# Probability question regarding 670 coin flips

1. Jan 4, 2012

### tuttlerice

Hi. Music theorist here. I'm working on a paper analyzing a certain number of events in a piece of music statistically. I've found that my form of analysis has determined that another theorist's form of analysis identifies 423 out of 670 events in this piece of music being what I call "upper tier" events. This is as opposed to "lower tier" events.

I'm trying to figure out the statistical significance of this. So what I'm trying to figure out is, what are the chances that, out of 670 coin flips, 423-or-more of them would come up heads?

I think this question sounds deceptively simple, but isn't.

You would think that pretty much coin flips converge toward halfsies the more often you flip them, right? 423/670 is 63%, which is 13% more than the 50% one would intuitively expect.

But I've read that to find statistical significance, one has to find the chances that this event could occur *randomly*. So what are the odds that I could get 423-or-more flips to come up heads out 670 *at random*? That number seems like it would be quite small (which is what I want--- a small chance this is random; therefore the outcome has significance).

The 50-50 intuition sometimes is counter-intuitive. If you flip a coin four times, there are sixteen possible outcome configurations: HHHH, HHHT... TTTH, TTTT. How many are half-heads? You'd think 8, right? But there are only 6 possible configurations: HHTT, HTHT, HTTH, THHT, THTH, TTHH.

So that's why I want to be careful about this. I think I'm a little in over my head and so I'd like some help from an expert! Thank you so much in advance.

2. Jan 4, 2012

3. Jan 4, 2012

### SW VandeCarr

I don't know how much probability theory you know, but this is a straightforward calculation based the normal approximation to the binomial distribution. The mean or expected value of a fair coin side for 670 trials is (n)(p)=(670)(0.5)=335 The standard deviation is a property of the normal distribution, not strictly speaking of the binonmial. However you can use the formula

$$SD= \sqrt {np(1-p)} = \sqrt {(335)(0.5)} = 12.942$$

To find how many standard deviations 423 or more is from the mean based on the normal approximation: (423 -335)/12.942 = 6.800

For all practical purposes, this is equivalent to zero probability unless you're a physicist.

Last edited: Jan 4, 2012
4. Jan 4, 2012

### tuttlerice

Thanks for the response. Sadly, I have not had much probability theory. But after reading the Gambler's Fallacy, here's what I got out of it:

-The probability of any one particular total outcome of 670 trials is 1 / (2^670).

-Therefore, the probability of getting 423 heads out of 670 trials is 1 / (2^670).
-The probability of getting 424 heads out of 670 trials is also 1 / (2^670).
... and so on, through observing that the probability of getting 670 heads out of 670 trials is 1 / (2^670).

These are 248 different outcomes I'm interested in. So the answer to my question would seem to be 248 * (1/2^670)) or 5.062378 * 10^200.

Since this is nowhere near 9.6159, I've obviously done something wrong in my logic.

How would I report the 9.6159 number? "There is a 9.6159% chance this would occur randomly"? 0.096159%?

Thanks again. Sorry to bother everyone with my lack of schooling.

5. Jan 4, 2012

### SW VandeCarr

You may not know any probability theory but hopefully you can follow my calculations. The standard deviation is a measure in terms of the integral of the probability density function. One standard deviation on the bell shaped curve takes in about 63% of the area under the curve. With three standard deviations you're around 99%, That means the probability of a value more than 3 SD is about 1% split between both tails of the curve. For over 6 sd the probability is extremely small. Off the top of my head for 6.8 sd I would guess the one tail $P =7.22561987 *10^{-13}$ but I could be way off (joke).

You report the outcome is almost surely not the result of a random process or the coin is biased if you get 432 heads out of 670 trials.

Last edited: Jan 4, 2012
6. Jan 4, 2012

### tuttlerice

Thanks so much. This is really helpful.

7. Jan 4, 2012

### SW VandeCarr

No problem. I was searching on the web for a calculator or table that would actually give me a value. No luck. The probability is based on the number of standard deviations (Z score) and is calculated from a formula containing the error function (erf). There's no general solution for integrating this function as far as I know. If you plug in a number like 7 into a calculator, you typically get 0 for a probability. I'd be curious to know if anyone can compute it for Z=6.760 (corrected) or do an exact binomial probability. By the way, I made an error in my calculation which I corrected, but the probability that the result is random for a fair coin is still effectively 0.

Last edited: Jan 4, 2012