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I was discussing a small experiment with a friend of mine this week. He said, "we can just do 3 trials."

I said, "sure, but the subject will have to get all 3 trials right to be better than chance."

We discussed it a bit and verified that you would indeed have a 50/50 chance of correctly guessing at least 2 out of 3 coin flips.

I started wondering how many trials you'd have to have to get statistical significance. I figured out that the probability of guessing at least $n-1$ correct trials is this formula:

$$\frac{(n + 1)}{ 2^n}$$

If I set that equal to $.05$, how do I out how many trials you'd have to have to get a 95% confidence if the subject misses no more than one trial?

id est, if $$.05 = \frac{(n + 1)}{ 2^n}\,,$$ what is $n$?

More importantly, how did you solve for $n$?

Thanks a lot!


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up vote 1 down vote accepted

At $7$ rolls the probability is $0.0625$, at $8$ it is $0.035156$. I just calculated them from 2 to 27 in a spreadsheet. There is no algebraic solution without invoking Lambert's W function.

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Thanks for the response. I'm pleased there wasn't something really simple I was overlooking. Edit: Evidently, hitting enter submits the form. Because I don't know any better, I'm going to leave it open for just a little longer to see if there are any conflicting opinions and then give credit where credit is due. Thanks Ross. – D. Patrick Jun 27 '11 at 20:38

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