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Suppose you have a loaded coin, and we assume a priori that when you flip it you get heads 20% of the time and tails 80%.

If I wanted to be 99% certain that this coin is a fraud, how many times do I need to flip?

I know this involves some advanced statistics, but I can't recall?

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You also need to assume an a priori probability that the coin is a fraud. The lower this probability is, the more flips will be needed. Consider that if this probability is 0 (i.e. you are certain the coin is fair), then no number of flips will convince you, because is is possible although improbable that a fair coin will produce 800 heads and 200 tails. Then I believe this can be solved with Bayes' theorem and the Law of the iterated logarithm? But I am not sure about the details. –  Dan Brumleve Sep 12 '11 at 2:28
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how does iterated algorithm step in here @dan ? –  Bhargav Sep 12 '11 at 2:42
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Also I was assuming that either the coin is fraudulent (with 80-20 odds), or else it is fair (with 50-50 odds) and that no other cases are possible, which may or may not be the interpretation you had in mind. More generally you could have some a priori probability distribution on the bias of the coin. –  Dan Brumleve Sep 12 '11 at 2:50
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168335 and Adel, I just reviewed LoIL on wiki and it doesn't step in here at all; my memory of it was faulty. All we need is the binomial CDF and Bayes' I suppose. I need to read some more books too! –  Dan Brumleve Sep 12 '11 at 3:02
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Adel, your edit makes the answer to the question trivially "0 flips" because you are already assuming the coin is fraudulent. What I meant is something like: "Suppose you have a coin, and there is a 2% chance it is fraudulent (meaning it lands on heads 80% of the time), and a 98% chance it is fair. If I wanted to be 99% certain that this coin is a fraud..." –  Dan Brumleve Sep 12 '11 at 3:29

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