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I asked this question on Physics SE and it was suggested that it should be posted here.

I will ask the question in terms of an observation on a hypothetical coin toss experiment.

If you like, you may answer the question based upon the coin toss description alone.

For little more curious and patient readers, I have provided link to a paper that I wrote about an observation on an actual quantum entanglement experiment data.

You are welcome to read that paper at http://vixra.org/abs/1609.0016 and answer based upon the observation noted in the paper if you prefer, your choice, but please do not judge this question based upon writing style of the paper.

Coin Toss experiment -

Let us gather the imbalance (Total tails - total heads) between outcomes over time. Or, let us look at difference between total number of tails and total number of heads at any time, in a coin toss experiment of say 280000 tosses. Assume the tosses are conducted by unbiased, untiring machines in a random manner.

Suppose the total number of tails remains higher than total number of heads for most of your experiment but eventually they become equal. The difference first increases to a peak and then decreases to make the outcome 50/50. The experiment is concluded at that point, instead of continuing to see how the imbalance behaves afterward.

To clarify, further - for example, we are halfway the experiment, say at trial number 140000. At this point, total number of tails is say 72000 and total number of heads is say 68000 and till this point, number of tails has consistently been higher than number of heads, and the gap has been widening with time. Then, starting with say trial number 140001, the gap starts narrowing down with time and by the time we reach trial 280000, the gap is erased completely for the first time. And the experiment is concluded at this point.

The number 280000 is arbitrarily chosen to indicate it is a large enough experiment.

Now, suppose you do this experiment with 4 different coins at the same time, in parallel and same trend is seen with all 4 coins. i.e. Number of tails takes a lead, the lead keeps accumulating and peaks and then starts to clear and outcome become 50/50 in the end, at the same time, for all 4 coins.

The magnitude of the peak imbalance is not very high - between 1% to 2.34%. Which can look like a probabilistic distribution.

But the question is about magnitude, plus same direction of imbalance in all 4 coins and clearing of the imbalance at the same time in all 4 coins.

The question is - does the observation described in the coin toss example, suggests role of anything other than independent probability? Something like - A mechanism that initially favors number of tails to a limit, and then tilts to opposite and starts to favor number of heads till the imbalance is cleared?

What explains the same direction/trend that was observed in all 4 coins. 1) Same direction/trend (tails lead first) of imbalance , 2) concurrent clearance of imbalance in all coins (same heads and tails total in the end).

In case you chose to read actual observation of entanglement data at http://vixra.org/abs/1609.0016 then -

The reason for probing experimental data this way, is that Bells inequality is often used to prove or disprove certain theories related to quantum entanglement.

If the data itself can be found not independent over time, then Bell's inequality would not be applicable to such data.

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  • $\begingroup$ Please make this question more concise. $\endgroup$ – usεr11852 Sep 4 '16 at 22:21
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Polite criticism: anything from vixra.org should put you on red alert.

The phenomenon of seeing heads lead over tails is formally encapsulated by the concept of ballot-sequences. According to Bertrand's theorem, suppose after $n$ steps we observe $p$ heads and $q$ tails, with $p>q$. Then the probability that the number of heads always exceeded the number of tails up to time $n$ is:

$$\frac{p-q}{p+q}.$$

Roughly speaking, this implies that if you force the number of heads/tails to equal each other at time $n$, then the chance of heads leading tails up till then is approximately $\frac{1}{n}$

More generally, if you want to calculate all the ways of there being at least as many heads as tails up to time $n$, where the number of heads equals the number of tails (note that the time must be even), then you're looking at Dyck paths, which are counted by Catalan numbers. Note that this allows for the number of heads equaling the number of tails at times prior to $n$. The typical average height of a Dyck path is $\sqrt{\pi n}$.

The "mechanism" here a result from 2D random walk theory, which says that the number of heads will equal the number of tails infinitely often with probability 1. This means that you will, perhaps after a very long time, always return to heads=tails. And then do so again and again. This result is also irrespective of conditioning. This means that if you're in a situation where you have $p$ heads and $q$ tails with $p>q$, then you will still return to heads=tails, again possibly after a very long time.

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  • $\begingroup$ Thank you for the details, and I have heard the vixra criticism earlier. Have no other option though. Now I want to clarify the question - Let us start experiment with 4 different coins. Obviously number of heads and tails is same in the beginning (0, 0). As we start tossing, number of tails starts becoming higher and higher to a peak difference. Then the difference starts clearing (meaning more heads start showing) till the experiment is concluded after a large number of trials when number of heads becomes = number of tails. This same trend happens in all 4 coins. $\endgroup$ – kpv Sep 5 '16 at 23:31
  • $\begingroup$ Meaning in all 4 coins, the number of tails leads most of the experiment duration (> 93% of duration) in all 4 coins. Question is - Can this be attributed to simple probability, or there is a possibility of some other mechanism? $\endgroup$ – kpv Sep 5 '16 at 23:33
  • $\begingroup$ @kpv: It depends on the duration. From the above reasoning, the chance of heads leading tails up to time $n$ is roughly $1/n$. That's just for one experiment and all these experiments are independent (unless I'm misunderstanding something). In which case the chance of seeing heads lead tails up to time $n$ for all 4 coins is $1/n^4$, which becomes exceedingly small as $n$ gets large. So with 280000 tosses it's something like $10^{-22}$. $\endgroup$ – Alex R. Sep 5 '16 at 23:36
  • $\begingroup$ AlexR, Total number of tails lead per question. You saying heads. It does not matter but just to clarify. So, per probability, it is a rare chance. Is that correct. $\endgroup$ – kpv Sep 5 '16 at 23:41
  • $\begingroup$ Yes, it should be quite rare. In fact, at the onset, you would need all four coins to generate tails, which already happens with probability 1/16. $\endgroup$ – Alex R. Sep 5 '16 at 23:43

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