--The Flip-Side (Continued)--

John Kruschke

via instagram-

  This is a continuation of an earlier post of mine, on Instagram,  in regards to an old math riddle I am working on…here goes…

The “flip-side” (continued)--.

   Well, if we look at 1 flip of the coin, if the expected variance goes the other way, on the scale ( see my earlier “flip-side” post), 1 flip should have an expected variance of 0.4 , and 0.6, heads, and tails…If the expected variance model is congruent with the rest of the scale... Meaning, 0.4 and 0.6, for 1 flip of the coin…and 4, and 6, heads, or tails, at 10 flips…with 40 and 60 for 100 flips, and 400 and 600 for 1000 flips...So, does this indicate 60,000 flips heads, or 40,000 tails, at 100,000 flips??...I have my doubts... It seems good, except for the 0.4 heads, and 0.6 tails, at one flip...Meaning, it must be a whole number, of a value of at least 1…since we can not have a “partial flip”… And so, I feel that the variance should decrease exponentially, if the “expected deviation value” is small( the “range of deviation”), at low numbers of flips... Refining “expected deviation”, and not increasing the “range of expected deviation”...with larger “expected variance” values--

Summary--

   Either way, moving the scale refines the “expected deviation”... Is it 0.4 and 0.6, heads and tails at 1 flip??...Again, I feel the value must be expressed in a whole number...Indicating to me, that the numbers must go up the scale, and not down, to refine "expected variance"...Meaning, my original notion in my first post about this problem, is most likely the correct one...

Best wishes, john kruschke--   :o)

   P. S. : My phone battery just died, and I lost what I was thinking about before this post could be posted...Sadly, good work is often lost, before the notion is saved by the person working the problem…oh now I remember...Can we deduce, that if the values of heads or tails, is greater than 20 percent, out of 100, that we then know, from the data collected, that the odds of getting heads or tails, is not “50, 50”??...or “50 percent”??...and is it then instead, an un-unequal value?? ( I feel that the data collected from the study, must reflect the “expected variance” of the “percent value” assigned to the study, or, the model has a different percentile, than the assigned value in the study) ...”Roll-it-around-awhile” everybody--