From Loading to Occupancy
A loading-based prediction begins with a simple question: if cells enter droplets at random, how often will a droplet receive two or more cells? This prediction is useful for understanding the scale of the multiplet problem before examining cell-level evidence. It is not, however, a direct count of detected multiplets. The model predicts how many droplets receive multiple cells under specified assumptions; an observed score-positive fraction also depends on how scores behave and which barcodes enter its denominator.
Let \(K\) be the number of cells entering a randomly selected droplet, and let \(\lambda\) be the mean number of cells per droplet across all droplets, including empty ones. Under the Poisson occupancy model, \(K\sim\operatorname{Poisson}(\lambda)\). Here \(\lambda\) is an expected occupancy, not a fraction of droplets and not necessarily the nominal concentration used to prepare a loading mixture.
The model assumes cells are allocated independently and randomly among comparable droplets, with a common mean occupancy \(\lambda\). In particular, it treats the occupancy opportunities as sufficiently homogeneous that one Poisson distribution is a reasonable description. It also assumes the relevant loading has been represented by the effective mean number of cells per droplet. If cell clumping, strong droplet-to-droplet variation, or substantial cell loss makes these assumptions implausible, the calculation can misrepresent actual occupancy.
Three Rates With Different Denominators
The probability of an empty droplet is \(P(K=0)=e^{-\lambda}\). The probability of exactly one cell is \(P(K=1)=\lambda e^{-\lambda}\). Therefore the fraction of all droplets expected to be occupied is \(P(K\geq1)=1-e^{-\lambda}\), while the fraction of all droplets expected to be multiply occupied is \(P(K\geq2)=1-e^{-\lambda}(1+\lambda)\). These are not the same rate.
If the denominator is restricted to occupied droplets, the relevant probability is conditional on \(K\geq1\). Because every multiply occupied droplet is occupied, the conditional multiplet fraction is the all-droplet multiplet probability divided by the occupied-droplet probability.
A further distinction is useful: \(P(K\geq2)\) counts droplets with at least two cells, not the fraction of individual cells that happen to be in such droplets. A droplet with three cells still contributes one multiplet droplet to the numerator. The model therefore predicts occupancy categories; the denominator and target quantity should be stated before interpreting a reported “doublet rate.”
Worked Example: Calculating Rates at a Mean Loading of 0.20
Suppose a hypothetical loading has mean occupancy \(\lambda=0.20\) cells per droplet. Calculate the empty, singly occupied, multiply occupied, and occupied-droplet probabilities, then find the multiplet fraction among occupied droplets.
The empty-droplet probability is \(P(K=0)=e^{-0.20}\approx0.8187308\), or about 81.87% of all droplets. The exactly-one-cell probability is \(P(K=1)=0.20e^{-0.20}\approx0.1637462\), or about 16.37%.
For all droplets, the multiplet probability is \(P(K\geq2)=1-e^{-0.20}(1+0.20)\approx0.01752310\), or about 1.752% of all droplets. The occupied probability is \(P(K\geq1)=1-e^{-0.20}\approx0.1812692\), or about 18.13% of all droplets.
Among occupied droplets, the multiplet fraction is \[ \frac{P(K\geq2)}{P(K\geq1)} = \frac{0.01752310}{0.1812692} \approx0.0966689. \] Thus about 9.667% of occupied droplets are predicted to be multiply occupied. The 1.752% and 9.667% figures are both correct for this model, but they use different denominators.
Why the Small-Loading Approximation Works
At low loading, the probability of receiving two or more cells is much smaller than the probability of receiving one cell. A short expansion makes the dependence on \(\lambda\) clear. Using the Taylor expansion \(e^{-\lambda}=1-\lambda+\lambda^2/2-\lambda^3/6+\cdots\), the all-droplet multiplet probability becomes \(1-e^{-\lambda}(1+\lambda)=\lambda^2/2+O(\lambda^3)\). Meanwhile, the occupied probability is \(1-e^{-\lambda}=\lambda-\lambda^2/2+O(\lambda^3)\). Dividing these expressions gives a conditional multiplet fraction of approximately \(\lambda/2\) for small \(\lambda\).
The quadratic leading term also explains why the all-droplet multiplet probability does not increase in direct proportion to loading near zero. Doubling a small \(\lambda\) approximately quadruples \(P(K\geq2)\), whereas the conditional rate among occupied droplets approximately doubles. These are local approximations, not exact scaling rules for every loading level.
Worked Example: Expected Multiplet Droplets in a Run
Suppose a hypothetical run produces \(D=10{,}000\) comparable droplets under the same \(\lambda=0.20\) model. Estimate the expected number of multiply occupied droplets and the expected number of occupied droplets.
For each droplet, the probability of multiple occupancy is \(0.0175230963\). By linearity of expectation, the expected number of multiplet droplets is \[ D P(K\geq2)=10{,}000(0.0175230963)\approx175.231. \] The expected occupied-droplet count is \[ D P(K\geq1)=10{,}000(0.1812692469)\approx1812.69. \] The ratio of these expected counts is approximately \(175.231/1812.69=0.0966689\), consistent with the conditional multiplet fraction. These are expected counts, not guarantees that a particular run will contain exactly those numbers.
The calculation concerns droplet occupancy. It does not imply that an analysis will identify precisely 175 multiplets: some may not be detected, and some singly occupied barcodes may be classified as positive by a score-based procedure.
