Scheduling meetings: are the odds in your favor?

Abstract

Polling all the participants to find a time when everyone is available is the ubiquitous method of scheduling meetings nowadays. We examine the probability of a poll with m participants and $\ell$ possible meeting times succeeding, where each participant rejects r of the $\ell$ options. For large $\ell$ and fixed $r/\ell ,$ we can carry out a saddle-point expansion and obtain analytical results for the probability of success. Despite the thermodynamic limit of large $\ell ,$ the ‘microcanonical’ version of the problem where each participant rejects exactly r possible meeting times, and the ‘canonical’ version where each participant has a probability $p = r/\ell$ of rejecting any meeting time, only agree with each other if $m\rightarrow \infty .$ For $m\rightarrow \infty ,$ $\ell$ has to be $O(p^{-m})$ for the poll to succeed, i.e., the number of meeting times that have to be polled increases exponentially with m. Equivalently, as a function of p, there is a discontinuous transition in the probability of success at $p \sim 1/\ell ^{1/m}$. If the participants’ availability is approximated as being unchanging from one week to another, i.e., $\ell$ is limited, a realistic example discussed in the text of the paper shows that the probability of success drops sharply if the number of participants is greater than approximately 4.

Graphical abstract

Many critical problems require building a consensus among stakeholders: A group of friends at a restaurant must agree on what appetizers to share. Nations at a climate summit must arrive at a consensus on emission goals. Scheduling meetings is a particularly simple problem of this kind: participants planning a meeting must agree upon a time to meet. Here, we study the combinatorics and statistics of the scheduling problem. We posit that this analysis can be fruitfully extended to more difficult and pressing problems that require consensus building such as the ones mentioned above.

When all responses of a scheduling poll such as Doodle are collated it usually turns out that there is no time that works for everyone. In this paper, we develop formulae that help estimate the likelihood that a poll will succeed if all participants make a good faith attempt to attend the meeting.

The paper is organized as follows. We begin with an analysis of the scheduling problem. In the conclusion, we discuss the generalization to broader consensus building problems and the relationship of our work to prior work on consensus building in the statistics [1] and statistical physics [2] literature.

1 Basic model of scheduling polls

We suppose that there are $\ell$ time slots available for the meeting and m respondents to the poll. It is assumed that each respondent has prior immovable commitments that conflict with r time slots on the poll. $g = \ell - r$ is thus the number of time slots each respondent can attend. The poll is taken to be arranged in the usual fashion. There are $\ell$ columns corresponding to the different time slots and m rows each corresponding to a different respondent. A given respondent can fill the $\ell$ columns of the poll in $C(\ell ,r)$ different ways and the total number of polls that can result is

$$\begin{aligned} N = [C(\ell ,r)]^m. \end{aligned}$$

(1)

We assume that all the different ways to fill the poll are equally likely.

We define failure as a poll in which there is no time slot that works for all respondents. Our objective is to calculate $\pi _0,$ the probability of failure. Evidently, if $m r < \ell$, the probability of failure is zero: no matter how the respondents distribute their responses they cannot block all the available time slots. When $m r = \ell$, failure becomes an option. In this case, it is easy to see that the number of distinct polls that fail, $n_0 = \ell !/[(\ell /m)!]^m$, and $\pi _0 = n_0 / N$, where N is given by Eq. (1) with $r \rightarrow \ell /m$. For $m r > \ell$, calculating the number of polls that fail is a more complicated problem of combinatorics. We find

$$\begin{aligned} \pi _0 = \sum _{j=0}^{\ell - r} \frac{ (-1)^j \ell !}{j! (\ell - j)!} \left[ \frac{ (\ell - r)! (\ell - j)! }{ \ell ! ( \ell - r - j )! } \right] ^m. \end{aligned}$$

(2)

The derivation of this formula is a bit lengthy and is relegated to Appendix A.

As an application of this formula with a remarkable outcome, suppose that the participants’ schedules are the same each week, a common circumstance in an academic workplace. Then, under reasonable assumptions, we find that it is risky to try to arrange a meeting of more than four participants! Because the schedule repeats from week to week, the maximum number of available slots is $\ell = 40$ and if we assume that each participant is free for only half of the available slots ($g = 20$), then the probability of failure $\pi _0 = 7.5\%$ for $m = 4$, but it rises precipitously to $\pi _0 = 28\%$ for $m = 5$.

Fig. 1

Histogram of probabilities $\pi _i$ that a poll will yield exactly i viable meeting times. In the plot $i = 1, \ldots , 7$. The histogram on the left (red) is for $\ell =15, m=9, r = 4.$ The histogram on the right (blue) is for $\ell =20, m = 9, r = 4.$ The failure probability is 0.34 for the first case and 0.027 for the second case, and the shape of the histogram for the two cases is also different. When the probability of failure is high, $\pi _i$ decreases monotonically; when it is low, $\pi _i$ rises to a peak corresponding to the most likely number of viable meetings

It is also of interest to know $\pi _i,$ the probability that the poll will yield exactly i viable times, where $i = 1, 2, \ldots , g$. This is shown in Fig. 1, which is easily generated using Eq. (A13) in Appendix A.

