How do I show that the two methods of permutation test are both valid? My main objective is to show the methods described below are really the same. However, I am having difficult both formulating the idea clearly and proving my assertion. Below is my attempt.
Suppose we are give a categorical dataset characterized by the following:


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*Treatment Group(T): Buy DVD(B), 40; Not Buy(N), 20.

*Control Group(C): Buy DVD(B), 60; Not Buy(N), 20.


Now, we wish to derive the distribution of the permutation resample. Note this will be a discrete distribution with four possible outcomes, namely $(T, B), (C, B), (T, N), (C, N)$. Now, I present two ways of deriving the permutation resample.
The Standard Approach
Let a deck of $140$ cards be given. Label 100 of them as $B$ and the rest 40 as $N$. Then shuffle the cards carefully but randomly, then distribute the first 60 of them to the $T$ pile and the rest 80 to the $C$ pile. Now, we investigate the probabilistic distribution of the resulting resample. 


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*$\mathbb{P}((T,B))=\mathbb{P}(B)\cdot\mathbb{P}(T\mid B)=\frac{100}{140}\cdot\frac{60}{140}=\frac{6000}{140^2}$ since the probability any card, regardless of $B$ or $N$, being in the top 60 cards in the shuffled deck  is 60 out of 140 (So the conditional probability is really the same as the unconditioned as since being $B$ or $N$ is independent from going to $T$ pile or $C$ pile). Similarly, we derive the following.

*$\mathbb{P}((T,N))=\mathbb{P}(N)\cdot\mathbb{P}(T\mid N)=\frac{40}{140}\cdot\frac{60}{140}=\frac{2400}{140^2}$

*$\mathbb{P}((C,B))=\mathbb{P}(B)\cdot\mathbb{P}(C\mid B)=\frac{40}{140}\cdot\frac{80}{140}=\frac{3200}{140^2}$

*$\mathbb{P}((C,N))=\mathbb{P}(N)\cdot\mathbb{P}(C\mid B)=\frac{100}{140}\cdot\frac{80}{140}=\frac{8000}{140^2}$


An easy check is to add up all the probability values, this gives us 1, a successful check. Now, if we can show the probability of each possible outcome using the below approach is the same as the standard approach, we are done.
An Alternative Approach
Let a deck of $140$ cards be given. Label 60 of them as $T$ and the rest 80 as $C$. Then shuffle the cards carefully but randomly, then distribute the first 100 of them to the $B$ pile and the rest 40 to the $N$ pile. Again, picking a random card from the resulting resample, the probability it is a $T$ is trivially $\frac{60}{140}$. Furthermore, the chance a random card is $B$ is simply $\frac{100}{140}$ since the chance any card is placed in the top 100 cards is $\frac{100}{140}$. Note that the $\mathbb{P}(B\mid T)=\mathbb{P}(B)$ since the regardless a card is $T$ or $N$, they are have the same probability of being in the top 100 of the shuffled pule. Therefore, we have that
$$\mathbb{P}((T,B))=\mathbb{P}(T)\cdot\mathbb{P}(B\mid T)=\frac{60}{140}\cdot\frac{100}{140}=\frac{6000}{140^2}.$$ Note that this is exactly what we had in the standard method. The probabilities of other outcomes then follows from the same logic and should each be equal to their corresponding ones in the previous approach.
To sum up, hopefully I showed the probabilistic distribution of the two resampling methods are identical by showing each outcome has the same probability in both of the methods.
 A: Some terminology will help here.  Basically you have a 2x2 contingency table, and your methods are trying to produce a series of independent replications when the "row" (T or C) and "column" (B or N) are independent.  I would think about setting up a joint distribution for the counts in each cell of the table, under each distribution.  To help start you off, I'll do the alternative method.  When you take a deck of cards and label then as T and C, then the distribution of the number of units that are "buy" and "treatment", $n_{TB} $, is a hypergeometric distribution:
$$p(n_{TB})=\frac{\binom {100}{n_{TB}}\binom {40}{60-n_{TB}}}{\binom{140}{60}}$$
To reason this out, note that there are $\binom {140}{60} $ ways to label the data $T $ and $C $.  Then we have $\binom {100}{n_{TB}} $ ways to pick how many of the $B $ assignments also get assigned to $T $, and for each way we also have $\binom {40}{60-n_{TB }} $ ways to assign the remaining $60-n_{TB} $ units as "TN".  Note this implies that $60\geq n_{TB}\geq 20$.
Also, once we have $n_{TB} $ the rest follows from what is known about the table.  We have $n_{TN}=60-n_{TB} $ and $n_{CB}=100-n_{TB} $ and finally $n_{CN}=n_{TB}-20$.  Note that these reproduce the constraints provided $60\geq n_{TB}\geq 20$.
Now you need to show that you get the same hypergeometric distribution for $n_{TB} $ if you apply the standard method.
hope this helps!
