Question about probabilities

[Kyriakos]

Let's assume that you have three empty boxes. You are given an infinite number of balls, but only of x number of types, and asked to present all possible arrangements (ball of type x1 in box + ball off type x2 etc). Eg, if you only have one ball type, the first arrangement is singular, given you can only fill the boxes in one way:
Box A has ball type x1, box 2 also type x1, same for box3.
If, on the other hand, you were given two types of balls, the full set of possible arrangements would have been 8, because each box could have 2 types and each of those types would form a set with each distinct arrangement for the remaining boxes, ie 2X2X2= 8. Likewise, for type=3, you have 3^3, and for type=n you have n^3.

Now, my question:

Someone asks you to do all these arrangements, up to nx (with x being an arbitrarily large integer). Then asks you to choose one case for each arrangement (so for n=2, you choose just one of the 8 possible cases), and add all those single cases as probabilities. What in essence you would be adding up is the probability of each single case, but for a continuum of cases (from n=1 to n=x). In practice what is that probability about, if expressed as a single effect instead of a continuum?
It obviously is not the probability of getting all those specific single cases if the arrangement was run for the entire continuum, given that this has an upper bound at the smallest singular probability of the individual cases.

Just asking for a stupid game I am making. Or I'll go with that anyway

edit: one idea I had (but I am not sure if it is correct...)

Basically, for large n, the sum of all singular case probabilities is roughly 1,20. I was thinking that this 1,20 would be the number of times you would certainly choose the correct case (the only correct case), if the experiment was run for all n. If n=2 (n=[1,2]), the sum is 1+1/8= 1.125, so a 1,125 guaranteed correct answer if the experiment runs for all n with n=1,2. But I am not confident this is a correct notion... I mean obviously the 1 comes from n=1, not from the continuum, and I want to express this as having to do with the continuum of all n.

 

[Kyriakos]

Nothing?
Fine, if I manage to think this through myself, I won't pay anything to you guys and also pretend I never posted here when I am famous

But first I have to move near the problematic area in Basel.

 

[red_elk]

If I understood you correctly, you are summing the unrelated probabilities (1,1/8,1/27,...).
The resulting number is obviously not a probability (as it's more than one) and I'm not sure if it does have much sense except as the sum of the sequence.

For n=1,000,000 the sum is ~1.20205690315

 

[Kyriakos]

Yes, and what I would like to do is express the sum in any of its parts as tied to the continuum n. Eg for n= (1,2) the sum is 1+1/8=1,125. For large n it nears 1.2.
Maybe logarithms can be used in some way?

If I understood you correctly, you are summing the unrelated probabilities (1,1/8,1/27,...).
The resulting number is obviously not a probability (as it's more than one) and I'm not sure if it does have much sense except as the sum of the sequence.

For n=1,000,000 the sum is ~1.20205690315

A related question:

Assuming this is a game, where one wins only if they guess correctly all the (single for each movement) cases to pass (eg there is only one case for n=1, so you pass anyway, but there is 1/8 chance to pass n=2 etc), wouldn't the probability to guess correctly up to n be probn1 X probn2 X ... probn? So 1/8 probability to pass n1,n2 etc.
So the summation of those x1,x2,x3 should mean something in relation to their multiplication. It is what I am asking.

 

[red_elk]

Assuming this is a game, where one wins only if they guess correctly all the (single for each movement) cases to pass (eg there is only one case for n=1, so you pass anyway, but there is 1/8 chance to pass n=2 etc), wouldn't the probability to guess correctly up to n be probn1 X probn2 X ... probn?

Yes, if you need the total probability to guess correctly in all cases, you need to multiply individual probabilities.
1 * 1/8 * 1/27 * 1/64...

So the summation of those x1,x2,x3 should mean something in relation to their multiplication. It is what I am asking.

Not sure what you mean.
You can calculate a sum of n numbers, but it won't help you if you need a product.

Logarithm of a product is a sum of logarithms, may that's what you are looking for. But it's not applied to the sum of your probabilities.