The basic model of scheduling analyzed here belongs to the class of mathematical problems called urn models. However, since scheduling polls are new, this particular model has not been analyzed in the literature [3].

2 Simplified model

We now describe a modification of the model that yields simpler expressions but is qualitatively and (under appropriate circumstances) quantitatively the same as the original model. In the simplified model, each respondent fills each entry by an independent toss of a biased coin that comes up “yes” with probability $p = g/\ell$ and “no” with probability $q = r/\ell$. In this model, the number of yes entries made by each respondent is equal to g on average, but it fluctuates from realization to realization. Evidently, in this case, the probability that a given column will be viable is $\mathcal{P} = p^m$ and the probability that the poll fails, i.e., no column is viable, is

$$\begin{aligned} \pi _0 = ( 1 - p^m )^\ell . \end{aligned}$$

(3)

Revisiting the two examples in Fig. 1, the failure probabilities for the simplified model are 0.38 and 0.055, respectively, in qualitative agreement with the earlier numbers. The relationship between the two models is similar to that between the microcanonical and canonical ensembles in statistical physics. It is an empirical question which one is a better model for real polls: there is no a priori reason to prefer one over the other.

A big advantage of the simplified model is that its behavior can be understood easily. For example, it is easy to see from Eq. 3 that $\pi _0\rightarrow 0$ if $\ell \rightarrow \infty$ with p and m fixed (unless $p = 0$, i.e., respondents reject every possible meeting time). In other words, a sufficiently large poll becomes too big to fail. Conversely, if the number of respondents $m\rightarrow \infty$ with p and $\ell$ fixed, then $\pi _0\rightarrow 1,$ i.e., the poll will always fail (except if $p=1$). If m and $\ell$ are both very large and $p < 1,$ then

$$\begin{aligned} \pi _0 \rightarrow \exp [- \ell p^m] \end{aligned}$$

(4)

which tells us that $\ell \sim (1/p)^m$ or larger if we want the poll to have a reasonable chance of succeeding, i.e., $\ell$ grows exponentially with m. If $\ell = K(1/p_c)^m$, where $K > 0$, then for large m

$$\begin{aligned} \pi _0 \rightarrow \exp [- K (p/p_c)^m] \end{aligned}$$

(5)

which has a sharp transition for $m\rightarrow \infty$ from $\pi _0 = 1$ for $p < p_c$ to $\pi _0 = 0$ for $p > p_c,$ and $\pi _0 = \exp (-K)$ for $p = p_c$ as shown in Fig. 2.

Fig. 2

Phase Transition. The probability for the poll to fail, $\pi _0$, plotted as a function of p, the fraction of time slots that each respondent is available. The plot is shown for $m= 3$, $m = 10$ and $m = 20$ where m is the number of respondents. We assume that $\ell = 2^m$, i.e., $p_c = 1/2$ and $K = 1$. Note that the transition in $\pi _0$ sharpens to an abrupt discontinuity in the thermodynamic limit $m \rightarrow \infty$ showing that the poll failure probability undergoes a first-order phase transition as p varies past $p_c$

As with the basic model discussed in the previous section, it is possible to obtain an expression for $\pi _i,$ the probability of ending up with i viable columns: $\pi _i = C(\ell , i) \mathcal{P}^i (1 - \mathcal{P})^{\ell -i}$ with Eq. (3) being the case when $i=0.$

Returning to our original model, one might hope that—as for the microcanonical and canonical ensembles of statistical physics—it will reduce to the simplified model when $\ell$ is large, even for finite m. In Appendix B, we have constructed a complex integral representation of the probability of failure $\pi _0$ (as well as $\pi _1, \ldots , \pi _g$) for the original model. Through a saddle-point analysis, we show that $\pi _0 \rightarrow \exp [-\ell f(m, r/\ell )]$ when $\ell \rightarrow \infty .$ Although the exponential decay of $\pi _0$ with $\ell$ is similar to the simplified model, rather surprisingly we find that the decay constant $f(m, r/\ell )$ is only equal to $-\ln (1-p^m)$ when $m\rightarrow \infty .$ Thus, for the two models to agree, it is not sufficient that $\ell \rightarrow \infty$; we also require $m\rightarrow \infty .$

Since the two models agree asymptotically in the limit that $\ell \rightarrow \infty$ and $m \rightarrow \infty$, the exponential scaling and first-order phase transition are exactly the same for both models.