Loading Predictions and Score-Positive Fractions
A score-positive fraction is an observed proportion defined by an analysis rule, usually among a specified set of cell-associated barcodes. It is not automatically equal to the model’s conditional multiplet probability. A score can fail to flag some true multiplets and can flag some non-multiplets. In addition, filtering or cell-calling can change which barcodes are included in the denominator.
For a simple illustration, suppose the target population is occupied droplets, the true multiplet fraction in that population is \(\pi=P(K\geq2\mid K\geq1)\), and a binary score has sensitivity \(Se\) and specificity \(Sp\) for that target. Sensitivity is the probability the score is positive for a true multiplet; specificity is the probability it is negative for a non-multiplet. If these rates apply to the same population and the same definition of a positive call, the expected score-positive fraction \(q\) is \(q=Se\pi+(1-Sp)(1-\pi)\). This identity separates true positives from false positives; it does not assume that score performance is perfect.
Worked Example: A Score-Positive Fraction Need Not Equal the Loading Rate
Continue the hypothetical \(\lambda=0.20\) example, so the predicted multiplet fraction among occupied droplets is \(\pi\approx0.0966689\). Suppose, solely for illustration, that a binary score has sensitivity \(Se=0.80\) and specificity \(Sp=0.98\) for identifying multiplets in this population. Find the expected score-positive fraction.
Among true multiplets, the positive fraction is \(Se\pi=0.80(0.0966689)\approx0.0773351\). Among non-multiplets, the false-positive fraction is \((1-Sp)(1-\pi)=0.02(1-0.0966689)\approx0.0180666\). Thus \[ q=0.80(0.0966689)+0.02(0.9033311)\approx0.0954017. \] The expected score-positive fraction is about 9.540% of occupied droplets, whereas the loading-based conditional multiplet fraction is about 9.667%. Their closeness here is specific to the hypothetical sensitivity and specificity; it does not make the quantities interchangeable.
Checking the Model and Choosing an Interpretation
A loading calculation is best treated as a model-based expectation or benchmark. Its usefulness depends on whether \(\lambda\) describes the effective cell occupancy of the droplet population being discussed. A nominal input count may not equal this value if cells are lost, retained unevenly, or not distributed as assumed. The model’s predictions can also be inconsistent with observed occupancy summaries; such disagreement is a reason to revisit assumptions and definitions rather than to force the observed data to match the formula.
When diagnosing a calculation, check the following points:
- Parameter meaning: \(\lambda\) is the mean cells per droplet, not the mean among only occupied droplets.
- Denominator: State whether the reported multiplet fraction is among all droplets or among occupied droplets.
- Definition: \(K\geq2\) includes droplets with two or more cells, including counts above two.
- Population alignment: A predicted fraction and an observed fraction should refer to the same type of droplets or barcodes before they are compared.
- Model fit: Clumping or heterogeneous occupancy can produce a distribution unlike a single Poisson model.
The main alternative to a loading-based prediction is an empirical fraction obtained from observed labels or score-positive calls. An empirical fraction reflects what the measurement and classification process recorded in its sample; the Poisson calculation instead predicts occupancy from a mean loading under stated assumptions. Neither quantity automatically substitutes for the other. Loading-based estimates are useful when direct labels are unavailable or as a check on plausibility, while observed fractions answer a question about the scored data and its denominator. Both have limitations: the Poisson model can be misspecified, and an observed score-positive proportion is not a true multiplet proportion unless classification errors and population selection are appropriately accounted for.
Common Mistake
A frequent error is to calculate \(P(K\geq2)\) and describe it as “the percentage of cell-containing droplets that are multiplets.” That probability uses all droplets, including empty ones, as its denominator. For the cell-containing subset, divide by \(P(K\geq1)\). A second error is to equate either occupancy probability with the observed score-positive fraction. That conclusion requires additional assumptions about score accuracy and matching denominators; the loading model alone does not provide them.
Check Your Understanding
Answer each question, then compare your work with the solutions below.
- For \(\lambda=0.10\), write the exact expressions for the all-droplet multiplet probability and the occupied-droplet probability. Which is the denominator for a conditional multiplet fraction among occupied droplets?
- At low loading, why is the all-droplet multiplet probability approximately proportional to \(\lambda^2\), while the conditional multiplet fraction is approximately proportional to \(\lambda\)?
- If the model predicts a 5% multiplet fraction among occupied droplets, can an observed score-positive fraction of 7% be used to conclude that the loading prediction is wrong? Explain.
- Name two conditions that could make a single Poisson occupancy model a poor description of actual droplets.
Solutions. (1) The exact probabilities are \(P(K\geq2)=1-e^{-0.10}(1+0.10)\) and \(P(K\geq1)=1-e^{-0.10}\). The conditional fraction divides the first by the second. Numerically, these are approximately \(0.00467884\) and \(0.0951626\), respectively. (2) The Taylor expansion gives \(P(K\geq2)=\lambda^2/2+O(\lambda^3)\), while \(P(K\geq1)=\lambda+O(\lambda^2)\). Their ratio is \(\lambda/2+O(\lambda^2)\). (3) No. The score-positive fraction may differ because of false positives, missed multiplets, or a different denominator. The two values are directly comparable only when the population and classification properties are appropriately accounted for. (4) Cell clumping and substantial variation in occupancy across droplets are examples; either can violate the random, homogeneous allocation assumption.