 

[Kyriakos]

Not sure what you mean.
You can calculate a sum of n numbers, but it won't help you if you need a product.

Logarithm of a product is a sum of logarithms, may that's what you are looking for. But it's not applied to the sum of your probabilities.

This is where you are supposed to help, Perelman

I mean, my uni degree is in philosophy, it means my math is basic. But I suppose I will have to think this through.

 

[uppi]

Yes, and what I would like to do is express the sum in any of its parts as tied to the continuum n. Eg for n= (1,2) the sum is 1+1/8=1,125. For large n it nears 1.2.

For n -> infinity it is the result of the Riemann zeta function of 3, also known as Apéry's constant
https://en.wikipedia.org/wiki/Apéry's_constant

Not sure how this is related to probability, though.

 

[innonimatu]

Let's assume that you have three empty boxes. You are given an infinite number of balls, but only of x number of types, and asked to present all possible arrangements (ball of type x1 in box + ball off type x2 etc). Eg, if you only have one ball type, the first arrangement is singular, given you can only fill the boxes in one way:
Box A has ball type x1, box 2 also type x1, same for box3.
If, on the other hand, you were given two types of balls, the full set of possible arrangements would have been 8, because each box could have 2 types and each of those types would form a set with each distinct arrangement for the remaining boxes, ie 2X2X2= 8.

You lost me at this point. If you can have combinations of two objects in each box you can have either 3 distinct combinations for the box (x1x1, x2x2, x1x2), or 4 if order matters. Let's assume it doesn't, from what you said that's reasonable.

Then you look at the possible combinations of boxes with these 3 possible types of box fillings. Those are simply 3 combinations of 3 different items:
x1x1 x1x1 x1x1
x1x2 x1x1 x1x1
x2x2 x1x1 x1x1
x1x1 x1x2 x1x2
x1x2 x1x2 x1x2
x2x2 x1x2 x1x2
x1x1 x2x2 x2x2
x1x2 x2x2 x2x2
x2x2 x2x2 x2x2

I can't follow you further because the basis for all the reasoning that comes after is flawed.

 

[red_elk]

I think only one object per box is assumed. That's the only way you can have n^3 combinations.

 

[Kyriakos]

For n -> infinity it is the result of the Riemann zeta function of 3, also known as Apéry's constant
https://en.wikipedia.org/wiki/Apéry's_constant

Not sure how this is related to probability, though.

I was trying to think of a way to present this as tied to something specific. Like with 2 it is π^2/6

Though in the game it is supposed to be a doomed attempt to calculate consciously, which is why the person uses a mask with electrodes (to have something else hopefully do the work)

@innonimatu yes, as Red said, this is for when you can only have one type of ball in each box.
So for, eg, two types of ball, it is 8 cause you have:

AAA
AAB
ABA
ABB
BAA
BBA
BBB
BAB

more simply, 2 x what you would have for two boxes:
AA
AB
BA
BB

 

[Kyriakos]

I suppose there are no ideas. A pity, I could really use the math money.

I am still going to upload the horror game - I estimate it will be complete near the end of the month, and will be free to play
I doubt many people will guess what is going on - there are some cryptic hints, such as the enigmatic "7c" - can you guess what it means?

@uppi : maybe the computer can have a small monkey toy on it, to refer to Apery.

The idea in the game is that the person gave up trying to solve the issue, and thought of using make-it-yourself brain scan to utilize his reaction to a custom computer game which is about guessing all correct probabilities for n up to a number.
Most likely result is just death - and brain damage may have set in prior to the start of the game anyway, cause the person seems to forget he has been in those rooms before.

 

[Ferocitus]

Yes, and what I would like to do is express the sum in any of its parts as tied to the continuum n. Eg for n= (1,2) the sum is 1+1/8=1,125. For large n it nears 1.2.
Maybe logarithms can be used in some way?

