# joint probability of two hypergeometric trial sets

Let's say I have one experiment where I'm drawing 50 balls from a bag of 1000 without replacement. The bag contains 50 white balls - the rest are black. It's trivial to calculate the cumulative probability of drawing at least x white balls. Now, I repeat the experiment again and draw another 50 balls from a new bag of 1000 (again, 50 white, 950 black). I'm interested in the probability of drawing at least 2x balls in both of these experiments. I thought that should be roughly equivalent to one pooled experiment, drawing 100 balls out of 2000 (basically combining the 2 bags together), but it doesn't appear to be the case from some simulations I ran. This seems like a pretty simple question but I haven't seen any relevant discussions online.

There is a little restoring force around the mean in a hypergeometric distribution. If your first few balls have been black more than average, then the probability that the next ball is white is a little higher than if the first few balls had been white more than average. This restoring force differs when you sample $100$ balls from $2000$ versus $50$ from $1000$ twice. As an extreme, you could sample $1$ from $20$ $100$ times, and then there would be no restoring force.

One measure of the spread of a distribution is the variance. The variance when you draw n balls out of w+b is

$$\frac{n (w+b-n) w b }{(w+b)^2(w+b-1)}.$$

For $n=100, w=100, b=1900,$ the variance is $\frac{9025}{1999} = 4.51476.$

For $n=50, w=50, b=950,$ the variance is $\frac{9025}{3996}$. When you do this twice independently and add the results, you get twice the variance, $\frac{9025}{1998} = 4.51702.$

For $n=1, w=1, b=19,$ the variance is $\frac{19}{400}.$ If you do this $100$ times independently and add the results, you get $100$ times the variance, $\frac{19}{4} = \frac{9025}{1900} = 4.75.$

The difference between $\frac{9025}{1999}$ and $\frac{9025}{1998}$ is not very large, and you won't easily tell the difference by eye-balling results. There is a larger difference if you look at the exact probabilities of events far from the mean.

$0$ is not very far from the mean ($5$), so the probabilities are not much different. The probability that you get $0$ white out of $100$ when everything is put together is $0.0051735.$ The probability that you get $0$ white out of $50$ twice is $0.0051707.$ The probability that you get $0$ out of $1$ $100$ times is $0.00592053.$

$40$ is farther from the mean, and there is a larger proportional difference between the probabilities. The chance that you get $40$ out of $100$ is $3.12262 \times 10^{-29}.$ The chance to get a total of $40$ when you take $50$ from each half is $3.35434 \times 10^{-29},$ about $7\%$ larger. The chance to get $40$ if you draw $1$ ball out of each group of $20$ is $5.75970 \times 10^{-26} = 5759.70 \times 10^{-29},$ almost $2000$ times as large. Even that far from the mean, you don't see a large difference between drawing $100$ out of $2000$ and drawing $50$ from $1000$ twice, but there is a larger difference with drawing $1$ out of $20$ $100$ times.

• Thanks, this was certainly helpful. However I'm still unclear which of the two answers (pooled vs separate) is more "correct". Unfortunately, the differences only arise when you are so far away from the mean that it's a bit hard to simulate. Commented Sep 10, 2012 at 19:00

Since you are not replacing the balls, you start the second bag with a full set of 1000 balls whereas continuing to draw from the first you would start the second set of draws with only 950 with the distribution of black and white determined by previous draws as compared to 1000 distributed 950 black and 50 white independent of the first 50 draws. See the difference? It has an effect on the probability of the various possible outcomes.

Consider the following scenarios to see why the probabilities you describe need not agree:

The Two Bag Scenario: We have two bags. Let's call them bag 1 and bag 2 for convenience. Each bag contains $$N$$ balls of which $$K$$ are white and $$N-K$$ are black. From each bag we draw $$n$$ balls without replacement. Let $$X_1$$ be the number of white balls drawn from the bag 1, and $$X_2$$ the number of white balls drawn from bag 2. The probability of drawing at least $$2x$$ white balls in total is given by

\begin{align*} P(X_1+X_2 \geq 2x) &= \sum_{k=2x}^{2n} P(X_1+X_2=k),\\ &= \sum_{k=2x}^{2n} \sum_{i=0}^{k} P(X_1=i, X_2=k-i),\\ &= \sum_{k=2x}^{2n} \sum_{i=0}^{k} P(X_1=i) P(X_2=k-i), & \text{by assuming independence}\\ &= \sum_{k=2x}^{2n} \sum_{i=0}^{k} \frac{\binom{K}{i} \binom{N-K}{n-i}}{\binom{N}{n}} \times \frac{\binom{K}{k-i} \binom{N-K}{n-k+i}}{\binom{N}{n}}, \end{align*}

where $$\binom{K}{i}\binom{N-K}{n-i} / \binom{N}{n}$$ is the probability of drawing $$i$$ balls in $$n$$ draws without replacement from bag 1, and $$\binom{K}{k-i}\binom{N-K}{n-k+i} / \binom{N}{n}$$ is the probability of drawing $$k-i$$ balls in $$n$$ without replacement from bag 2.

The One Bag Scenario: Now consider a third bag, call it bag 3, containing $$2N$$ balls of which $$2K$$ are white and $$2N-2K$$ are black. Let $$X_3$$ be the number of white balls drawn from bag 3 in $$2n$$ draws without replacement. The probability of drawing at least $$2x$$ white balls is given by

\begin{align*} P(X_3 \geq 2x) &= \sum_{k=2x}^{2n} P(X_3=k),\\ &= \sum_{k=2x}^{2n} \frac{\binom{2K}{k} \binom{2N-2K}{2n-k}}{\binom{2N}{2n}}, \end{align*}

where the summand is the probability of drawing $$k$$ white balls in $$2n$$ draws without replacement from bag 3.

A Simple Example: To get an intuition for why these two formulas need not generally agree, consider the simpler setting in which we are only interested in comparing $$P(X_1 + X_2 = 2x)$$ with $$P(X_3=2x)$$ when $$n=1$$ (i.e. this is like your $$x=1$$ case). In other words: Is the probability of drawing a white ball from bag 1 and drawing a white ball from bag 2 the same as drawing two white balls from bag 3?

The probability of drawing a white ball from bag 1 and drawing a white ball from bag 2 is given by

$$P(X_1+X_2 = 2) = \frac{K}{N} \times \frac{K}{N} = \frac{K^2}{N^2}$$

since, assuming independence, the probability of drawing a white ball from either bag is just $$K/N$$. On the other hand, the probability of drawing two white balls without replacement from bag 3 is given by

\begin{align*} P(X_3 = 2) &= \frac{2K}{2N} \times \frac{2K-1}{2N-1},\\ &= \frac{K(2K-1)}{N(2N-1)}. \end{align*}

So $$P(X_1+X_2 = 2)$$ and $$P(X_3 = 2)$$ will differ in general. The intuitive reason being that it makes a difference when one neglects to put the first drawn ball back into bag 3 after having drawn it. This exemplifies why one cannot combine hypergeometric trials in the manner described in the question.