Re: Calculus XOR Probability
- From: Tony Orlow <aeo6@xxxxxxxxxxx>
- Date: Mon, 17 Apr 2006 11:36:30 -0400
Matt Gutting said:
Tony Orlow wrote:
Matt Gutting said:
Tony Orlow wrote:
Matt Gutting said:If you have finitely many elements, then I'm fine with what you're saying.
Tony Orlow wrote:First of all, discussing the naturals is a digression and a straw man. We have
Matt Gutting said:I'm not assuming a *uniform* probability distribution over the naturals. I'm
Tony Orlow wrote:Somethign tells me you're still trying to deal with a uniform probability
Matt Gutting said:It must hold, even if we don't know what probability distribution we're talking
Tony Orlow wrote:Not true. If you can establish a uniform probability distribution, then you can
cbrown@xxxxxxxxxxxxxxxxx said:<snip>
Tony:Han seems unbothered by the fact that 1 is not the value you get from
the sum of a countable number of 0's; he simply "wants" it to be 1. And
lo and behold, that is what his argument affirms, just as mine affirms
that the length of the diagonal is exactly what I "want" it to be: 2.
No, he doesn't just "want" it to be 1. That is the probability of somethingOnly if you can define a probability distribution, which is the assumption in
which is definitely true. If one of the elements is to be chosen from the set,
then the probability of one being chosen is 1. If there are multiple choices,
then that 1 representing the fact that one of those will be chosen is defined
to be the sum of the probabilities of each. That's how probability works,
question here.
say each possibility has the same probability. If you can formulate the
distribution at all, it will allow you to calculate which fraction of the
overall probability is assigned to each possiblity. But whether you can
determine what the distribution is or not, the overall probability that one
will be chosen, given the fact that you're choosing one, is 1, and the
probabilities of each mutually exclusive alternative contributes to that sum
equal to 1. That's a very basic concept that must hold, even if we don't know
what probability distribution we're looking at.
about, *assuming* that we're looking at a probability distribution. As I said,
the question is whether one can be defined here. You appear to be assuming it
can. I'm not.
distribution over tha naturals. That's impossible, but given any set of n
equally likely outcomes, if exactly one of them will occur, each has a 1/n
chance of being the one to occur. To assume you have a set of possibilities
WITHOUT a probability distribution is just to say you have no idea what any of
the individual probabilities is. That has no effect on the fact that if exactly
one is to occur, the probabilities of each sum to that 1.
saying that what you are asserting is only true when one can define *some*
probability distribution over the naturals. In effect, I'm not saying that I
"have no idea what any of the individual probabilities is"; instead, I'm saying
that referring to "individual probabilities" is meaningless in this case,
so that saying "the individual probabilities sum to 1" is likewise meaningless.
n, the upper bound of the set. There are n elements. The probability of each
being chosen has meaning if we are choosing one. If we know how many we have,
if we can calculate and average of n/2, then we can assign a probability,
albeit infinitesimal. With the naturals, you can't. We agree. Forget the
naturals.
But if you have infinitely many, then there's a problem.
Only in the context of "countable infinity". Throw a dart at [0,1]. I saw a
nice uniform probability distribution a while ago, when it started to sprinkle
outside...
I'm not sure what you mean here. If there is no n such that one can say "we have
n elements", then how can one define the probability 1/n of each element?
You can't. That's why "countably infinity" is problematic in this regard, as
has been amply demonstrated. If you have a specific n, you can assign a
probability distribution, even if n is infinite. If I have Big'un elements with
uniform probability distribution, then each has 1/Big'u=Lil'un chance of being
selected. That's the whole point of a unit infinity.
There are a bunch of issues with "scale" and "sub-infinitesimal numbers" thatWhat inconsistencies do you detect?After reading through numerous posts on the subject, I still don't see that thisThe difficulty here is that you haven't really shown an explicit and consistentWhat do you want, a number system? By all means, use the T-riffics. If we have
way of defining and working with infinitesimals. That must come first, before
you begin using them to explain things.
1:000...000 equally likely possibilities one of which must happen, each has a
1/1:000...000=0:000...001 chance of being chosen.
is a consistent treatment of infinitesimals.
don't seem to fit with the way order relationships work on the real numbers.
:) It's an extension. It's not consistent with the way the reals are dealt with
standardly. The sub-infinitesimals provide for the Archimedean principle on all
scales, even though continuity on one scale does not require it on a relatively
infinitesimal scale.
So, what, exactly, do you see as the main problems with it?
Order at one scale implies order at any scale, if all you're doing is scaling.
That doesn't seem to be the case with your infinitesimals.
Order doesn't go away at any scale. x<y & y<z -> x<z no matter what. The
Archimedean principle, x<z -> x<y<z, is almost compromised by considering
discrete infinitesimal reals, except that no contiguous reals x and z on that
scale can be distinguished on the finite scale. Since there is no finite
distance between them, x=z, and y is not required to exist. On the
infinitesimal scale, if continuity presists on that level, then we can apply
subinfinitesimals, but that doesn't affect the standard finite picture.