3 Discussion

The finding that the size of a poll $\ell$ must grow exponentially with the number of respondents m raises the question whether scheduling a meeting is an exponentially hard problem in the sense of algorithmic complexity. It is true that the brute force method of conducting an exhaustive poll to schedule a meeting is exponentially hard, but it is possible that there is a more clever way to search the space of possible meeting times that only scales polynomially in the number of respondents. We leave this as an open question.

There are several simple generalizations of our model of scheduling polls that are possible. For example, one could imagine that there are two populations of respondents with different numbers of conflicting commitments. However, there may be diminishing returns to such generalizations. Real polls fail due to complex dynamic phenomena (e.g., a respondent stalls the poll, while the schedules of the early responders shift and fill up) and it may be desirable to incorporate these processes in the model instead.

Within our model, the only fail safe strategy for a given number of respondents m, each with r conflicting commitments, is to increase the number $\ell$ of available time slots, so that $\ell > m r$. This may not always be practical. Indeed, as we argued under the common circumstance that the schedule is the same from week to week under reasonable assumptions, it may be difficult to schedule a meeting of more than four individuals.

We conclude by framing some other consensus building problems in a similar way to the scheduling problem. A college department chair who wishes to make fair teaching assignments might create a poll in which the columns correspond to the courses that need to be taught and the rows to faculty members available to teach them. Success in this case requires not only that there is someone willing to teach every course, but that the teaching assignments can be made in a way that each instructor is assigned their proper teaching load. As a less academic problem with a similar structure, consider the group of friends choosing appetizers to share at a restaurant. The columns would correspond to the different appetizers on the menu and the rows to the names of the diners. A high bar for success would be to only consider appetizers that make the yes list for all the diners; in practice, it may be necessary to set a lower bar. For the problem of creating climate agreements, the rows of the model are the nation states that might sign the agreement and the columns correspond to different versions of the agreement ranging from the most stringent to more watered down variants. Each box of the model would be characterized by two numbers: the probability that the corresponding nation state would sign onto that variant of the agreement and a score corresponding to the climate impact of the nation state signing onto that particular variant. Within the climate field, even the process of writing the summary for policy makers of an IPCC Report [4] is a large problem in consensus building: every sentence (the columns) has to be approved by all the stakeholders (the rows). We see that these are all closely linked problems, but in each case, the statistical and combinatoric analysis required is somewhat different. Apart from mathematical analysis, it may be useful to analyze these problems from the perspective of behaviorial economics [5]. For example, one could ask whether a better outcome is achieved by offering the full menu of choices from the outset or by repeating the poll, offering options that correspond to less favorable outcomes only on later iterations. It is relevant to note here that there is a class of prior studies on consensus building [1, 2] that is concerned with modeling how stakeholders influence each other leading to an evolution of their opinions. Here, by contrast, we are interested not in time evolution but in the complex combinatorics that arise from having a multiplicity of both stakeholders and choices.

Appendices

Combinatorics

We define $n_i$ as the number of polls in which column i is viable and no other column is viable; $n_{ij}$ as the number of polls in which columns i and j are viable and no other columns are viable; $n_{ijk}$ as the number of polls in which columns i, j and k are viable and no other columns are viable; and so on to $n_{i_1 i_2,\ldots , i_g}$ as the number of polls in which the columns $i_1, i_2, \ldots , i_g$ are viable and no other columns are viable. On the other hand, we define $N_i$ as the number of polls in which column i is viable irrespective of what is happening in the other columns; $N_{ij}$ as the number of polls in which columns i and j are viable irrespective of what is happening in the other columns; $N_{ijk}$ as the number of polls in which columns i, j and k are viable regardless of what is happening in the other columns; and so on to $N_{i_1, i_2, \ldots , i_g}$ as the number of polls in which the columns $i_1, i_2, \ldots , i_g$ are viable regardless of what is happening in the other columns.

It is easy to see that

$$\begin{aligned} N_1= & {} [ C(\ell - 1, g - 1)]^m; \nonumber \\ N_{12}= & {} [ C(\ell - 2, g - 2)]^m; \nonumber \\&\ldots&\nonumber \\ N_{1, \ldots , g-1}= & {} [ C(\ell -g+1, 1) ]^m; \nonumber \\ N_{1,\ldots , g}= & {} 1. \end{aligned}$$

(A1)

However, we are more interested in evaluating $n_1, n_{12}, \ldots , n_{1,\ldots ,g}$. We can accomplish this, because there are a set of simple recursion relations that relate the two sets of quantities. Starting from the top, we note that

$$\begin{aligned} N_1= & {} n_1 + (n_{12} + \cdots + n_{1n}) + \sum _{2 \le i< j \le n} n_{1ij} \nonumber \\{} & {} +\cdots + \sum _{2 \le i_1< \cdots < i_{g-1} \le n} n_{1 i_1,\ldots , i_{g-1}}. \end{aligned}$$

