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silentOpen

0.5005?

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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?

Ask yourself a question: can you use the data (the 10 coin tosses) to update the probability of the current coin being two-headed?

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.

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?

That seems like a completely different problem to me, since all randomness is out of the system the moment you see the first flip.

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.

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reasoning, but there is some justification for your instincts here http://www.stat.columbia.edu/~gelman/research/unpublished/ph...

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?

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.

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...

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