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(As Promised) Why Statistics Are Pseudoscience

Author
Akita T
Caldari Navy Volunteer Task Force
#81 - 2012-08-06 23:09:16 UTC  |  Edited by: Akita T
100 (or more) in a row are unlikely to ever be seen by a single person due to the infinitesimal chance of it happening with a realistic number of flips.

1 : (1*1,267,650,600,228,229,401,496,703,205,376) = 1 : 2^100 odds for a single set of 100 flips
3 : (2*1,267,650,600,228,229,401,496,703,205,376) = 3 : 2^101 odds for a single set of 101 flips
8 : (4*1,267,650,600,228,229,401,496,703,205,376) = 8 : 2^102 odds for a single set of 102 flips
20 : (8*1,267,650,600,228,229,401,496,703,205,376) = 20 : 2^103 odds for a single set of 103 flips
...and so on and so forth.

Notice however how the first number in the odds of actually getting at least 100 heads in a row quickly increase with added flips, while the second number multiplier grows at a slower rate.
The overall odds become better and keep getting better the more throws are made.

Eventually, the first number becomes more than 1,267,650,600,228,229,401,496,703,205,376 times larger than the second number used as the multiplier, and the odds become better than even to throw 100 or more heads in a row.
Granted, the number of throws is a lot larger than the number of throws likely to be possible for a single person to ever make during his expected lifetime, but that's a different story (and one that explains why you don't see anybody claiming he made 100 heads in a row).

Then the odds keep going up afterwards.
Eternum Praetorian
Doomheim
#82 - 2012-08-06 23:49:16 UTC  |  Edited by: Eternum Praetorian
Quote:
The odds of getting 7 heads in a row if you throw it 7 times is 1:128 (0.78125%).


Isn't it 1:128 (.0078125)?

(1/2)^7 = .0078125

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Eternum Praetorian
Doomheim
#83 - 2012-08-07 00:27:51 UTC
After some consideration with regards to your explanation of regression to the mean, it seems like it is saying that after a streak of 10 in a row, there is an an inclination for the streak to end, although that is not definite. The static 50:50 does not change, but this principle of distribution will tend to apply.


Is this not so?

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Akita T
Caldari Navy Volunteer Task Force
#84 - 2012-08-07 01:03:03 UTC  |  Edited by: Akita T
Eternum Praetorian wrote:
Quote:
The odds of getting 7 heads in a row if you throw it 7 times is 1:128 (0.78125%).

Isn't it 1:128 (.0078125)?
(1/2)^7 = .0078125

0.0078125 = 0.78125 / 100 = 0.78125%
0.5 = 50 / 100 = 50%
1 = 100 / 100 = 100%
That's what the % sign means - "percent", from the latin "per centum" meaning "from one hundred".

Eternum Praetorian wrote:
After some consideration with regards to your explanation of regression to the mean, it seems like it is saying that after a streak of 10 in a row, there is an an inclination for the streak to end, although that is not definite. The static 50:50 does not change, but this principle of distribution will tend to apply. Is this not so?

To regress towards some mean you'd first need a fixed mean to regress towards in the first place, but the average expected streak length changes with number of throws made, it doesn't stay the same as you make more throws.
Using an arbitrary fixed average streak length taken out of nowhere as a justification for guessing the likelihood of a streak of a fixed length to end in an ever-increasing number of throws is getting it completely backwards.

There's a 50% chance the streak will end on the very next throw, a 25% chance it will end in exactly two more throws, a 12.5% chance it will end in exactly three more throws, a 6.25% chance it will end in exactly 4 more throws, a 3.125% chance it will end in exactly 5 more throws, a 1.5625% chance it will end in exactly 6 more throws and so on and so forth.
That applies to the ending of a streak of ANY previous length, be it 2 or 10 or 20 or 30 throws.
A streak of length 2 is not more likely nor less likely to end than a streak of length 30.
Eternum Praetorian
Doomheim
#85 - 2012-08-07 01:16:46 UTC  |  Edited by: Eternum Praetorian
Ok now this is what I mean by applying relativity to statistics. If I get this wrong I will just let it go Lol so thx for bearing with me. You surely know your stuff.


But…


The odds of a streak ending in 3 more throws is 12.5 % from the perspective of the observer BEFORE he ever tosses the coin. This applies to the ending of a streak of any previous length, or any size and for any duration if you take into account general relativity before the coin is tossed. The static 50:50 of each toss does not have to change.


