in the question, it's just a math problem that you either do or do not know how to solve. This isn't really about intelligence as much as it is about whether you have taken a course on probability. If you flipped 10 heads in a row the probability of the coin you have being the double heads coin increases dramatically, so you have to take that into account for the next flip. For intuitive understanding it often helps to go to extremes. Suppose you do 1 billion flips and all come up heads. What is the probability that the next flip comes up heads? Because we had 1 billion heads it is virtually certain that we are dealing with the double heads coin, so the probability that the next flip will come up heads is close to 1.

[joezydeco]

"If you flipped 10 heads in a row the probability of the coin you have being the double heads coin increases dramatically..."

I disagree. The coin is the coin. It didn't magically transport itself or change state after flipping it 10, 100, or a billion times.

Lets change the puzzle to the simplest state: you pull a coin from the 1000-coin jar and flip it just once. What's the probability of heads?

This is why roulette and baccarat tables in Vegas have those signs showing previous outcomes. It's meant to mess with your head. Previous history has no effect on future outcomes. A fair coin could come up heads a billion times in a row as well. The next flip will still be 0.5.

[Bakkot]

The coin is the coin, but your information about the coin has changed. Probability is a fact about you, not a fact about the coin.

Here is perhaps a more visceral example: On any given day, your car has a 10% chance of having blown a wheel the previous night. If it has blown a wheel, every bump will feel jarring. If it hasn't blown a wheel, any given bump will feel jarring with 10% probability. You drive over a hundred bumps, and all feel jarring. Do you think the next bump will also feel jarring?

Yes, right? Because the fact that the last hundred bumps have been jarring means you probably blew a wheel last night. Even though 'previous history has no effect on future outcomes', previous history does give you information about the state of the world.

Edit: Or a better example, for this community. You're on a slightly flakey wifi connection, which sometimes drops a packet at random. Any given packet is dropped with probability 1/100. Also, any given time you connect to the router, the modem is down with probability 1/10, and all of your packets will get dropped. You connect, and the first hundred packets you send are dropped. What is the probability the next packet will get dropped? Very high, because now you know that the modem is probably down. The history of the connection gives you information about the state of the connection. Similarly, the history of coin toss results gives you information about the state of the coin.

(Why is this different from roulette? Because there the previous history doesn't give you information.)

[phazmatis]

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?".

[baddox]

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.

[phazmatis]

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

[joezydeco]

Probability is a fact about you, not a fact about the coin.

Huh? It's a coin. You flip it. It has no knowledge of the past and no idea about the future. It lands and it's either heads 50% of the time and tails 50% of the time. It's all about the coin and nothing about what you've observed in the last N trials.

That's the simple logic that the OP is trying to see if you understand.

[Bakkot]

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.

[harpastum]

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%

[mag487]

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])

[jules]

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...

[crntaylor]

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

[baddox]

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.

[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.

[jules]

The probability of heads is

    0.999*0.5 + 0.001*1 = 0.5005

The important thing is that as you observe heads from the coin, you learn something about the coin. As you observe more heads it is less likely to be a fair coin and more likely to be the coin with double heads. This doesn't change anything about the coin, but it changes something about what you know about the coin. See here for the correct answer: https://news.ycombinator.com/item?id=7000523

[joezydeco]

Since the question simply asked what p(heads) was on the next flip, it seems our answers match. Thanks!

[jules]

Our answers to your question may match, but I very much doubt that our answers on the original question match. The crux is that your question is not equivalent to the original question. If you are interested in learning why that is I can explain it further, but it doesn't look like you are?

[joezydeco]

Okay yeah, I'm stupid for pressing on while crntaylor already gave his answer. I'll shut up now.

[baddox]

Not at all. The answer is about 0.75, and stated in the answer jules linked to.

[bluepen44]

I think you're missing the point of the question which seems to get at whether the interviewee is familiar with/understands bayesian probability. For an excellent explanation you should see: http://yudkowsky.net/rational/bayes.

[baddox]

I don't see why knowledge of Bayesian probability is necessary for this question. I don't have anything more than a passing knowledge of Bayesian probability, but I was able to produce the correct answer and a reasonable explanation using what I believe to be classical probability: https://news.ycombinator.com/item?id=7001288

[patmcc]

>>>Previous history has no effect on future outcomes. A fair coin could come up heads a billion times in a row as well. The next flip will still be 0.5.

You hit the nail on the head here - a fair coin would. But if there are unfair coins, things start to change. If we know there's an unfair coin, and we see an unusual run, we should consider that it may be that coin.

Think of this: increase the flips to 1000, and you're still getting heads each time. Would you bet on the next being heads or tails?