Have you practiced a lot, or did being that dumb come naturally?
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This is false. Example:
Let P(a)=P(b)=p. Let event b occur if and only if a occurs, i.e. P(a|b)=P(b|a)=1. Then P(a\cap b)=P(a|b)\cdot P(b)=P(b|a)\cdot P(a)=1\cdot p=p\neq p^2=p\cdot p=P(a)\cdot P(b).
That doesn’t sound like it’s on topic, but go on…
This sentence is a mess. What on earth are “the total equally probable positions of both a and b”? Also, we started with a and b being events. Do events have positions? What on earth is “a\times b” when a and b are events?
It’s not my proof, and I’m not the one who is going to build it. You made the claim, you were challenged to present a proof for it, you build it from what ever straw you think can withstand the slightest summer breeze.
Okay, perhaps I’m unfamiliar with the professional vernacular of the subject. Please, enlighten me. What is a “letter” and how does it “contain elements” of any sort?
This still makes no sense for the reason stated above. However, I want to point out something. You say a=b=c=d. But if a=b, then P(a\cap b)=P(a\cap a)=P(a)\neq (P(a))^2=P(a)\cdot P(b). Maybe it is uncharitable of me to not just intuitively gather that you changed the meanings of the variables you use mid-proof without mentioning it, but at the same time, you were the one worried that an overly rigorous proof would fly over our collective heads here, so I figure maybe expecting some amount of care of expression or consistency wouldn’t be too much. Evidently, it is.
Cool. If only that was the actual subject, oh, if only.
Again, I’m not after generating a proof, I’m just asking for one I can read. Remember, this was the challenge:
You keep straying off topic and producing this incoherent mess instead. I say you lied when ever you said
Do with that what you will. I’m sure the image of your competency has been well beyond recovery for far longer than I have been a user on this board, but perchance some may think more of your character, should you at least apologize for lying to their faces so unashamedly.
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No its basic statistics.
a and b both occurring is the condition. They are independent events.
Combinatorics in this case is about events with a given probability happening in sequence. This is why I use base 2 for 50% chance of drawing a neutral residue.
Sorry for my cryptic equations. a x b is the probability of both events a and b happening. The probability of a and b is the same. I can see I caused confusion. The probability of the individual events are the same. This is what I mean A=B=C=D.
The letters I used are ABCD which represents 4 positions in a sequence. In the example they contain an element from the same group of elements with the same probability of a single element occurring.
The probability of ABCD are all the same in this case.
We have an esteemed group here that believe in an enormous series of statistical miracles. Their admiration is not expected since I am the cold water guy. 
Thanks for trying to work through this.
No, it is not. I provided a trivially constructed counter-example, proving that it is not. Your “nuh-uh”-ing at it makes no difference.
The condition for what?
You did not say that in the beginning.
Alright. Weird proprietary notation, but fine.
You did not say that in the beginning. If it matters, you should have. If no, then there is no need to mention it now either.
Okay, so more weird notation. I guess the capital letter denotes the probability of the event denoted by the lowercase one. Fair enough. Now, for some reason we assume that the probabilities of the four pairwise different events a, b, c, and d is identical. You did not say that in the beginning. I guess this is just natural to assume for no reason, or whatever.
“The example” being what example? Also, I guess we are going to just immediately re-purpose the notation from the last passage for something else now, are we? a, b, c, and d were events, then A, B, C, and D are their respective probabilities, but now the latter represent positions in a sequence that can with equal probability (for some reason) be occupied by any of a common set (for some reason) of elements. Not sure why probabilities are equal between the different elements either. Knowing where you are going I understand that you want to also based on nothing assume that what element the four positions are occupied by are pairwise independent events, but I guess that additional assumption is going to be implicit for now, because that’s how best to write rigorous mathematical proofs.
“This case” being what case? Also, I thought ABCD were sequence positions. Now they are events in their own right and have probabilities assigned to them?
Please, just get back on topic. This was the challenge:
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I can’t get past even that sentence. First of all, why do you begin by writing “From basic statistics”? and then just declare an assumption that isn’t an assumption in “basic statistics”? Nothing you write makes any sense. It’s just stupid blather. Word-salad written to sound like it is scientific. Something made to merely appear as something else. The modern equivalent of a cargo cult. A dummy.
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Chat GBT was not much help on the proof but here it is helpful.
The probability of getting heads on a single toss of a fair coin is 0.5, as there are two equally likely outcomes: heads (H) and tails (T).
When you toss a coin twice, the two events are independent of each other. To find the probability of getting heads on both tosses, you can simply multiply the probability of getting heads on the first toss by the probability of getting heads on the second toss:
Probability(Heads on both tosses) = Probability(Heads on first toss) × Probability(Heads on second toss) = 0.5 × 0.5 = 0.25
So, the probability of getting heads on both tosses of a coin is 0.25, or 25%.
Could be many things above it is getting a heads on both tosses of a coin.
Common ground
The letter is a position in a sequence. My area code is 510 it would be A=5. B=1 etc each position with equal probably of being generated of 1/10 with randomly generated numbers.
Again they are positions in a sequence.
Again this is better demonstrated with the search simulator I provided.
Rum thanks for asking for clarification.
It is a method of calculating two independent events occurring such as two heads in a row. This is first week of Stat 1 at UCB. Given this is a standard method used in statistics I think it is a safe assumption.
See chat GBT above.
Chat GBT obviously an abbreviation for Generic Bill Tripe
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That’s Chat-GPT. And I have zero interest in what it has to say on the subject. You are wrong and I have proven it. Anyone who’s spent an hour on any introductory course to statistics would understand it and correct their error. A clown would choose to consult a chat bot instead.