General Relativity being applied to statistical theory.
I guess you explained my idea for me. Shocked

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Akita T
Caldari Navy Volunteer Task Force
#86 - 2012-08-07 01:26:11 UTC  |  Edited by: Akita T
Eternum Praetorian wrote:
The odds of a streak ending in 3 more throws is 12.5 % from the perspective of the observer BEFORE he ever tosses the coin.

Well, actually, if you want to be accurate, the total chance to end the streak in 3 throws or less is always 87.5% regardless of streak length.
But as soon as he makes one extra throw, it either turns into 100% (if he just broke it) or into 75% to end it in 2 throws or less.
Making the second throw (if he didn't break it earlier), it turns into 100% (if he broke it now) or into 50% to end it in the very next throw.

Similarly, the total chance to end any streak of any previous length anywhere in the next 4 throws is always 93.75%.
Or the chance to end any streak in 5 throws or less, always 96.875%.
And so on and so forth.

But as soon as you make a throw, it's either 100%=broken and you revise probabilities for breakage in the remaining X-1 tosses, or keep the probability and re-extend the toss count for the expected failure probability.
The past has no relevance on future fixed probability events. It would only matter if the past could chance the odds of the very next coin toss, which it can't.

That's nothing new, that's old hat in statistics.
It not statistic's fault you didn't know Blink
Eternum Praetorian
Doomheim
#87 - 2012-08-07 01:31:07 UTC  |  Edited by: Eternum Praetorian
Your think about after you toss Big smile
I am thinking about before you ever even started.


Maybe there is a reason we are not communicating. The past has no relevance to the future if you are still standing within it.

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Akita T
Caldari Navy Volunteer Task Force
#88 - 2012-08-07 01:32:40 UTC
I think about both before and after a toss at the same time.
You think of either one or the other, never both at the same time.
That's the reason we're not properly communicating.
Eternum Praetorian
Doomheim
#89 - 2012-08-07 01:36:23 UTC  |  Edited by: Eternum Praetorian
Let me put it another way, I am thinking about before you even tried to flip it X times (relative to the observer) you are thinking of it before and after each toss.

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Akita T
Caldari Navy Volunteer Task Force
#90 - 2012-08-07 01:37:11 UTC  |  Edited by: Akita T
Quick example, I give you one single coin toss.

BEFORE you make the toss, what's the chance that single toss you make is heads ?
The correct answer assuming a perfectly unbiased coin is obviously 50:50.
50% heads, 50% tails. It didn't happen yet. Anything can still happen.

AFTER you make the toss and the toss came up heads, what's the chance the singe toss you make is heads ?
Obviously, it's 100% heads, 0% tails. It already happened.
You can't undo the past. You collapsed the wave function. You picked one of two alternate universes.
Pick whichever explanation is more convenient, fact remains, for you, that particular toss changed from 50% heads before the toss to 100% heads after the toss was made.
Eternum Praetorian
Doomheim
#91 - 2012-08-07 01:39:32 UTC  |  Edited by: Eternum Praetorian
Akita T wrote:
Quick example, I give you one single coin toss.

BEFORE you make the toss, what's the chance that single toss you make is heads ?
The correct answer assuming a perfectly unbiased coin is obviously 50:50.
50% heads, 50% tails. It didn't happen yet. Anything can still happen.

AFTER you make the toss and the toss came up heads, what's the chance the singe toss you make is heads ?
Obviously, it's 100% heads, 0% tails. It already happened. You can't undo the past.



But before I toss anything what are the odds of me generating a 3 heads or tails streak? from where I am standing relative to the observer? According to you it has an odds of ending of 87.5%

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Akita T
Caldari Navy Volunteer Task Force
#92 - 2012-08-07 01:45:37 UTC
Remember your earlier musings about the 70 flips.
What were the odds to make a particular set of 70 flips before you threw it ?
It was negligible. Yet you threw the exact sequence you actually threw.

Whether you go with the waveform collapse interpretation or with the multiple universe interpretation or any other interpretation you can come up with, the past is fixed and unchangeable, what happened happened with 100% certainty to have happened, the past can't change.
Akita T
Caldari Navy Volunteer Task Force
#93 - 2012-08-07 01:48:10 UTC  |  Edited by: Akita T
Eternum Praetorian wrote:
But before I toss anything what are the odds of me generating a 3 heads or tails streak?

It depends on how many throws you want to make in total.
And the chance to get a 3-heads streak in the remaining throws constantly changes as each subsequent throw is made.