(A2)

This is just the obvious statement that if the first column is viable that could be because it is the only viable column, or it could be one of exactly two columns that are viable, or one of exactly three $\ldots$ or one of exactly g columns that are viable. Next, we observe that $n_{12} = n_{13} = \cdots = n_{1n}$ and the multiplicity of such terms is $C(\ell -1,1)$. Similarly, $n_{1ij}$ are all the same independent of i and j and the multiplicity of these terms is $C(\ell -1,2)$. Reasoning in this way, we obtain

$$\begin{aligned} N_1= & {} n_1 + C(\ell -1,1) n_{12} + C(\ell -1, 2) n_{123} \nonumber \\{} & {} +\cdots + C(\ell -1, g-1) n_{1, \ldots , g}. \end{aligned}$$

(A3)

At the other extreme, obviously

$$\begin{aligned} N_{1, \ldots , g} = n_{1, \ldots , g} = 1, \end{aligned}$$

(A4)

because if g columns are viable, then that fixes the remaining columns to be not viable. Proceeding in this way, we can write down an inverted triangle of g relations of which Eq. (A3) is the base and Eq. (A4) is the apex.

Assembling these relations from apex to base, we have

$$\begin{aligned} N_{1,\ldots , g}= & {} n_{1,\ldots , g} \nonumber \\ N_{1,\ldots , g-1}= & {} n_{1,\ldots , g - 1} + C(\ell -g+1,1) \; n_{1,\ldots , g} \nonumber \\ N_{1,\ldots , g-2}= & {} n_{1,\ldots , g - 2} + C(\ell - g + 2, 1) \; n_{1,\ldots , g - 1} \nonumber \\{} & {} + C (\ell - g + 2, 2 ) \; n_{1,\ldots , g} \nonumber \\&\ldots&\end{aligned}$$

(A5)

To these relations, we can add the following:

$$\begin{aligned} N = n_0 + C(\ell ,1) n_1 + C(\ell , 2) n_{12} + \cdots + C(\ell ,g) n_{1,\ldots , g}. \nonumber \\ \end{aligned}$$

(A6)

Here, N is the total number of possible polls given by Eq. (A1). Equation (A6) simply says that the total number of polls consists of the sum of the number of polls that have zero, one, two, $\ldots , g$ viable columns. Inverting these relations successively yields

$$\begin{aligned} n_{1,\ldots , g}= & {} N_{1, \ldots , g} \nonumber \\ n_{1,\ldots , g-1}= & {} N_{1, \ldots , g-1} - C( \ell - g + 1, 1 ) \; N_{1, \ldots , g} \nonumber \\ n_{1, \ldots , g - 2}= & {} N_{1, \ldots , g - 2} - C(\ell - g + 2, 1) \; N_{1, \ldots , g-1} \nonumber \\{} & {} + C(\ell - g + 2, 2 ) \; N_{1, \ldots , g} \nonumber \\&\ldots&\nonumber \\ n_{12}= & {} N_{12} - C(\ell -2,1) N_{123} + C(\ell - 2, 2) N_{1234} + \cdots \nonumber \\{} & {} (-1)^{g-2} C(\ell - 2, g-2) N_{1, \ldots , g} \nonumber \\ n_1= & {} N_1 - C(\ell -1, 1) N_{12} + C(\ell - 1, 2) N_{123} + \cdots \nonumber \\{} & {} + (-1)^{g-1} C(\ell - 1, g - 1) N_{1, \ldots , g} \nonumber \\ n_0= & {} N - C(\ell ,1) N_1 + C(\ell ,2) N_{12} + \cdots \nonumber \\{} & {} + (-1)^g C(\ell ,g) N_{1, \ldots , g}. \end{aligned}$$

(A7)

Thus, we are led to the final expression

$$\begin{aligned} n_{1, \ldots , \gamma } = \sum _{j=0}^{g - \gamma } ( - 1)^j C( \ell - \gamma , j ) N_{1,\ldots , \gamma + j}, \end{aligned}$$

(A8)

where $\gamma = 1, \ldots , g$ and

$$\begin{aligned} n_0 = N + \sum _{j=1}^g (-1)^j C(\ell , j) N_{1,\ldots , j}. \end{aligned}$$

(A9)

The probabilities can then be worked out as follows:

$$\begin{aligned} \pi _1 = C(\ell ,1) \frac{n_1}{N}. \end{aligned}$$

(A10)

Here, $\pi _1$ is the probability that the poll is viable and that exactly one column works. Recall that N is the total number of polls given by Eq. (1) and $n_1$ given by Eq. (A8) is the number of polls in which the first column is viable and no other column is viable. The combinatoric factor in Eq. (A10) takes into account that any one of the other columns besides the first column might be the viable one. Similarly

$$\begin{aligned} \pi _2 = C(\ell , 2) \frac{n_{12}}{N}. \end{aligned}$$

(A11)

