Supplementary Online Appendices
Appendix A. Comparing actual distributions of ants at the corners of experimental food items to expected
distributions
The aim here is to create a series of models to describe how many distinct corners are occupied when 4 ants carry
a square food item when they have no preference for occupying the corners. We then compare the results of
these models with the results of experiments using real ants and determine whether they are significantly
different, and thus whether real ants are more or less likely to occupy a corner than would be expected if they had
no preference for occupying corners.
Description of the problem
The sides of the square food item were divided in 5 sections of equal length (figure A1). During an experiment, one
can record whether an ant chooses a sector belonging to a corner (sectors A, B, C and D) or not (sectors O).
Figure A1
We define k as number of corners occupied by at least one ant. In terms of probability, our problem is similar to
describing the probability of picking ‘k’ distinct letters (excluding O) in the following set:
{A,A,B,B,C,C,D,D,O,O,O,O,O,O,O,O,O,O,O,O}.
We can define three situations:
Letter picked are immediately replaced (e.g. sectors can be occupied by more than one ant).
Letter picked are removed (e.g. sectors can be occupied by only one ant).
Letter picked are removed and if a corner letter is picked (A, B, C or D) the other corner letter is
removed (e.g. sector and corner can be occupied by only one ant).
Situation with replacement
-
This situation is equivalent to having n=4 independent trials with each trial resulting in one of 5 possible outcomes
(picking either A, B, C, D or O). This is equivalent to describing a multinomial distribution. For notation, event ‘A’
(picking the square at one of edge A) occur ‘a’ times (and respectively for other letter)
Knowing:
Event A, B, C and D occur with the same probability pc=0.1 (pc : probability of joining one of the
corners).
- Event O occurs with probability qc=1-4pc=0.6 (qc : probability of joining any non-corner sector).
We have:
-
𝑃1 (𝐴 = 𝑎, 𝐵 = 𝑏, 𝐶 = 𝑐, 𝐷 = 𝑑, 𝑂 = 𝑜) =
𝑛!
𝑝 𝑎+𝑏+𝑐+𝑑 𝑞𝑜
𝑎! 𝑏! 𝑐! 𝑑! 𝑜! 𝑐
The equation can be interpreted as the number of permutations where the same event happens (in this situation
order is not important) times the probability of such an event happening.
The goal is to determine the probability density function of observing k distinct corners sectors occupied. As an
example for 1 corner occupied we are looking for: 𝑃(𝐴 ≠ 0, 𝐵 = 0, 𝐶 = 0, 𝐷 = 0) which can be viewed as a
sum of multinomial with = [1 𝑡𝑜 4] . If we define 𝜙1 (𝑘) as the probability of observing k corner occupied, we
derived 𝜙1 (𝑘) for each values of k:
𝜙1 (0) = 𝑞𝑐 4
𝜙1 (1) = 4 [(𝑝𝑐 + 𝑞𝑐 )4 − 𝑞𝑐 4 ]
𝜙1 (2) = 6 [(2𝑝𝑐 + 𝑞𝑐 )4 − 2(𝑝𝑐 + 𝑞𝑐 )4 + 𝑞𝑐 4 ]
𝜙1 (3) = 4 [(3𝑝𝑐 + 𝑞𝑐 )4 − 3(2𝑝𝑐 + 𝑞𝑐 )4 + 3(𝑝𝑐 + 𝑞𝑐 )4 − 𝑞𝑐 4 ]
𝜙1 (4) = (4𝑝𝑐 + 𝑞𝑐 )4 − 4(3𝑝𝑐 + 𝑞𝑐 )4 + 6(2𝑝𝑐 + 𝑞𝑐 )4 − 4(𝑝𝑐 + 𝑞𝑐 )4 + 𝑞𝑐 4
If ants do display a preference we expect the value of pc to differ from 0.1 .
Situation without replacement and no preferences for corner sector
In this situation when one sector is occupied it will not be available for subsequent individuals. Because there is no
replacement, the probability of one happening will not follow an exactly multinomial distribution.
If we define:
- K: number of distinct corner occupied
- C: total number of sectors occupied at a corner
We have:
4!
12!
1
𝑃(𝐾 = 𝑘, 𝐶 = 𝑐) = (4−𝑐)!2𝑐𝑘 2𝑘 (12+𝑐−4)! 𝑇
with 𝑇 = 20 ∗ 19 ∗ 18 ∗ 17
Because each corner is equivalent, this probability must be multiplied by the number of permutation resulting in
the same outcome:
4!
.
