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Question:

I've been trying to find an answer to this for months (to be used in a machine learning application), it doesn't seem like it should be a terribly hard problem, but I'm a software engineer, and math was never one of my strengths.

Here is the scenario:

I have a (possibly) unevenly weighted coin and I want to figure out the probability of it coming up heads. I know that coins from the same box that this one came from have an average probability of **p**, and I also know the standard deviation of these probabilities (call it **s**).

(If other summary properties of the probabilities of other coins aside from their mean and stddev would be useful, I can probably get them too.)

I toss the coin **n** times, and it comes up heads **h** times.

The naive approach is that the probability is just **h/n** - but if n is small this is unlikely to be accurate.

Is there a computationally efficient way (ie. doesn't involve very very large or very very small numbers) to take **p** and **s** into consideration to come up with a more accurate probability estimate, even when **n** is small?

I'd appreciate it if any answers could use pseudocode rather than mathematical notation since I find most mathematical notation to be impenetrable ;-)

**Other answers:** There are some other answers on SO that are similar, but the answers provided are unsatisfactory. For example this is not computationally efficient because it quickly involves numbers way smaller than can be represented even in double-precision floats. And this one turned out to be incorrect.

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Solution:1

You can use `p`

as a prior on your estimated probability. This is basically the same as doing pseudocount smoothing. I.e., use

`(h + c * p) / (n + c) `

as your estimate. When `h`

and `n`

are large, then this just becomes `h / n`

. When `h`

and `n`

are small, this is just `c * p / c = p`

. The choice of `c`

is up to you. You can base it on `s`

but in the end you have to decide how small is too small.

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Solution:2

Unfortunately you can't do machine learning without knowing some basic math---it's like asking somebody for help in programming but not wanting to know about "variables" , "subroutines" and all that if-then stuff.

The better way to do this is called a Bayesian integration, but there is a simpler approximation called "maximum a postieri" (MAP). It's pretty much like the usual thinking except you can put in the prior distribution.

Fancy words, but you may ask, well where did the h/(h+t) formula come from? Of course it's obvious, but it turns out that it is answer that you get when you have "no prior". And the method below is the next level of sophistication up when you add a prior. Going to Bayesian integration would be the next one but that's harder and perhaps unnecessary.

As I understand it the problem is two fold: first you draw a coin from the bag of coins. This coin has a "headsiness" called theta, so that it gives a head theta fraction of the flips. But the theta for this coin comes from the master distribution which I guess I assume is Gaussian with mean P and standard deviation S.

What you do next is to write down the total unnormalized probability (called likelihood) of seeing the whole shebang, all the data: (h heads, t tails)

L = (theta)^h * (1-theta)^t * Gaussian(theta; P, S).

Gaussian(theta; P, S) = exp( -(theta-P)^2/(2*S^2) ) / sqrt(2*Pi*S^2)

This is the meaning of "first draw 1 value of theta from the Gaussian" and then draw h heads and t tails from a coin using that theta.

The MAP principle says, if you don't know theta, find the value which maximizes L given the data that you do know. You do that with calculus. The trick to make it easy is that you take logarithms first. Define LL = log(L). Wherever L is maximized, then LL will be too.

so LL = h*log(theta) + t*log(1-theta) + -(theta-P)^2 / (2*S^2)) - 1/2 * log(2*pi*S^2)

By calculus to look for extrema you find the value of theta such that dLL/dtheta = 0. Since the last term with the log has no theta in it you can ignore it.

dLL/dtheta = 0 = (h/theta) + (P-theta)/S^2 - (t/(1-theta)) = 0.

If you can solve this equation for theta you will get an answer, the MAP estimate for theta given the number of heads h and the number of tails t.

If you want a fast approximation, try doing one step of Newton's method, where you start with your proposed theta at the obvious (called maximum likelihood) estimate of theta = h/(h+t).

And where does that 'obvious' estimate come from? If you do the stuff above but don't put in the Gaussian prior: h/theta - t/(1-theta) = 0 you'll come up with theta = h/(h+t).

If your prior probabilities are really small, as is often the case, instead of near 0.5, then a Gaussian prior on theta is probably inappropriate, as it predicts some weight with negative probabilities, clearly wrong. More appropriate is a Gaussian prior on log theta ('lognormal distribution'). Plug it in the same way and work through the calculus.

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Solution:3

You don't have nearly enough info in this question.

How many coins are in the box? If it's two, then in some scenarios (for example one coin is always heads, the other always tails) knowing p and s would be useful. If it's more than a few, and especially if only some of the coins are only slightly weighted then it is not useful.

What is a small n? 2? 5? 10? 100? What is the probability of a weighted coin coming up heads/tail? 100/0, 60/40, 50.00001/49.99999? How is the weighting distributed? Is every coin one of 2 possible weightings? Do they follow a bell curve? etc.

It boils down to this: the differences between a weighted/unweighted coin, the distribution of weighted coins, and the number coins in your box will all decide what n has to be for you to solve this with a high confidence.

The name for what you're trying to do is a Bernoulli trial. Knowing the name should be helpful in finding better resources.

Response to comment:

If you have differences in p that small, you are going to have to do a lot of trials and there's no getting around it.

Assuming a uniform distribution of bias, p will still be 0.5 and all standard deviation will tell you is that at least some of the coins have a minor bias.

How many tosses, again, will be determined under these circumstances by the weighting of the coins. Even with 500 tosses, you won't get a strong confidence (about 2/3) detecting a .51/.49 split.

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Solution:4

In general, what you are looking for is Maximum Likelihood Estimation. Wolfram Demonstration Project has an illustration of estimating the probability of a coin landing head, given a sample of tosses.

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Solution:5

Well I'm no math man, but I think the simple Bayesian approach is intuitive and broadly applicable enough to put a little though into it. Others above have already suggested this, but perhaps if your like me you would prefer more verbosity. In this lingo, you have a set of mutually-exclusive hypotheses, H, and some data D, and you want to find the (posterior) probabilities that each hypothesis Hi is correct given the data. Presumably you would choose the hypothesis that had the largest posterior probability (the MAP as noted above), if you had to choose one. As Matt notes above, what distinguishes the Bayesian approach from only maximum likelihood (finding the H that maximizes Pr(D|H)) is that you also have some PRIOR info regarding which hypotheses are most likely, and you want to incorporate these priors.

So you have from basic probability Pr(H|D) = Pr(D|H)*Pr(H)/Pr(D). You can estimate these Pr(H|D) numerically by creating a series of discrete probabilities Hi for each hypothesis you wish to test, eg [0.0,0.05, 0.1 ... 0.95, 1.0], and then determining your prior Pr(H) for each Hi -- above it is assumed you have a normal distribution of priors, and if that is acceptable you could use the mean and stdev to get each Pr(Hi) -- or use another distribution if you prefer. With coin tosses the Pr(D|H) is of course determined by the binomial using the observed number of successes with n trials and the particular Hi being tested. The denominator Pr(D) may seem daunting but we assume that we have covered all the bases with our hypotheses, so that Pr(D) is the summation of Pr(D|Hi)Pr(H) over all H.

Very simple if you think about it a bit, and maybe not so if you think about it a bit more.

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