--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--  

--The Flip Side--

John Kruschke

via Instagram--

  I'm working on an old probability riddle... Here goes...

The question--

  The question… is about if you flip a coin, over and over…can It always be heads?? I say, no... Yes, it's a 50, 50, percent chance, each flip…yet, If we keep flipping, I feel it can not always be heads... And I feel we can utilize a math technique I use often, to check...if we crunch the problem a new way--

The gimbals--

    I believe we can prove/disprove the 50 percent value in the problem, via increasing the number of flips, meaning, if we increase the number of flips, the 50 percent chance each flip model should remain congruent...So, if we examine 100 flips, we should likely see 42 heads, and 58 tails, or a variation of heads and tails, being more, or less, occurring...and if not... I suggest that the coin is “weighted”, more on one side... Even though the odds should be 50, 50 each flip, each time…and the numbers generated should reflect that fact, if the percentage expected is correct...to put an even finer point on it, the "range of variance" is likely between 40 and 60, for heads and tails values... And we can then flip 100 times, 100 times...to gain the "percentage of variance", per-hundred flips...as we can see, the range of expected variance is 20, heads or tails...and half of the value is 10...and this would bring the "mean" of the "curve", at an even 50 percent of the time heads, and tails... So, I suggest, that if the odds are truly "50, 50"...the numeric "percentage" of the odds, must be expressed, long term, or the value is not an accurate expression of the physical model being studied...meaning... It can't always be heads…Unless it's heads on both sides...?? So, In my opinion, At 10,000 flips, the variance should be 5,000 heads, and 5,000 tails...Unless, the original expected variance of 20, is still present... Except expressed in the ten's column of the values...Like 4,952 heads, and 4,948 tails... ?? 

The “flip-side”( summary)--

   It could be, that the variance increases??...Which is an interesting thought, if we flip ten times, we should get between 4, and 6, heads, and tails... And if we flip 100 times, we should get between 40, and 60, heads, and tails... And if we flip 1000 times, we should get between 400, and 600, heads, or tails...So, how do we deduce if my first notion, of the variance going exponentially down, with more flips, is correct??…or this last one??…With the value increasing, per-flip??

  I will post more about that another time( I’m out of free time to post with my cell-phone)…

Best wishes, john kruschke--