From some work I have been doing on calculating the strength of the (composite) singularity that exists at the tips of circular wings in ideal flow, I know that there is a way to relate logarithms (or polylogarithms) to Apery's constant.
(I've just snipped these out my LaTeXed notes for a paper.)

where
psi is the digamma function.
gamma is the Euler-Mascheroni constant.
(You can change the sum for L_n from 0 to n-1 to 1 to n if it is better suited to your problem.)

Then

where
zeta(3) is Apery's constant.

This is one example of more general Double Euler sums.
They arose from a question in a letter Goldbach wrote to Euler in 1742. Euler must have misplaced his quill (my conjecture) because it took until 1775 for him to reply to Goldbach.

I suppose I should show my working, or rather how I first got onto this stuff.

I had just taken my ADHD meds and was watching a video by my favourite English super-group, The Go Go Cult.

Aha!
The Lerch Transcendent!

And that's how an applied mathematician's mind works.

The Enb.

Late Edit:
Edit: it was remiss of me to not give references. I know you guys are sticklers for that kind of thing.

Borwein, J.M, I.J. Zucker, I.J. and Boersma, J.,
The evaluation of character Euler double sums,
https://research.tue.nl/en/publications/the-evaluation-of-character-euler-double-sums

Euler, L., Meditationes circa singulare serierum genus, Novi commentarii acad.
scientiarum Petripolitanae 20 (1775), pp. 140--186.

Jordan, P.F., On lifting wings with parabolic tips, ZAMM, Vol. 54, 1974, pp. 463--477.

 

[Kyriakos]

From some work I have been doing on calculating the strength of the (composite) singularity that exists at the tips of circular wings in ideal flow, I know that there is a way to relate logarithms (or polylogarithms) to Apery's constant.
(I've just snipped these out my LaTeXed notes for a paper.)

View attachment 562257
where
psi is the digamma function.
gamma is the Euler-Mascheroni constant.
(You can change the sum for L_n from 0 to n-1 to 1 to n if it is better suited to your problem.)

Then
View attachment 562258
where
zeta(3) is Apery's constant.

This is one example of more general Double Euler sums.
They arose from a question in a letter Goldbach wrote to Euler in 1742. Euler must have misplaced his quill (my conjecture) because it took until 1775 for him to reply to Goldbach.

I suppose I should show my working, or rather how I first got onto this stuff.

I had just taken my ADHD meds and was watching a video by my favourite English super-group, The Go Go Cult.

Aha!
The Lerch Transcendent!

And that's how an applied mathematician's mind works.

The Enb.

Late Edit:
Edit: it was remiss of me to not give references. I know you guys are sticklers for that kind of thing.

Borwein, J.M, I.J. Zucker, I.J. and Boersma, J.,
The evaluation of character Euler double sums,
https://research.tue.nl/en/publications/the-evaluation-of-character-euler-double-sums

Euler, L., Meditationes circa singulare serierum genus, Novi commentarii acad.
scientiarum Petripolitanae 20 (1775), pp. 140--186.

Jordan, P.F., On lifting wings with parabolic tips, ZAMM, Vol. 54, 1974, pp. 463--477.

See, @red_elk ? I was intuitively correct about logarithms. Only issue now is that between the place of that general intuition and my consciousness there is an abyss of the kind described by Hesiod for the Underworld.

 

[red_elk]

See, @red_elk ? I was intuitively correct about logarithms. Only issue now is that between the place of that general intuition and my consciousness there is an abyss of the kind described by Hesiod for the Underworld.

Glad that your question was finally answered!

 

[Ferocitus]

See, @red_elk ? I was intuitively correct about logarithms. Only issue now is that between the place of that general intuition and my consciousness there is an abyss of the kind described by Hesiod for the Underworld.

I also gave you every opportunity to promote modern Greek literature and you didn't take the bait.
Are you not feeling well, K? Or is the Doxiados novel not worth your trouble?

 

[Kyriakos]

I also gave you every opportunity to promote modern Greek literature and you didn't take the bait.
Are you not feeling well, K? Or is the Doxiados novel not worth your trouble?

I haven't read Doxiades, I am afraid... I do know of his existence.

There's a nice quote by Borges: no writer wants to owe anything to his contemporaries.