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I'm looking for a formula to find the probability of exactly $k$ successes in $n$ trials where the initial probability of success is $\frac1x$, with $x\ge1$, and $x$ is incremented on each trial. The resulting probability of success for each trial is $\left[\frac{1}{x}, \frac{1}{x + 1}, \frac{1}{x + 2}, \dots, \frac{1}{x + n - 1}\right]$.

For example, $n=5, k=2, x=4$. The probability of 2 successes in 5 trials where the initial probability of success is $\frac14$ is the sum of the individual probabilities of the ${5 \choose 2}$ combinations.

$$\left(\frac14\cdot\frac15\cdot\frac56\cdot\frac67\cdot\frac78\right) + \left(\frac14\cdot\frac45\cdot\frac16\cdot\frac67\cdot\frac78\right) + \left(\text{8 other combinations}\right) = \frac{1175}{6720}$$

Is there a formula shortcut as with other probability distributions? Below I've attempted to use similar problems to arrive at a solution.

The characteristics are similar to the Pólya urn model, where an urn initially contains $w$ white balls and $b$ black balls. On each trial, a ball is drawn, its color is observed, and then it is returned to the urn along with another ball of the same color. The probability of drawing $k$ white balls in $n$ trials is

$$P(w,b,n,k)={n \choose k} \frac{\prod_{i=0}^{k-1}(w+i) \prod_{i=0}^{n-k-1}(b+i)}{\prod_{i=0}^{n-1}(w+b+i)}=\frac{{w+k-1 \choose k}{b+n-k-1 \choose n-k}}{{b+w+n-1 \choose n}}$$

My question can be described as finding the probability of drawing $k$ white balls in $n$ trials in a Pólya urn model where $w=1$ and $b=x-1$, and regardless of which color ball is drawn, it is always returned along with another black ball.

Also similar is this question about summing combinations of subsets, which applies to the case where $x=1$, i.e. a set which contains the numbers from $[1, n]$. But here, the set would be offset by $x-1$, to contain the numbers from $[x, n+x)$.

The set would be taken from the failures, as those are the quantities that build the numerator, to be the numbers from $[x-1, x+n-1)$. Using my example the set is $\{3, 4, 5, 6, 7\}$ and the result is the sum of the products of the ${5 \choose 3}$ combinations of failures, which is $1175$ as given before.

Based on the answer, the sum of the products of the $k$-combinations of the set of $[1, x+n-1) = {x+n-1 \brack x+k-1}$, where $n \brack k$ is the notation for unsigned Stirling numbers of the first kind. Using my example ${8\brack5} = 1960$, and subtracting the sum of the products of the $[2{5\choose2}$ combinations which contain 1 OR 2] $-$ $[{5\choose1}$ combinations which contain 1 AND 2] = $1960 - 735 - 50 = 1175$

Those other sums also happen to be given by Stirling numbers: ${8\brack5} - {7\brack4} - {5\brack2} = 1960 - 735 - 50 = 1175$, though this seems to be a coincidence.

alec
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1 Answers1

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Not really an answer.

We want the distribution of the random variable

$$Y = \sum_{i=0}^{n-1} X_i$$

where $X_i$ are independent Bernoulli variables with $p_i = P(X_i=1)=\frac{1}{x+i}$.

This is a special case of a Poisson binomial distribution. In general, there is no simple closed form formula, except for the first moments (and characteristic function).

Le Cam's theorem can be useful, for large values of $n,k,n$. The bound is not relevant for small values such as your example ($n=5$, $k=2$, $x=4$), but even then the Poisson approximation is not too bad: it gives $0.1615270$ against the exact $\frac{1175}{6720}=0.1748512$

leonbloy
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