Okay. So it is meaningless to say that they have probabilities, then. Unless we want to spew word salad, that is.
I’ll be the judge of what is more convincing of the truth of a mathematical theorem to me, thank you very much.
Yes, correct. But you chose to omit that part. The statement as you presented it is not a theorem of statistics, basic, or otherwise. “Two events” is not the same as “two independent events”, we don’t get to assume the latter merely from hearing the former, and anybody who spent any hours at all studying the subject would know this. Heck, had you said “of course I meant independent events, my bad, I should have said so”, at least there would be some reasonable doubt as to whether or not you can even tell the difference. You chose, however, to double down on the mistake, making it clear that far from an innocent mistake and meaning the right thing, you simply are completely out of your depth, because that coffin wasn’t nailed quite tightly enough shut yet. The way you responded here, demonstrating that you cannot even spot the error, much less correct it, would be embarassing to you, if only you understood half the nonsense you spew. If only…
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Gee really? So glad you told me. What isn’t making sense is your incoherent blather surrounding it. “From basic statistics.” Or why you think it’s relevant. There isn’t anyone here who doesn’t already know how to calculate probabilities of compounding chance events.
No it’s like 7th or 8th grade at primary school. It’s basic probability theory. You should have learned this shit when you were 12-14 years old?
The fact that you multiply the probability of each event, in order to get their compound probability, isn’t making it a safe assumption that the events are independent in biology. No. That needs a different sort of justification(measurements, data analysis, empirical experiments) than merely waving your hand in the direction of how to compute the probabilities of compounding chance events.
Whether the events truly are independent is not affected by how the compound probability of independent events is calculated. And I’d be tempted to write ‘obviously’, yet it’s clear nothing is obvious to you.
And still, none of it matters because what matters isn’t the size of the space. It’s the fraction that is functional combined with how functional sequences are connected, and then what kind of search strategy is employed to move around in that space. Gee it’s a big space isn’t an argument. Neither is taking N to the Lth power.
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Fair enough I should have clarified they were independent events. Thanks for the the proof that showed that this must be clarified.
Bill it’s your explanations that are cryptic. We know how exponents work. And truly, they’re not cryptic. Cryptic implies your “explanations” could be somehow unencrypted with the right algorithm and cipher, and they’d then suddenly make sense. But we know they wouldn’t. They’re just you being incoherent.
You have some sort of vague notion of something you are trying to say and you sort of assemble words into sentences that seem almost randomly scattered with words you appear to think are somehow relevant but aren’t sure how.
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What is this but the Stockholm syndrome of stupidity?
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I agree and will try to improve.
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You mean because I did not clarify the events were independent? Wouldn’t someone who knows statistics be able to understand this from the equation?
Someone who knows statistics would have either not made the mistake in the first place, or not doubled down on it when unambiguously shown exactly what the mistake was at the top of the first reply, with the eventual correction coming only after the third time or so, with more people chiming in, and not before consulting ChatGPT, of all sources.
Someone who knows any maths at all would know to lay out all their premises on the outset before commencing a proof. It’s that “rigor” you were so worried would fly over our heads if you brought it. Someone who knows maths would likely also pick a consistent vernacular and notation (if not necessarily one common in the field, though in that case they would explicate theirs, too), and not switch it up multiple times in every post, nor condescend to their readers for their silly heads getting all confused by their persistent failure to keep things straight.
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I was trying to work with you to develop the proof. I did learn from the exercise as I had not thought about the connection between combinatorics and a basic probability calculation.
If you did understand that my intent was that the equations were based on independent events why did you not communicate this?
This would have been a reasonable discussion if having a real discussion was your intent.
And despite that, he seems to be certain that he understands biology more than biologists do. Based on math!!!
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What, instead of just presenting yours? Why should I take any part in you proving your claims like any upstanding interlocutor would, when prompted, after boasting that they absolutely could?
How? How did that link not even occur to you? You said:
How on earth could it escape your notice that you were talking about matters pertaining to probability theory when the very first sentence of your “proof” began with
I did. I said:
Now, clearly – or obvious, anyway, to anybody who’s been studying any stochastics for as long as two week’s worth out of a semester’s introductory class in it – what’s crucial in my counter-example is that a and b are not independent, stated explicitly by “Let event b occur if and only if a occurs” and by the hard set conditional probabilities. Anybody with qualifications like ones you claim you have would immediately understand what’s going on, but you did not. You doubled down, because you had no idea that statistical events may sometimes be anything other than statistically independent. This is why I say you lied when you claimed you studied this. Because someone who had would not be as completely clueless about the subject as you make every effort to look like you are.
The only response this warrants is:
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This was not my first stab at this. I started with A x B x C x D as an assumption but called the proof not very rigorous as I knew I was not explaining why you multiply these units.
After thinking about it and during our conversation I realized you could use basic probability as an equivalence to each position of the sequence. Each sequence position has an equal probability of a letter, number or chemical arriving by an independent random draw. I had not thought about this clearly in the past and I appreciate that this insight arose from our conversation.
I flagged the error in your assumption. What do think the Chat GBT was about?
Your intention here appears to be to try and establish authority and then argue from that position. The problem is you have been arguing in bad faith as you admit which your emojis. You are not alone here as this tactic is used by many other materialists on this blog.
This is unfortunate as you have shown some ability to think outside of the box which is rare on sites like this one.
The error was yours.
Your failure to construct a rational explanation.
I don’t see that. I see that @Gisteron understands stats and math. You understand neither.
Please stop projecting your massive arrogance onto others.
I only see evidence of bad faith on your part.