I am interested in modeling the following experiment:

A binomial trial with $n$ Bernoulli experiments is run. For the k positive outcomes a second (independent) binomial trial with $k$ runs with a different success probability is run.

For example: I throw $n$ darts with probability $p_1$ of hitting the bull's eye. The number of hits is $k$. My friend is then allowed to throw $k$ times and has a different skill set than I do (probability $p_2$). I want to model the probability that he will hit the bull's eye $l$ times. Let us call the random variable $Y_{n}$ if n trials are run. The first binomial trial is $X^{1}_{n}$ and the second is $X^{2}_{k}$.

Question 1:

What is the distribution of this random variable?

I argue that an ugly form is:

$P \left(Y_{n}=l\right) = \sum_{i=l}^{n}P\left(X_{n}^{1}=i\right)\cdot P\left(X^{2}_i = l\right)$.

Is this correct and is there a nice way to determine confidence intervals for the success probability $p$ of $Y$? I am especially concerned with the quality of the confidence intervals for small $n$ so I am not feeling well with using a normal approximation which would I guess lead to a sum of product normal probabilities. Using Jeffrey's prior (which afaik leads to better results for smaller $n$) I would with a wild guess receive a sum of products of beta probabilities which seems to be numerically difficult (Product of beta distributions).

Question 2:

Finally I am interested for $Y^{i}\sim D \left(n_{i},p_{i}\right)$, with $D$ being the (for me) unknown distribution with overall success probability $p_{i}$ to calculate (approximately)

$P\left(p^{1}_{n_1}\gt p^2_{n_2}\right)$ given sample data. Is there a general way to attack this problem?

Interpretation: there are two teams of dart player pairs. I observe a series of outcomes and want to calculate the probability that one team has a higher sucess probability (success being hitting the bull's eye in the second step which requires at least one hit in the first step).

Question 3:

Is there a nice generalization if I have not only 2, but $k$ successive binomial experiments of this kind? Nice in the sense not simply extending my formula above and summing/integrating approximately.

  • $\begingroup$ I lost you at the first reference to $Y_n$: How many trials are run in the experiment--two as you state ("a binomial trial" and "a second" apparently corresponding to $X^1$ and $X^2$) or $1+k$ as you imply? In either case, obviously there is more than one random variable around: to what does "$Y_n$" refer? Are $X^1$ and $X^2$ actual "trials" or are they outcomes? If outcomes, then it sounds like they may be random variables too, so which of these three is "this random variable" in your first question? In the second question, what are you even talking about--"unknown distribution" of what? $\endgroup$
    – whuber
    Commented Aug 30, 2013 at 18:56
  • $\begingroup$ I edited the question. $Y_n$ is the random variable that models the whole experiment whereas the X's are the binomial random variables of each step. Since I don't know how to call the distribution of $Y_n$ I said "unknown" (potentially "sum of product of binomials"). $\endgroup$
    – Arthur G
    Commented Aug 30, 2013 at 19:20

1 Answer 1


Imagine a Binomial trial of size one with success probability $p_1$. Conditional on success, another trial of size one with success probability $p_2$ is run independently. Because independent probabilities multiply, the chance of two successes is $p_1p_2$. Thus this two-step procedure is the same as a one-step procedure in which "success" is declared with probability $p_1p_2$. Your experimental outcome $m$ (I use this instead of "$l$" which is easily misread) is the number of such pairs of successes out of $n$ independent replications of this two-step procedure, which makes it a Binomial experiment of size $n$ with success probability $p_1p_2$.

Mathematically this implies the theorem

$$\sum _{k=m}^n {p_1}^k {p_2}^m \binom{k}{m} \binom{n}{k} (1-{p_2})^{k-m} (1-{p_1})^{n-k} = \binom{n}{m} ({p_1} {p_2})^m (1-{p_1} {p_2})^{n-m}$$

which can be checked formally.

It should be clear how this generalizes.

In terms of the darts analogy, I am arguing that if you let your friend throw one dart each time you hit in the mark, then--although the sequence in which your throws alternates with her throws is different--in the end it's the same thing, because you have thrown $n$ times and your friend has thrown as many times as you had successes. At each one of your throws your friend will hit in the mark with probability $p_1p_2$: that's a Binomial experiment.

  • 1
    $\begingroup$ Thanks. This solution didn't seem intuitive for me when I first thought about the problem since the confidence interval for the overall p for me hitting 1 out of 100 and my friend hitting 1 of 1 and me hitting 50 out of 100 and him hitting 1 out of 50 are the same. Subjectively I would give less certainty to the 1% success probability in the first scenario. $\endgroup$
    – Arthur G
    Commented Aug 30, 2013 at 20:11

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