Re: Probability of chain of n 'same side' deviations.

On Apr 13, 9:43 am, "Chalky" <chalkys...@xxxxxxxxxxxxxxxx> wrote:
On Apr 13, 9:19 am, "John (Liberty) Bell"





<john.b...@xxxxxxxxxxxxxxxxxx> wrote:
On Apr 13, 8:41 am, "Chalky" <chalkys...@xxxxxxxxxxxxxxxx> wrote:

I am examining the 'goodness of fit' of statistical data to theory
curves, and reckon that the probability of a sequence of n data points
all lying on the same side of the curve, should be 1/2^n-1, (basic
heads/tails binary probability theory, if the data deviations from
theory are supposed to be random). That, at least, should be the case
if n is the total number of available data points.

However, I am not so sure how to calculate probabilities (more
precisely, improbabilities), if you have, say, 7 highs in a row, out
of a total set of 10 data points,

First example, at least, seems to be easy.

Probability of next 6 coin flips after the first one, giving same
result, is 1/2^(n-1).
Else, probability of next 6 coin flips after the second one, giving
same result, is 1/2^(n-1).
Else, probability of next 6 coin flips after the third one, giving
same result, is 1/2^(n-1).
Else, probability of next 6 coin flips after the 4th one, giving same
result, is 1/2^(n-1).

Hence P = (q-n)/2^(n-1), where q is total size of set.

Your next example might be a bit trickier.
or, say, 7 highs in a row, followed
by 6 lows in a row, out of a total set of 16 data points.

Perhaps not trickier, after all.

1st chain thus has probability of 9/2^6
2nd chain thus has probability of 8/2^5=1/4

Total probability (product of the two) is thus 9/2^8 =~ 3.5%

Theory thus disproven by data with ~ 96.5% confidence!

Can anyone find any weakness/error in this logical conclusion?

Actually, I already have found a weakness, myself.

John said:

Probability of next 6 coin flips after the first one, giving same
result, is 1/2^(n-1).
Else, probability of next 6 coin flips after the second one, giving
same result, is 1/2^(n-1).
Else, probability of next 6 coin flips after the third one, giving
same result, is 1/2^(n-1).
Else, probability of next 6 coin flips after the 4th one, giving same
result, is 1/2^(n-1).

Hence P = (q-n)/2^(n-1), where q is total size of set.

The final statement is wrong, as it gives > 100% probability when q >>
n.
If a is the first probability, shouldn't the total probability be:
a + (1 - a)a + (1 - (1 - a)a)a + ....?

.

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