Here, $\pi _2$ is the probability that the poll is viable and exactly two columns work. $n_{12}$ is the number of polls in which the first columns rows are viable and no other columns work. The combinatoric factor is to take into account that polls in which a different pair of columns are viable also contribute to $\pi _2$. Generalizing we arrive at the main result of this section

$$\begin{aligned} \pi _\gamma = C( \ell , \gamma ) \frac{n_{1,\ldots ,\gamma }}{N}, \end{aligned}$$

(A12)

where $\gamma = 1, \ldots , g$. Making use of Eqs. (A1) and (A8), we can write Eq. (A12) more explicitly as the sum

$$\begin{aligned} \pi _\gamma = \sum _{j=0}^{\ell - r - \gamma } \frac{ (-1)^j \ell !}{j! \gamma ! (\ell - \gamma - j)!} \left[ \frac{ (\ell - r)! (\ell - \gamma - j)! }{ \ell ! ( \ell - r - j )! } \right] ^m.\nonumber \\ \end{aligned}$$

(A13)

Finally $\pi _0$, the probability that a poll will fail, can be computed from

$$\begin{aligned} \pi _0 = \frac{n_0}{N}. \end{aligned}$$

(A14)

Making use of Eqs. (A1) and (A9), we see that $\pi _0$ is also given by Eq. (A13) with $\gamma = 0$.

Although we derived $\pi _0$ as part of the recursive procedure that yields the complete set of probabilities, it is in fact possible to directly write down the expression for $n_0$ in Eqs. (A7) and (A9) based on the following interpretation of the individual terms. The first term is just the total number of polls. The second term subtracts the number of polls in which one column is viable, but it also subtracts the number of polls in which two columns are viable and it does so twice. The third term fixes this by adding back the polls in which two columns are columns are viable. Next, we observe that the first term also subtracts off the polls in which three columns are viable thrice; the second term then adds back the number of polls in which three columns are viable thrice. We do want to exclude the number of polls in which three columns are viable; hence the third term. Proceeding in this way, we arrive directly at Eq. (A9).

Asymptotics

1.1 Integral representation

In this appendix, we analyze the circumstances under which the simplified and original model are equivalent. To this end, we start by developing a complex integral representation of the probability of failure for the original model. The integral representation can then be used to derive the asymptotics of the model as $\ell \rightarrow \infty$ and identify the circumstances under which it matches the simplified model.

We consider a poll with $\ell$ columns and m rows, each column corresponding to a distinct time slot and each row to a different respondent. As a warm-up, we derive an integral representation for the total number of possible polls. Consider the expression

$$\begin{aligned} {[}(1 + e^{i \beta _1}) (1 + e^{i\beta _2}), \ldots , (1 + e^{i\beta _m}) ]^\ell . \end{aligned}$$

(B1)

Expanding the expression in square brackets above, we get $2^m$ terms corresponding to whether we pick 1 or $e^{i \beta _i}$ from the $i^{\textrm{th}}$ factor. If we associate choosing 1 with “accept” and $e^{i \beta _i}$ with “reject”, then each of the $2^m$ terms corresponds to a distinct filling of a column of the poll by the m respondents. If we raise the expression in square brackets to the power of $\ell$, each term in the resulting expression can be interpreted as a distinct way that the respondents could fill the entire poll if each respondent were at liberty to accept or reject every time slot that was offered to them. However, in our model, each respondent is only allowed to reject r time slots and must accept the remainder. Thus, we only wish to retain those terms in the above expansion that multiply out to the following expression:

$$\begin{aligned} e^{i r \beta _1} e^{i r \beta _2}, \ldots , e^{i r \beta _m}. \end{aligned}$$

(B2)

Evidently, we can impose this constraint by multiplying Eq. (B1) by the conjugate of Eq. (B2) and integrating over the phases $\beta _1, \beta _2, \ldots , \beta _m$. Thus, we arrive at the following integral representation of the total number of polls N:

$$\begin{aligned} N= & {} \frac{1}{(2\pi )^m} \int _0^{2\pi } d\beta _1 \int _0^{2\pi } d\beta _2\nonumber \\{} & {} \ldots \int _0^{2\pi } d\beta _m [(1 + e^{i \beta _1}) (1 + e^{i\beta _2}) \nonumber \\{} & {} \ldots (1 + e^{i\beta _m})]^\ell e^{-i r (\beta _1 + \beta _2 + \cdots \beta _m)}. \end{aligned}$$

(B3)

The integral is easily evaluated

$$\begin{aligned} N = \left[ \frac{1}{2\pi } \int _0^{2\pi } d\beta (1 + e^{i\beta })^\ell e^{-i r \beta }\right] ^m = [C(\ell , r)]^m \end{aligned}$$

(B4)

and yields the obvious result for the total number of possible polls in our model.

