I am not understanding the explanation here in regards to expectations and probabilities for sampling without replacement. I am not seeing why we don't treat each draw as a unique random variable. For example, if there is only one ball left, then Pr(Xn = red) can either only be 1 or 0. We don't know which one it is until we get there, but clearly Pr(Xn = red) does not equal p. (Let's assume originally we have 3 reds and 7 blues.)

The below excerpt implies that Pr(Xi = red) = p for all i. And thus, it implies Pr(Xn = red) = 0.30, not 1 or 0.

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    $\begingroup$ You are mixing up different probabilities. At the last draw, the probabilities you give (zero or one) are necessarily conditional on all the preceding draws. You can analyze the situation this way, but it requires analyzing the entire tree of all possible sequences of draws and how the probabilities change after each draw--and that's a huge and complex endeavor. It pays, then, to study the clever methods people have found to achieve simple answers to such complex questions. $\endgroup$ – whuber Feb 25 at 16:10
  • $\begingroup$ Ok thanks. I mean I understand the way he's describing it. I just don't understand why it is viewed that way. Like if I were to say, before we even drew any balls, that the Pr(X7 = red) = p, that I agree with. But why wouldn't we view Pr(X7 = red) given X1-6? I don't understand when to view things in which way and when to use the different probabilities. $\endgroup$ – confused Feb 25 at 16:32

Your statement that "if there is only one ball left, then Pr(Xn = red) can either only be 1 or 0" isn't quite right. The value of Xn itself can only be 1 or 0, but the probability that Xn=1 is p, unless you have looked at all the other balls. When doing a sequential sampling without replacement, you do get more information about later draws, but that would require modeling the changing conditional probabilities at every step. It's much easier to view sampling N times without replacement as simply ordering all samples and taking the first N, as described in the text. In this view, every individual element has the same likelihood of being red, which is simply p.

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  • $\begingroup$ Is one more correct than the other or do they come to the same answer? Either just using the relationship above or modeling probs as they come? $\endgroup$ – confused Feb 25 at 16:34
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    $\begingroup$ @confused They give the same answer, but the sequential conditional probability modeling quickly becomes cumbersome. For binary-outcome trials, at the Nth trial, you're effectively marginalizing over 2^(N-1) possible outcome sets over the previous N-1 trials that will dictate the odds for your current trial. Much easier to view each trial in absence of the others - you can draw the balls one by one, but if you don't look at them, p never changes. Instead of asking what's the odds of X7 given X1-6, you can just look at X7 alone. $\endgroup$ – Nuclear Wang Feb 25 at 16:48

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