3 ms·

0.5005?

joezydeco 13y ago
That's what I got. There's a .999 chance you have a fair coin and a .001 chance you have the rigged coin. (0.999 * 0.5) + (0.001 * 1) = 0.5005. Seems too simple, but a coin is a coin, right?
- 11001 13y ago
  Ask yourself a question: can you use the data (the 10 coin tosses) to update the probability of the current coin being two-headed?
  - joezydeco 13y ago
    I wouldn't use the data. The coin hasn't changed since I picked it out of the jar. If I flip it 1, 10, or 10e100 times, the coin would still be the same coin. So figure the p(heads) for the coin and ignore the previous history. Overthinking it is why this makes a good FizzBuzz problem.
    - dmurray 13y ago
      An example that should show this approach is wrong: Suppose that the jar contains 500 double-head coins and 500 double-tail coins. You pull a coin from the jar, flip it 10 times, and get 10 heads. What is the probability it will come up heads next time?
      - joezydeco 13y ago
        That seems like a completely different problem to me, since all randomness is out of the system the moment you see the first flip.
        gjm11 13y ago
        OK, so now imagine that there are 1000000 double-headed coins, 1000000 double-tailed coins, and one fair coin. Now (1) there's still (potentially) randomness present, so it's not "completely different" from the original problem, but (2) the ignore-the-data approach gives an obviously wrong answer whereas using the data gives a believable answer.
    - thwest 13y ago
      You are getting downvotes because you didn't follow Bayesian reasoning, but there is some justification for your instincts here http://www.stat.columbia.edu/~gelman/research/unpublished/philosophy.pdf http://www.stat.columbia.edu/~gelman/research/unpublished/ph...
    - dustyleary 13y ago
      Let's consider that it might be a fair coin, or it might be a double-headed coin. Let's also say that every time you flip the coin and it comes up tails, you win $5. And every time you flip the coin and it comes up heads, you lose $1. Clearly, this would be a great game to have the opportunity to play, if the coin is fair. Every time you flip you either win $5 or lose $1, so your profit, on average, is $4 per flip. You've flipped it 10 times so far, and it's come up heads every time, and you've lost $10. After you're $10, $100, $1000, or $10e100 in the red, without ever seeing a win, when do you change your mind about playing this game?
    - bonobo 13y ago
      Yes, it's still the same coin, but you don't know which coin you got. You know that you got 10 consecutive heads though. How improbable this is if you got a fair coin? How probable this is with the double-head coin? This is the data you can use to update the probability.
    - chrismcb 13y ago
      So if you flipped the coin twice, once it was tails, and once it was heads... you'd ignore that info? Or look at it another way, if you flipped it a million times and it always came up heads...

0.5005?

That's what I got. There's a .999 chance you have a fair coin and a .001 chance you have the rigged coin. (0.999 * 0.5) + (0.001 * 1) = 0.5005. Seems too simple, but a coin is a coin, right?

11001 13y ago
Ask yourself a question: can you use the data (the 10 coin tosses) to update the probability of the current coin being two-headed?
- joezydeco 13y ago
  I wouldn't use the data. The coin hasn't changed since I picked it out of the jar. If I flip it 1, 10, or 10e100 times, the coin would still be the same coin. So figure the p(heads) for the coin and ignore the previous history. Overthinking it is why this makes a good FizzBuzz problem.
  - dmurray 13y ago
    An example that should show this approach is wrong: Suppose that the jar contains 500 double-head coins and 500 double-tail coins. You pull a coin from the jar, flip it 10 times, and get 10 heads. What is the probability it will come up heads next time?
    - joezydeco 13y ago
      That seems like a completely different problem to me, since all randomness is out of the system the moment you see the first flip.
      - gjm11 13y ago
        OK, so now imagine that there are 1000000 double-headed coins, 1000000 double-tailed coins, and one fair coin. Now (1) there's still (potentially) randomness present, so it's not "completely different" from the original problem, but (2) the ignore-the-data approach gives an obviously wrong answer whereas using the data gives a believable answer.
  - thwest 13y ago
    You are getting downvotes because you didn't follow Bayesian reasoning, but there is some justification for your instincts here http://www.stat.columbia.edu/~gelman/research/unpublished/philosophy.pdf http://www.stat.columbia.edu/~gelman/research/unpublished/ph...
  - dustyleary 13y ago
    Let's consider that it might be a fair coin, or it might be a double-headed coin. Let's also say that every time you flip the coin and it comes up tails, you win $5. And every time you flip the coin and it comes up heads, you lose $1. Clearly, this would be a great game to have the opportunity to play, if the coin is fair. Every time you flip you either win $5 or lose $1, so your profit, on average, is $4 per flip. You've flipped it 10 times so far, and it's come up heads every time, and you've lost $10. After you're $10, $100, $1000, or $10e100 in the red, without ever seeing a win, when do you change your mind about playing this game?
  - bonobo 13y ago
    Yes, it's still the same coin, but you don't know which coin you got. You know that you got 10 consecutive heads though. How improbable this is if you got a fair coin? How probable this is with the double-head coin? This is the data you can use to update the probability.
  - chrismcb 13y ago
    So if you flipped the coin twice, once it was tails, and once it was heads... you'd ignore that info? Or look at it another way, if you flipped it a million times and it always came up heads...