Now, let us derive an integral representation for $n_0$, the total number of ways that the poll can fail. To this end, it is useful to consider the expression

$$\begin{aligned} {[}(1 + e^{i \beta _1}) (1 + e^{i\beta _2}), \ldots ,(1 + e^{i\beta _m}) - 1]^\ell . \end{aligned}$$

(B5)

Now, the expression in square brackets represents all the different ways a column can be filled except for the filling corresponding to all acceptances. When the expression in square brackets is raised to the power $\ell$ individual terms in that expansion are in one to one correspondence with the different ways in which an unconstrained poll can fail. We can impose the constraint that respondents in our model must reject exactly r time slots and accept the remainder exactly as in the previous paragraph. Thus, we arrive at the integral representation

$$\begin{aligned} n_0= & {} \frac{1}{(2\pi )^m} \int _0^{2\pi } d\beta _1 \int _0^{2\pi } d\beta _2 \ldots \int _0^{2\pi } d\beta _m [(1 + e^{i \beta _1}) (1 + e^{i\beta _2}) \nonumber \\{} & {} \ldots (1 + e^{i\beta _m}) - 1]^\ell e^{-i r (\beta _1 + \beta _2 + \cdots \beta _m)}. \end{aligned}$$

(B6)

In the next section, we will use this integral representation to analyze the $\ell \rightarrow \infty$ asymptotics of $n_0$. In order to verify the equivalence of the integral representation to our previous exact result for $n_0$, it is helpful to use the binomial expansion

$$\begin{aligned}{} & {} {[}(1 + e^{i \beta _1}) (1 + e^{i\beta _2}) \ldots (1 + e^{i\beta _m}) - 1]^\ell \nonumber \\{} & {} \quad = \sum _{k=0}^\ell (-1)^k C(\ell , k) [ (1 + e^{i \beta _1}) (1 + e^{i\beta _2}) \ldots (1 + e^{i\beta _m}) ]^{\ell - k}. \nonumber \\ \end{aligned}$$

(B7)

Substituting Eq. (B7) in Eq. (B6), one can then perform the multiple integrals term by term to obtain

$$\begin{aligned} n_0 = \sum _{k=0}^{\ell - r} (-1)^k C(\ell , k) [C(\ell - k, r)]^m. \end{aligned}$$

(B8)

Note that the upper limit of the sum is $\ell - r$, because for terms with $k = \ell - r + 1, \ldots , \ell$, the integral vanishes identically. Equation (B8) exactly matches our previous exact result in Eq. (A9) supplemented with Eq. (A1).

By similar reasoning, one can write down the following integral expression:

$$\begin{aligned} n_{1 \ldots \gamma }= & {} \frac{1}{(2 \pi )^m} \int _0^{2 \pi } d \beta _1 \ldots \int _0^{2 \pi } d \beta _m \left[ (1 + e^{i \beta _1} ) \right. \nonumber \\{} & {} \quad \left. \ldots (1 + e^{i \beta _m}) - 1 \right] ^{\ell - \gamma } e^{- i \beta _1 r} \ldots e^{- i \beta _m r}. \nonumber \\ \end{aligned}$$

(B9)

Making use of the binomial expansion and evaluating the integral as above, we recover the expression for $n_{1, \ldots , \gamma }$ derived by a recursive procedure in the last section and given by Eq. (A8) supplemented with (A1). In principle, the integral representation Eq. (B9) can be used to derive the large $\ell$ asymptotics of $\pi _1, \pi _2, \ldots , \pi _g$, but in this paper, we focus exclusively on the asymptotics of $\pi _0$.

1.2 Saddle-point approximation

In this section, we will use the integral representations in Eqs. (B3) and (B6) and the method of steepest descents [6] to derive the large $\ell$ asymptotic behavior of the probability of the poll failing

$$\begin{aligned} \pi _0 = \frac{n_0}{N}. \end{aligned}$$

(B10)

In both the numerator and denominator, if we replace $e^{i\beta _j}$ with $z_j,$ the integral is over the unit circle for each $z_j$. We deform the closed contour of integration to pass through the saddle point of the integrand, which is obtained by requiring that the derivative of its logarithm with respect to $\beta _j$ should be zero. For the denominator, we have

$$\begin{aligned} \ell \frac{i e^{i\beta _j}}{1 + e^{i\beta j}} - i r = 0, \end{aligned}$$

(B11)

that is

$$\begin{aligned} \frac{z_j}{1 + z_j} = \frac{r}{\ell }. \end{aligned}$$

(B12)

