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I am confused with the concept of completely random actions. I was thinking of a very common statistical experiment in which we have a device or black box which randomly chooses between 1 and -1. If we infinitely do this processes, we will have set of randomly chosen infinite 1's and -1's. If we sum up all the elements of this randomly generated set we will probably get 0 every time we perform this experiment. This implies that it is a biased random as it is following the constraint that sum of elements of randomly generated infinite set is 0 every time.

But I was wondering, since the above procedure of choosing between 1 and -1 is completely random, then the sum of all the elements of randomly generated set must be a random number instead of being zero every time. If the sum of elements is any random number, then only it must be called as a unbiased random number and hence confirming the performed experiment to be random.

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closed as unclear what you're asking by WillO, sammy gerbil, Jon Custer, Kyle Kanos, coconut Dec 18 '17 at 16:38

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ You might find it worth asking this question on Phil.SE rather than physics as it looks likely that you're confused conceptually about what constitutes randomness; simply because a sequence is random does not mean that there are not determined quantities about it; as you've already pointed out, the mean or average is a determined number; another, very simple fact, is that we only see 0 and 1s, but that of course is tautological. $\endgroup$ – Mozibur Ullah Dec 15 '17 at 21:31
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    $\begingroup$ I'm voting to close this question as off-topic because it is about mathematics not physics. $\endgroup$ – sammy gerbil Dec 15 '17 at 21:53
  • $\begingroup$ The conceptual confusion arises because we're using determinate in at least two different ways here. $\endgroup$ – Mozibur Ullah Dec 15 '17 at 22:01
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That's why it's preferable to be precise and say that these numbers are not simply "random numbers", but rather random variables that follow a probability distribution, which, in the black box example, is the sum of two Dirac delta functions centered on $-1$ and on $+1$.

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