Chance to get 3 heads in a row in the next 3 throws if previous throw was tails or this is the very first throw ? 1:8.
Chance to get 3 heads in a row the next 2 throws if previous throw was heads ? 1:4.
Chance to get 3 heads in a row the next throw if previous two throws were heads ? 1:2.
Eternum Praetorian
Doomheim
#94 - 2012-08-07 02:01:10 UTC  |  Edited by: Eternum Praetorian
I meant to say ending.
I know the odds by now Big smile

According to you it is "if you want to be accurate, the total chance to end the streak in 3 throws or less is always 87.5% regardless of streak length"



I am not sure how you got to that number but I will take you at your word for now. Is this statement true or false?

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Akita T
Caldari Navy Volunteer Task Force
#95 - 2012-08-07 02:10:17 UTC  |  Edited by: Akita T
Eternum Praetorian wrote:
According to you it is "if you want to be accurate, the total chance to end the streak in 3 throws or less is always 87.5% regardless of streak length"
I am not sure how you got to that number but I will take you at your word for now. Is this statement true or false?

50% chance to end in very next throw, 50%*50%=25% additional chance to end in exactly 2 more throws, 50%*50%*50%=12.5% additional chance to end in exactly 3 more throws -> 50%+25%+12.5% = 87.5% combined chance to end in 3 throws or less.
And a 12.5% chance to continue the streak at least 3 more throws.
And a 6.25% chance to continue the streak at least 4 more throws.
And a 3.125% chance to continue the streak at least 5 more throws.
And so on and so forth.
Regardless of streak length so far.

Limit of sum(0.5/x) for x going from 1 to infinity equals 1, so a p=1 (or if you prefer 100% chance) for any streak to end in "infinity" throws or less.
Anything less than infinity additional throws and the chance for a streak to end is BELOW 100%.
Not much below, but still below.
Eternum Praetorian
Doomheim
#96 - 2012-08-07 02:18:22 UTC
And there you have mathematical proof telling of a coin that has a static 50:50 probability at each toss and a sequence that has a descending probability relative to the observer before any event occurs. The two things do not contradict.


Can you not see it?

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Akita T
Caldari Navy Volunteer Task Force
#97 - 2012-08-07 02:22:53 UTC  |  Edited by: Akita T
Descending probability to continue an existing streak in increasing number of tosses, but NOT reaching 100% chance to end a streak in any FINITE number of tosses.

Also

Descending probability to get a streak given a FIXED number of tosses and an increasing streak length.
Which nobody is disputing, but is hardly relevant.

And

ASCENDING probability to get a streak given increasing number of tosses and FIXED streak length.
Which is telling you that a streak of any fixed length X is possible (even if very improbable) if you make at least X tosses, and it gets increasingly likely with additional tosses (never reaching 100% chance, but always increasing with additional tosses).
Eternum Praetorian
Doomheim
#98 - 2012-08-07 02:29:06 UTC
Akita T wrote:
Descending probability given a FIXED number of tosses and an increasing streak length.



That would be the definition of relative to the observer yes. A fixed number of tosses relative to the observer before a single coin was thrown. There is a definite descending probability to generate any long streak or long reoccurring pattern.

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Akita T
Caldari Navy Volunteer Task Force
#99 - 2012-08-07 02:30:07 UTC  |  Edited by: Akita T
Eternum Praetorian wrote:
That would be the definition of relative to the observer yes. A fixed number of tosses relative to the observer before a single coin was thrown.

Not really relevant to the discussion.

Quote:
There is a definite descending probability to generate any long streak

A probability that never reaches zero for any finite number.
Which means that there's still a chance to continue a streak for any finite number of additional tosses.

Quote:
or long reoccurring pattern.

Also not very relevant to the discussion.
Xenuria
#100 - 2012-08-07 03:16:41 UTC  |  Edited by: Xenuria
Hey Hey!
Hey guis

2=3

I can prove it

"Le equation with small hardly noticeable logical fallacy"

/ thread


This is basically what you have done.



Step 1: Make whacky Claim
Step 2: Support said claim with a lengthy argument with small hardly noticeable logical fallacy(s)
Step 3: Watch everybody but Xenuria fall for it



There is no static 50/50, by function of logic its not even a coin you are talking about its a 2 dimensional circular plane. Seriously. Stop FEEDING it. Although the op gets points for explicitly mentioning one of the fallacies he uses in his op.