The equation is the same for all j. We call the solution to this equation as $z_c$. Hence, at the saddle point

$$\begin{aligned} z_j = z_c = \frac{r}{\ell - r}. \end{aligned}$$

(B13)

for all j. We now approximate the integral in Eq. (B3) by the saddle-point value of the integrand to obtain

$$\begin{aligned} N{} & {} \sim \left[ \left( \frac{\ell }{\ell - r}\right) ^\ell \left( \frac{\ell - r}{r}\right) ^r\right] ^m \nonumber \\{} & {} = \left[ \frac{1}{(1 - r/\ell )^{1-r/\ell } (r/\ell )^{r/\ell }}\right] ^{m \ell }. \end{aligned}$$

(B14)

Note that we have arranged the final expression to be in terms of $\ell$ and the ratio $r/\ell$, since we plan to take the limit $\ell \rightarrow \infty$ while holding $r/\ell$ fixed.

(Strictly, the saddle-point approximation includes an additional prefactor obtained by integrating a Gaussian with a width determined by the second derivatives of the logarithm of the integrand with respect to the $\beta _j$s. However, the dominant $\ell$ dependence is captured by our approximation. Moreover, because we are ultimately evaluating the ratio of two saddle-point integrals, the ratio of subdominant prefactors will become independent of $\ell$ and will be essentially a factor of order unity. In the limit $m \rightarrow \infty$ as we will see, the ratio actually tends to unity further justifying our neglect of this factor.)

Turning to the numerator $n_0,$ a similar analysis tells us that the saddle point occurs at

$$\begin{aligned} \frac{e^{i\beta _j} \prod _{p\ne j} (1 + e^{i\beta _p})}{\prod _p (1 + e^{i\beta _p}) -1} = \frac{r}{l}. \end{aligned}$$

(B15)

By symmetry, all the $z_j$’s are equal at the saddle point, so this simplifies to

$$\begin{aligned} \frac{z (1 + z)^{m-1}}{(1 + z)^m - 1} = \frac{r}{\ell }. \end{aligned}$$

(B16)

It is convenient to rewrite Eq. (B16) in the form

$$\begin{aligned} 1 + z - \frac{1}{(1 + z)^{m-1}} = \frac{\ell }{r} z. \end{aligned}$$

(B17)

We denote the solution to Eq. (B17) as $z_0$. Now, we approximate the integral in Eq. (B6) by its saddle-point value and make use of Eqs. (B10) and (B14)

$$\begin{aligned} \pi _0 \approx [(1+z_0)^m - 1]^\ell z_0^{-rm} \left[ (1 - r/\ell )^{1-r/\ell } (r/\ell )^{r/\ell }\right] ^{m \ell }. \nonumber \\ \end{aligned}$$

(B18)

Equation (B18) is the main result of our saddle point analysis. For fixed $r/\ell$ and m and therefore $z_0$, we see that $\pi _0$ is of the form $\exp [\ell f(m, r/\ell )].$

Further progress depends on the value of m. There are three cases to consider: (1) The first case is $m \le \ell / r$. For $m < \ell /r,$ we know that $\pi _0 = 0$ exactly and we are able to reproduce this finding with the saddle-point approximation. For $m = \ell /r$, the probability of failure $\pi _0$ is non-zero and can be computed exactly and we can compare the saddle-point approximation to the exact result. (2) The second case is $m \rightarrow \infty$. In this case, the saddle-point equations simplify greatly and we find in this limit that $\pi _0$ exactly matches the simplified model. Thus, we establish that the two models are equivalent in the limit $\ell \rightarrow \infty$ and $m \rightarrow \infty$. (3) Finally, we have to contemplate the case that $m > \ell / r$, but m is not necessarily very large. In this case, the saddle-point Eq. (B17) has to be solved numerically. We demonstrate that the saddle-point approximation agrees well with the exact result Eq. (A13). However, we find surprisingly that in this case, the basic model and the simplified model do not agree quantitatively. In the following subsections for each case, we demonstrate the conclusions summarized here.

1.2.1 $m \le \ell /r$

It is easy to verify that if $m < \ell /r,$ the only solution to the saddle-point Eq. (B17) is at $z_0=0.$ As $z_0\rightarrow 0,$ Eq. (B18) for $\pi _0$ simplifies to

$$\begin{aligned} \pi _0 \approx \lim _{z_0\rightarrow 0} m^\ell z_0^{\ell - r m} \left[ (1 - r/\ell )^{1-r/\ell } (r/\ell )^{r/\ell }\right] ^{m \ell }. \end{aligned}$$

(B19)

If $\ell > m r,$ this is equal to zero, in accordance with the exact result.