(4−𝑘)!(𝑐−𝑘)!(2𝑘−𝑐)!
Finally, the probability of having ‘k’ distinct corners occupied is given by:
min(2𝑘,4)
𝜑(𝑘) = 𝑃(𝐾 = 𝑘) =
∑
𝑐=𝑘
4!
4!
12!
1
2𝑘
𝑐−𝑘
(4 − 𝑘)! (𝑐 − 𝑘)! (2𝑘 − 𝑐)! (4 − 𝑐)! 2
(12 + 𝑐 − 4)! 𝑇
The ‘simplicity’ of this formulation is allowed by the restriction that each corner can be carried by at most 2
individuals.
Situation without replacement and a preference for some sectors
We consider the same situation as above but with a preference for either corner sectors or non-corner sectors
such that pc ≠ .1 .
Given the complexity of finding a general analytical formulation, but the limited number of possible outcomes (K
ranging between 0 and 4 ), we simply derived all the probabilities, depending on p, of each event. To illustrate the
logic behind this, we present an example:
We define:
pc: probability of carrying one of the two sectors of a corner (if there is no preference pc=0.1)
x=20pc and y=20-4x
We then derive the probability of 2 corners being occupied:
First let consider the probability of the distribution: 𝑃(𝐴 = 1, 𝐵 = 1, 𝐶 = 0, 𝐷 = 0)
-
one possibility of obtaining such a result is by drawing {a,b,o,o} in this order:
𝑃({𝑎, 𝑏, 𝑜, 𝑜}) =
-
𝑦
𝑥 2 𝑦 (𝑦 − 12)
𝑥
2𝑥
2𝑥 𝑦
20 × (20 − 2) × (20 − 2 ) × (20 − 2 − 12)
alternatively the same outcome with different order {a,o,b,o}
𝑃({𝑎, 𝑜, 𝑏, 𝑜}) =
𝑦
𝑥 2 𝑦 (𝑦 − 12)
𝑥
𝑥 𝑦
2𝑥 𝑦
20 × (20 − 2) × (20 − 2 − 12) × (20 − 2 − 12)
And so forth for each order.
Then we consider: 𝑃(𝐴 = 2, 𝐵 = 1, 𝐶 = 0, 𝐷 = 0) using the same logic. And finally we count all the possibilities
of having the same outcome but with different corners occupied.
The process is tedious but gives us the possibility of drawing an analytical expected distribution 𝜙2 for each value
of pc ranging between 0 and 0.25 (total avoidance of corner, up to total avoidance of non-corner). We also note
that 𝜑 is a special case of 𝜙2 for 𝑝𝑐 = 0.1 .
Situation without replacement, a preference for some sectors, and the two sectors of each corner can only be
occupied by one ant
This situation is similar to the previous one, but corners cannot be co-occupied. Using the same example:
First let us consider the probability: 𝑃(𝐴 = 1, 𝐵 = 1, 𝐶 = 0, 𝐷 = 0)
-
one possibility of obtaining such a result is by drawing {a,b,o,o} in this order:
𝑃({𝑎, 𝑏, 𝑜, 𝑜}) =
-
𝑦
𝑥 2 𝑦 (𝑦 − 12)
𝑦
20 × (20 − 𝑥) × (20 − 2𝑥) × (20 − 2𝑥 − 12)
alternatively the same outcome with different order {a,o,b,o}
𝑦
𝑥 2 𝑦 (𝑦 − 12)
𝑃({𝑎, 𝑜, 𝑏, 𝑜}) =
𝑦
𝑦
20 × (20 − 𝑥) × (20 − 𝑥 − 12) × (20 − 2𝑥 − 12)
And so forth for each order.
In this situation, 𝑃(𝐴 = 2, 𝐵 = 1, 𝐶 = 0, 𝐷 = 0) cannot happen (same corner occupied twice). Finally we count
all the possibilities of having the same outcome but with different corners occupied.
Again the process is tedious but gives us the possibility of drawing an analytical expected distribution 𝜑3 for each
value of pc ranging between 0 and 0.25 (total avoidance of corner, up to total avoidance of non-corner). Having an
analytical expectation for the distribution of occupied corners, we can perform a maximum likelihood estimation
of the parameter pc. Finally, having estimated pc , given that there are 2 corner sectors per corner and there are 12
non-corner sectors and 8 corner sectors, we can show that the preference ratio for corner sectors over side sectors
𝑝
𝑝
equal: 𝑟 = 12( 𝑐⁄2)⁄(1 − 8( 𝑐⁄2)). Thus r = 5.4.