# Proof that random assignment guarantees the same probability for the individual to be selected?

Consider a completely randomized design, where every unit is randomly assigned to a treatment group. Let's say we have 30 observations and 3 treatment groups. When we choose one observation at random, it has 1/30 of chance of being selected, then the second one has 1/29, and so on. However, I've been told that there is a mathematical proof showing that, in the end, every single observation has the same chance of being selected to be part of a treatment group.

Is this true?

• Since the procedure does not distinguish groups or observations, suppose on the contrary you had a proof that a particular unit had a greater chance of being selected for a particular group than some other unit. The same proof would apply with the units swapped, demonstrating their chances must have been the same all along.
– whuber
Oct 31, 2018 at 1:21
• @whuber: An excellent point. However, there are some algorithms for simple random sampling (e.g., one-at-a-time sampling below) that do distinguish the observations (e.g., by taking them in an arbitrary chosen order) and therefore fail the symmetry property you are relying on.
– Ben
Oct 31, 2018 at 21:33
• @Ben But even the algorithm you describe does not distinguish among observations: all have equal chances of being selected first, etc.
– whuber
Nov 1, 2018 at 0:00
• @whuber: In the algorithm I describe, you can take the observations in any order, so you are free to distinguish them. There is no requirement to randomise the order (since you are randomising the allocation).
– Ben
Nov 1, 2018 at 0:34
• @whuber: For example, you could choose to take the observations in a specified order based on a covariate value, and so long as the allocation was done using the described randomisation, you would still get SRSWOR.
– Ben
Nov 1, 2018 at 0:36

Your description of the sampling method is a little unclear, but I think I understand what you are referring to. Simple random sampling is defined by the property that every sample allocation of a given size is equally likely, which implies that each individual has the same marginal distribution of being sampled into a particular group. It is possible to conduct simple random sampling using various algorithms, and one method to do this is to allocated objects to groups on a one-at-a-time basis, where you allocate one object to a group at random, then another, etc., until you have allocated everyone. I am not sure if this correctly corresponds to the sampling method you are describing, but if so, it appears that you are asking for proof that this allocation method meets the requirements of simple random sampling.

Simple random sampling: Suppose we want to allocate a population of $$n$$ objects into $$m$$ groups of sizes $$n_1,...,n_m$$. (Since all objects will be allocated to exactly one group we have $$\sum n_i = n$$.) For each object with index $$i$$ we let $$S_i = 1,...,m$$ denote the group that object is allocated to and we let $$\boldsymbol{S}=(S_1,...,S_m)$$. The allowable allocations leading to the required group sizes $$\mathbf{n}$$ are:

$$\mathscr{S}(\mathbf{n}) \equiv \Bigg\{ \boldsymbol{s} \in \{ 1,...,m \}^n \Bigg| \sum_{i=1}^n \mathbb{I}(s_i = s) = n_s \text{ for all } s=1,...,m \Bigg\}.$$

Under simple random sampling we require that every allowable allocation is equally likely:

$$\mathbb{P}(\boldsymbol{S}=\boldsymbol{s}) = 1 \Big/ {n \choose \mathbf{n}} \quad \quad \quad \text{for all }\boldsymbol{s} \in \mathscr{S}(\mathbf{n}) .$$

Any sampling procedure that obeys this probability condition constitutes a valid method for simple random sampling. One consequence of this condition is that:

$$\mathbb{P}(S_i=s) = \frac{n_s}{n} \quad \quad \quad \text{for all } s = 1,...,m.$$

Note that the above mathematical requirement only gives us a defined requirement for this type of sampling - it does not give us an algorithm by which to accomplish this. Simple random sampling can be accomplished in various ways, and is often implemented using iterative methods, such as allocating objects using the one-at-a-time method below.

One-at-a-time simple random sampling: In this sampling method we allocate the objects one-at-a-time to the groups. Each object is allocated to the groups so that its probability of allocation to a group is proportional to the number of unfilled spots in that group. Taking the objects in order of our (arbitrary) labeling, we use the recursive probabilities:

$$\mathbb{P}(S_{i} = s | S_1=s_1,...,S_{i-1}=s_{i-1}) = \frac{n_s - \sum_{j=1}^{i-1} \mathbb{I}(s_j=s)}{n-i+1}.$$

(The numerator is the number of remaining spots in group $$s$$ after allocation of the previous objects, and the denominator is the total number of remaining spots in all groups.) Breaking down the joint probability of the allocation vector into its conditional parts, for any $$\boldsymbol{s} \in \mathscr{S}(\mathbf{n})$$, we have:

\begin{aligned} \mathbb{P}(\boldsymbol{S}=\boldsymbol{s}) &= \prod_{i=1}^n \mathbb{P}(S_{i} = s | S_1=s_1,...,S_{i-1}=s_{i-1}) \\[6pt] &= \prod_{i=1}^n \frac{n_s - \sum_{j=1}^{i-1} \mathbb{I}(s_j=s)}{n-i+1} \\[6pt] &= \frac{\prod_{i=1}^n (n_s - \sum_{j=1}^{i-1} \mathbb{I}(s_j=s)) }{\prod_{i=1}^n (n-i+1)} \\[6pt] &= \frac{\prod_{s=1}^m \prod_{i=1}^{n_s} (n_s-i+1) }{\prod_{i=1}^n (n-i+1)} \\[6pt] &= \frac{\prod_{s=1}^m n_s! }{n!} \\[6pt] &= 1 \Big/ {n \choose \mathbf{n}}. \end{aligned}

This confirms that the one-at-a-time method is a valid method for simple random sampling. Note that this holds regardless of the order in which the objects are allocated to the groups. Hence, it remains a valid algorithm for simple random sampling even if the order of allocation of the objects is chosen by some biased process. Each allowable sample allocation is equally likely, and so the marginal probability of allocation is the same for each object.