If $\ell = m r,$ we have

$$\begin{aligned} \pi _0 \approx m^\ell \left[ (1 - r/\ell )^{1-r/\ell } (r/\ell )^{r/\ell }\right] ^{m\ell }. \end{aligned}$$

(B20)

This can be compared numerically with the exact result

$$\begin{aligned} \pi _0 = \frac{\ell !}{(r!)^m}[C(\ell , r)]^{-m}. \end{aligned}$$

(B21)

In the bottom row of Table 1, we have computed $\pi _0$ using the exact result Eq. (B21) in the first column and using the saddle-point approximation to it using Eq. (B20) in the second column for $\ell = 20, r = 5, m = 4$ corresponding to a large value of $\ell$ that satisfies $\ell = m r$. We find not surprisingly that both results are very small.

1.2.2 $m\rightarrow \infty$

For $m\rightarrow \infty ,$ Eq. (B17) for the saddle point in $n_0$ reduces to

$$\begin{aligned} 1 + z = \frac{\ell }{r} z, \end{aligned}$$

(B22)

i.e., $z_0 = r/(\ell -r).$ This is the same saddle point as in the denominator Eq. (B13). In other words, $z_0 = z_c$. From Eq. (B18), we then obtain

$$\begin{aligned} \pi _0 \approx \frac{[(1+z_c)^m - 1]^\ell z_c^{-rm}}{(1+z_c)^{m\ell } z_c^{-rm}} = \left[ 1 - \left( \frac{\ell - r}{\ell }\right) ^m\right] ^\ell .\quad \end{aligned}$$

(B23)

This limiting form exactly matches Eq. (3) for the simplified model discussed in the main paper, in which each respondent either accepts or rejects each meeting slot by an independent biased coin with probability probability $q = r/\ell$ for rejection and probability $p = (\ell - r)/\ell$ for acceptance.

1.2.3 $m > \ell / r$

Finally, we consider the case that $m > \ell / r$, but m is finite. As noted above, the basic model and the simplified model agree quantitatively for $\ell \rightarrow \infty$ and $m \rightarrow \infty$, but for fixed finite m, the agreement is only qualitative. This is surprising and contrary to the usual situation in statistical physics where the microcanonical and canonical ensembles generally agree in the thermodynamic limit.

Table 1 tabulates the value of $\pi _0$ for $\ell = 20$ and $r = 5$ for values of m ranging from 4 to 14. The first column gives the exact $\pi _0$ values for the basic model evaluated using Eq. (2) and the third column gives the exact $\pi _0$ values for the simplified model using Eq. (3). As claimed above, we see that for the largest value of m, the two models are in reasonable quantitative agreement but not for the smaller values.

Table 1 Table of values of $\pi _0$ computed exactly ($\pi _{0;{\textrm{ex}}}$) and with the saddle-point approximation ($\pi _{0;{\textrm{sp}}}$) for $\ell =20$ and $r=5$, for various values of m

These conclusions are borne out by the saddle-point analysis. Recall that $z_0$ is the solution to the saddle-point Eq. (B17) and $z_c = r/(\ell - r)$ is the solution to the saddle-point Eq. (B13). For finite m, the two are not equal. It follows from Eq. (B18) that:

$$\begin{aligned} \pi _{0;{\textrm{sp}}} \approx \left( \frac{[(1+z_0)^m - 1]^\ell z_0^{-rm}}{[(1 + z_c)^m - 1]^\ell z_c^{-rm}}\right) \pi _{0;{\textrm{can}}}, \end{aligned}$$

(B24)

where $\pi _{0;{\textrm{sp}}}$ denotes our saddle-point approximation to $\pi _0$ for the basic model and $\pi _{0;{\textrm{can}}}$ is the exact result for the simplified model. Equation (B24) shows that to the extent that the saddle-point expression is an accurate approximation to the basic model, $\pi _0$ for the basic model does not tend to the result for the simplified model even as $\ell \rightarrow \infty$ for finite m.

To quantitatively compare the saddle-point approximation for the basic model to the exact result, we solve Eq. (B17) numerically to determine $z_0$ and use Eq. (B24) to calculate the saddle-point approximation. The results are tabulated in the middle column of Table 1 and also in Fig. 3.

Fig. 3

Plot of $\ln (-\ln (\pi _0))$ as a function of m using the exact result Eq. (2) for the basic model (red circles), the saddle-point approximation to it (blue triangles), and the results for the simplified model using Eq. (3) (black squares). Note that the saddle-point approximation to the exact result for the basic model is good on the scale of this plot for all m and that the basic model and the simplified model agree for large m but deviate markedly for smaller values of m

As we see from the table, the saddle-point approximation deviates from the exact result for small m,  but is of the correct order of magnitude, and is much closer to the exact result for the basic model than the simplified (canonical) model. Although a better saddle-point approximation could be obtained if we were to carry out the Gaussian integral around the saddle point instead of only taking the integrand, such accuracy is not needed here. For large m, the saddle-point approximation and the exact result for the basic model are virtually indistinguishable, and approach the canonical result as $m\rightarrow \infty$ as expected.