You do get one guarantee, though: All the experiments are Bernoulli processes. In particular, the order of the trials is irrelevant.
I think those aren’t quite equivalent statements? If I pick my favorite string of bits, and shuffle it by a random permutation, then the probability of each bit being 1 is equal, the order is totally irrelevant (it was chosen at random), but it’s not Bernoulli because the trials aren’t independent of each other (if you know what my favorite string of bits is, you can learn the final bit as soon as you’ve observed all the rest.)
Correct, they are not equivalent. The second statement is a consequence of the first. I made this consequence explicit to justify my choice later on to bucket by the number of Rs but not their order.
The first statement, though, is also true. It’s your full guarantee.
I think those aren’t quite equivalent statements? If I pick my favorite string of bits, and shuffle it by a random permutation, then the probability of each bit being 1 is equal, the order is totally irrelevant (it was chosen at random), but it’s not Bernoulli because the trials aren’t independent of each other (if you know what my favorite string of bits is, you can learn the final bit as soon as you’ve observed all the rest.)
That’s what “in particular” means, i.e. the “the order of the trials is irrelevant” is a particular feature
Correct, they are not equivalent. The second statement is a consequence of the first. I made this consequence explicit to justify my choice later on to bucket by the number of Rs but not their order.
The first statement, though, is also true. It’s your full guarantee.