If 3 observations are selected from 30 observations for 3 treatments then we will get a sample containing 3 from a population containing 30.

In this case, the probability of each of the 30 observations to be selected in the Ist draw is 1/30.

The probability of each of the remaining 29 observations to be selected in the 3rd draw is 1/28.

Now, an observation will be selected in the sample if it is selected in any of the 3 draws.

The probability that it is selected in the sample is given by

P(it is selected in 1st draw) + P(it is not selected in 1st draw).P(it is selected in 2nd draw) + P(it is not selected in 1st draw & 2nd draw).P(it is selected in 3rd

draw)

which becomes

1/30 + (29/30).(1/29) + (29/30). (28/29). (1/28)

which is equal to 3/30 =1/10

Therefore, the probability that each of the 30 observations is selected in the sample is 3/30 =1/10.

This is the required result.

Random assignment is a theoretical concept. It is defined as follows:

" If an assignment is such that each individual has equal probability of being selected then the assignment is called random."

The following result is an immediate consequence of this concept / definition:

" Random assignment guarantees the same probability for the individual to be selected. "

Thus, there is no other mathematical proof (and / or there is no necessity of other mathematical proof) to show that random assignment guarantees the same / equal probability for the individuals to be selected.

If 3 observations are selected from 30 observations for 3 treatments then we will get a sample containing 3 from a population containing 30.

In this case, the probability of each of the 30 observations to be selected in the Ist draw is 1/30.

The probability of each of the remaining 29 observations to be selected in the 3rd draw is 1/28.

Now, an observation will be selected in the sample if it is selected in any of the 3 draws.

The probability that it is selected in the sample is given by

P(it is selected in 1st draw) + P(it is not selected in 1st draw).P(it is selected in 2nd draw) + P(it is not selected in 1st draw & 2nd draw).P(it is selected in 3rd

draw)

which becomes

1/30 + (29/30).(1/29) + (29/30). (28/29). (1/28)

which is equal to 3/30 =1/10

Therefore, the probability that each of the 30 observations is selected in the sample is 3/30 =1/10.

This is the required result.

If you identify all possible sample units and then select a subset at random then any one sample unit has the same probability of being selected as any other. That is, if you have 100 possible sample units and you withdraw three, and then you put them back and try again, and again, and again, and 100,000 more "again," and then plot the distribution of the number of times each unit was selected, then you will get a uniform distribution. Each unit has the same probability of selection as any other.

Sometimes the problem is not this easy. Say you are sampling people in a grocery store. In this case, there will be specific demographics that tend to shop at specific times. So a random sample of people who go to this store might miss a specific demographic. Then you get into things like purposive sampling, stratified sampling.

The interesting term is to "select a subset at random". I have nowhere seen any practical explanation of what this means that was not circular or simply nonsensical.

Since the word "random" gets used a lot (sometimes correctly) it is worth questioner stating if the question means SRS (simple random sampling). The standard definition is what Timothy gave (that a SRS of size n (of N) means that each possible set of size n has equal probability). I believe this is a definition of SRS, so will sound circular if trying to prove it. It is worth also adding that this is different from a sampling process that means each element is equally likely to be chosen (e.g., in the question, assume the 30 people are split into 7 classrooms, then each classroom is randomly allocated a condition).

I wonder if by "observation" you mean participant?

You do not need a mathematical proof, random assignment implies each participant will have a 1/3 chance of being chosen for any given group.

I wrote earlier "If you identify all possible sample units and then select a subset at random"

Jochen made a fair point that this statement is not really clear. So please replace with: You take all possible sampling units. From this you want to take a sample of size n. Whatever value you take for n is the subset. Each element in n is taken from the population at random without replacement. Once you have chosen n then n is fixed for repeating this experiment.

As David indicated this is simple random sampling. All units in the population are given a number and a random number is generated to select the next individual taken. Most people will accept a pseudorandom number generator, though it is possible to get pure random values.

https://www.random.org/

https://cran.r-project.org/web/packages/random/vignettes/random-intro.pdf

I do agree with the opinion of Lawrence Patihis that there is no need of a mathematical proof of the the statement

" Random assignment guarantees the same probability for the individual to be selected "

since it is an immediate consequence of the concept / definition of random assignment as mentioned below:

" Random assignment, which is a theoretical concept, is defined to be an assignment such that each individual has equal probability of being selected ."

I also agree with Lawrence Patihis; you do not need a mathematical proof, the concept "random assignment" employed has made all your observation to have a 1/30 probability of been selected.

Proof by definition. To select individuals from a population at random I select individuals in such a way that all individuals have an equal probability of selection.

I'm still not sure if this is a "proof by definition" or just a circular statement.

Negation: Can we find an example of a unequal probability of sampling some units versus other units given that the sample was withdrawn randomly? Or, if I withdraw a sample such that the probability of selecting one unit over another is different is there any way that I can claim the sampling was random?

Can I get a sample that most statisticians would agree is random that violates one or more of the following:

1) I cannot control the outcome (I select the unit if it is green regardless).

2) The outcome is not linked to the sample (the random number generator always gives a value of 7 if the unit is square).

3) The selection of one unit does not influence the probability of drawing other units. If I draw the queen of spades (probability 1/52) from a deck of cards that cannot influence the probability of drawing the two of diamonds in the next draw. Every card must have a probability of 1 in 51 for the second draw, 1 in 50 for the third draw, and so forth.

4) The random value generator is actually random.

5) The outcome of withdrawing one sample of n units does not influence the outcome of subsequent samples of n units.

Even a negative example that follows from arbitrary selection but satisfies rules 1 through 3 could be instructive. Drawing balls from a hat is arbitrary, but weighting some balls so that others rise to the top and thereby increases their probability of selection violates rule 2.

The example must survive repetition. A single outcome is uninteresting because we know that it is possible to randomly withdraw all the cards from a deck in order by suit. It is simply highly unlikely without using a trick that violates one or more of the conditions.

Attached the proof in the picture!

"The random value generator is actually random."

What do you mean with this, precisely?

2nd thought: using a random generator is just shifting the problem to another instance, but not solving it. If I could specify what makes a number generator "random", I could directly select subjects/specimens "randomly". Or the other way around: why don't one use a random number generator to tell us the number of the card to be drawn from a deck of cards?

And a twist on that: consider the deck is sorted (as it is packed after production and unshuffled). Person A knows this, person B doesn't. If both draw a card, do both, one or none of them draw "at random"? Consider the deck was shuffled, but incidently it happend that the shuffled cards just have the same order than an unshuffled new pack. What now?

And one more thing to think about: consider a person draws a card. It happened to be the k-th card from the deck (let k be a specific number between 1 and 52). If you knew, that this person always draws the k-th card, would this change your mind about if this was a random draw?

Actually random as opposed to pseudorandom. I am not quite sure how a sufficiently bad pseudorandom number generator would affect the results. In a true random number generator, knowing the current value does not provide any information about another value no matter how many values are selected.

Does it make a difference?

1) Number each card and have a random number generator select a value from 1 to 52 and take that card.

2) Always select the 5th card, but put the cards in order based on a set of 52 random numbers. So type out in Excel all the cards in column 1, in column 2 enter =rand(), and then sort on column 2 with column 1. Place the cards in this order and withdraw the kth card. Reduce N by 1, and rerandomize. There will be a problem if N-n < k.

3) Combine approach 1 and 2.

If the person always draws the kth card, and I always put the deck in the same order then this would not be random.