Here's an off the wall question. How did you generate that first sequence?
http://www.psychicscience.org/random.aspxGot in on the 12th sequence. Different sequence than the original I posted, but a max of three sets of three.
That doesn't answer the question.
I'm familiar with the Gambler's Fallacy (the erroneous belief that an outcome is "due" based on a prior sequence (e.g., some gamblers believe that "black" is due on a roulette wheel after seeing four "reds")), but "GF patterns" sounds like something you made up.
...
In any case, it isn't widely discussed:
You literally searched for the exact phrase "gambler's fallacy patterns"?
Try "clustering illusion". The term "gambler's fallacy pattern" isn't technical--I'm using it simply because I was talking about the gambler's fallacy earlier, so to refer to the pattern left by this, I say gambler's fallacy pattern.
And if you want to see the test I'm using somewhere else, try
this (random.org links to this site too... by the way).
Random.Org wouldn't display a single column of 55 to fit the screen. PsychicScience does.
It would if you read my instructions.
But that's okay... I can make this so easy to follow, even you can do it!
I guess I missed the label. Where's the fail?
The fail isn't merely in your error--it's in the fact that we've been through this before talking about this very thing.
Given P(A) and P(B), here's how you get [LATEX]$$P\left(A \cap B \right)$$[/LATEX]:
[LATEX]$P\left(A \cap B\right)=P\left(A \right)P\left(B \middle| A \right)$[/LATEX]
This equation is always true. The
product of P(A)P(B) means absolutely nothing useful, unless A and B are mutually independent, in which case [LATEX]$$P\left(B \right)=P\left(B \middle| A \right)$$[/LATEX].
So let's see if these events are mutually independent...
- [2] is impossible if there are 6 or more 1's in the first 15. [1], coincidentally, can only possibly happen if that's the case. So [1] and [2] are dependent.
- [1] and [2] together are a special case of [4], so they are dependent.
And as for [3], that deserves an entire new paragraph. The only way to get those successive "runs of 2" is to put a 1 in between. That
would leave ambiguity of where to add the final 1, except for the fact that the last run has to be two 1's, so if you put it at that end, it would not match (that makes the last a run of 3). If you pair it with any other 1, it wouldn't be "the runs, in order" either (whichever 1 you pair it with suddenly creates a run of 2). You can't stick it inside one of the runs of 2 either, because you said exactly how long those were. The only place left is to stick that 1 on the left...
Meaning... that your description of [3] is simply another way to state the exact sequence in the first place. So [3] is a special case of [1], [2], and [4], and is identical with what you gave anyway. Its probability is exactly 2
-30.
Your meaningless product is going to be smaller than the probability that you get the exact sequence by chance.
Um, the improbability of all four patterns occuring in the same set?
Not even in the same ball park.
...and what are you trying to show exactly? Step back a moment.
I analyzed 2,000 samples of 55 head flips using random.org, showing you the results. I didn't go and click over and over again trying to figure out just the right one to show you, I just did one run. You, on the other hand, are running individual runs, trying to
look for interesting patterns, and showing them to me one at a time.
Do you somehow think that by providing me anecdotal single samples that you personally ran through your confirmation bias filters, you somehow will nullify the results of my 2,000 samples?
But this is all a needless derail. The original question was, which coin is biased, based on the two sequences I originally gave. Let's stick to that, shall we?
This thread was started by Robin, and the topic of this thread is "The Fine Tuning argument". Therefore, I propose an alternative... let's just stick to things applicable to
that topic as opposed to whatever points you were making, shall we?