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If i have a known population ($N$ marbles of which $M$ are black) and draw $n$ samples without replacement the probability to draw $x$ black marbles is given by the hypergeometric distribution.

Is there a way to get probabilities for the total number of blacks $M$ from the number of black marbles $x$ in my sample?

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Under suitable assumptions one can. Assume for example that $N$ is fixed, and Alicia decided on how many of these will be black, $0$ to $N$, using a uniform distribution, or any other known distribution. Then based on sample proportion, we can calculate the probabilities Alicia decided to put in $k$ black. – André Nicolas Dec 5 '12 at 21:37
Can you get any quantitative results without assumptions on the distribution of the $M$ (i.e. Alicia's choice)? I want to find a maximum number of the remaining black marbles (i.e. $M-x$), that holds except some small probability. – Tom S Dec 5 '12 at 21:47
You always have to make at least a tacit assumption, despite what Frequentist statisticians sometimes say. If I were doing it, I would set it up as a Bayesian problem with a conjugate Beta-Binomial prior on M. – Jonathan Christensen Dec 5 '12 at 21:56
I do not have any insight on how to attack the problem without a prior. – André Nicolas Dec 5 '12 at 21:56
up vote 1 down vote accepted

The probability of extracting exactly $x$ black marbles in a sample of size $n$ from a population of $N$ marbles of which $M$ are black can be calculated as:

$$P(X=x|n,N,M) = \frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}$$

If you assume that all values of $M$ compatible with your result are equally likely (this would be André's prior distribution assumption, I believe), you can use the exact same formula with a different twist, considering the total number of black marbles, $M$, as the independent variable, and the number of black marbles in your sample, $x$, as a parameter instead:

$$f(M) = P(X=x|n,N,M) = \frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}$$

The above function computes the relative likelihood of a value of $M$, and the value that maximizes it, is the maximum likelihood estimator of $M$ for the population. If you plot the above function for all possible values, $M\in[x,N-n+x]$, after normalization you'll get a probability distribution which can be used to compute the probabilities you are after.

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Right, this corresponds to a uniform prior on M. – Jonathan Christensen Dec 5 '12 at 22:23

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