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Flipped coin, 3 possibilities?


Capiert

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Thanks for your tip.

Try green coffee extract (400 mg/day before your meal),

it (=the chlorogenic acid, .. helps, &)

burns fat for more energy (Joules, but that's biochem).

 

(You'll have to toss a many coins away for it's price. :) )

 

 

+1

 

:)

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I disagree (but I'm only saying that

to prevent confusion).

 

But you are wrong.

 

 

It's made of 2 decisions: a toss, & then specifying which.

 

???????

 

You have a roughly 1 in 6000 chance of landing the coin on its edge on the first toss, as well as any other toss of your choosing. 6000^2 is incredibly wrong.

 

 

I agree.

That is the 1st decision.

It is biased (so to speak)

to find "any" stander.

If we collect data

that's what we get.

(Let's say we've tossed for multi_billion times,

as rediculous as it sounds,

so we have enough stander counts.)

However if we become more biased

 

(please remember

we have (biasedly) selected

for (any) edge_standers;

because heads & tails have been excluded (by our bias=decision(s) rules),

 

we can take that "same data"

& ask how many standers

happened only on the 1st tosses.

(=That is more bias, &)

we will find

that new number is much less.

 

How do you explain

that difference

in those 2 numbers?

It's quite significant.

 

This makes no sense to me. I'll give you the benefit of the doubt and say it's a language barrier.

 

 

It's ok,

(maybe my wrong (informal) vocabulary?)

you just need a bit of time

(to let things settle & clear)

to understand

because you are logical

you will find the answer.

But as you see it is not

because I have a different perspective

than you.

But I'm confident you will get it.

 

No. You need to understand. I appreciate that you think I'm logical but you must understand that I am using actual confirmed mathematics. What you are saying either makes no sense to me or it's just plain wrong.

 

 

(1/6000)*(1/6000) is for a selected toss

of "your" choice;

& you can choose

from all possibilities

within the 6000 (virtually speaking).

 

NO!

 

1/6000 is for a specified toss of your choice and for a random toss as well. You choosing a specific toss has no impact on the outcome.

 

Whether you choose a toss or not, the yielded result will always have been 1 in 6000. This is what you need to understand.

 

 

Have you at least understood

that the results

are from a selection (=choice, bias)

(of (the total) data)?

I.e. a specific percent

(that had selection rules).

No selection rules means

all the data (E.g. multi_billion 100%;

but (uselessly) NOT sorted

into H, T, E.)

Completely (=100%) neutral

means you weren't looking

for anything,

& found everything

that told you nothing

but the tossing's grand total.

 

Doesn't that ring a bell?

How do you show that mathematically?

Answer: with the extra *(1/N).

 

No, that makes no sense. There is no need to multiply anything. You would destroy mathematics with your ways. Specifying a toss will ALWAYS have a chance of 1 in 6000.

 

To put it as simply as possible:

 

Let's say you have just one toss and you need to choose either heads or tails. There will always be a 50% chance for either heads or tails, regardless of whether you specify which. You must understand this.

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