I don't know that many (or any, for that matter) other people know what you<snip>Sure, I know, "undefined". So far, undefined. Undefinable? Is sqrt(-1)
The equation (lim x->:a f(x)g(x)) = (lim x->a:f(x))(lim x->a:g(x)) is only trueIf we are discussing an infinite set of possiblities, and I defineImmaterial. If n is infinite, then 1/n is infinitesimal. It's smaller than anyThat's still circular. If you haven't defined infinitesimal carefully, and only
finite value and yet nonzero, just like the probability we're discussing.
in terms of other well-defined concepts (and "infinite" isn't such a concept,
because you haven't really defined it well except in relation to
infinitesimals), then you can't use the term to explain or define anything else.
infinitesimal in terms of that infinity, it's not circular. If you want to have
a rigorous notion, start with infinity, but I rather doubt your "injections
into a proper subset" definition is going to help a whole lot here. This isn't
rocket science. n*1/n=1 and lim(n->oo: n*1/n)=1 and lim(n->oo:n)*lim(n->oo:
1/n)=1. All is One, Cricket.
when both limits exist. It's explicitly proved that way. You can't use that
theorem here, since (lim n->oo:n) does not exist.
undefinable? It used to be undefined. What happened? Did we define it and find
it useful? Gee, could that happen here? Sure, that's what's going on. (lim n->
oo:n)=oo. Now, if you used a SPECIFIC oo, you'd have a specific answer,
wouldn't you?
mean when you say "a SPECIFIC oo". The problem goes back, again, to your
insistence that there exist infinitely large natural numbers. There don't,
and defining them into existence puts them in conflict with basic axioms
defining and describing numbers. You haven't worked out any of these conflicts
(at least, I haven't seen a consistent list of axioms), and even if you
had, you would be working in a number system entirely different from the
standard, thus making your conclusions inapplicable to what we're talking
about.
Hmmmm....how to respond to this...think i'll reread and go for a smoke.....okay
I'm back.
This goes back to a very basic tenet of set theory which the transfinite
version violates: the addition of elements to a set increases its size. When I
think of completed infinities I think of the points on a finite line segment,
an internal infinity. There is a beginning and an end, and between them is a
length. Now, standard theory holds that the number of reals in [0,1) is the
same as the number of reals on [0,2), even though the first is a proper subset
of the other. I cannot agree. Every real in the first is contained in the
second, plus all the reals in [1,2), an infinite number more. In my mind, the
way this is resolved is by postulating some specific infinity of points per
unit of length, a density to the continuum, which leads to the intuitive result
that there are twice as many reals in [0,2) as in [0,1).
Who says this is a "basic tenet of set theory"?
I do. You add stuff, you get more. You take stuff away, and then there's less.
I feel like a cave man saying that, it's so basic. I think you can see that
it's trivial that the notion that a proper subset is always smaller than the
superset, coupled with the fact that a bijection is possible between an
infinite set and a proper subset, leads to the conclusion that a bijection does
NOT mean equal size for infinite sets. You can see how that would follow, can't
you?
The problem I see with your assumption is that you can't really define this
"specific infinity of points"; I haven't seen it done in a way that makes sense
to me. And I'm not quite sure what you mean by "a density to the continuum".
I mean the number of points per unit of distance. The number of real numbers in
[0,1). The number of unit intervals on the infinitely long real line. Big'un is
this number, and it's reciprocal unit infintiesimal is Lil'un. Big'un's a
primitive, and primordial.
Now, you say that postulating the infinitely large whole numbers which would
represent this infinite count of reals leads to contradictions. It certainly
contradicts some notions of transfinite set theory, but that couldn't bother me
less, since I see that whole area as based on unsound assumptions to begin
with. Does it violate any other mathematics? Only to the extent that numbers
are defined to be finite, but in general, dropping this assumption, I don't see
any contradictions. If you could be so kind as to specifically point out
contradictions that arise with anything besides transfinite set theory or the
*assumption* that infinite values don't exist, I'd be happy to entertain them.
For one thing, dropping the assumption that (natural) numbers must be finite
means that there may exist at least one number which is infinite.
Tautologically true.
But that number must have an infinite number of predecessors.
Okay, that jibes with the ide of an infinite set containing elements with
infinite numbers of predecessors.
If it has an infinite
number of predecessors, one of its predecessors must be 0, since 0 is a
predecessor of every number.
Okay.
Further, since the number of predecessors is
infinite, that means that it is never-ending; thus, every predecessor of this
infinite number has a predecessor.
Now you have made a bit of a leap. The set of reals in [0,1] is infinite but
has definite endpoints. If you want to consider the infinite set of positive
naturals, that infinite set can have an end, just like a ray is infinite in
length with one endpoint. Of course, if you want to include the entire real
line, 0 could be said to have a predecessor as does every other natural, and
then you get the integers.
But then the natural number 0 must have a
predecessor, which it very specifically doesn't.
It doesn't in the Peano formulation, but it does on the real line. Consider the
rules I put forth:
Order: x<y<z ->x<z
Internal oo: x<z ->x<y<z
External oo: y -> x<y<z
Notice in the External oo rule, every element has both successor and
predecessor. I don't see that as a problem.
No, the problem with the naturals is an infinite amount of elements.No, it must be infinite, yet bounded, to be able to define the probability ofIt doesn't matter what set we're talking about, as long as it's infinite (or,We're no longer talking about the problematic set of naturals, as far as I'mWhether one can determine a distribution or not, there is never 110% chance ofAgain, this is only true if there exists a probability distribution (determined
something happening. If you have a set of mutually exclusive possiblities, and
one of them is going to happen, the probabilities of each happening must sum to
1. If it sums to less than 1, then you have left out some possibliites, and if
it sums to greater than 1, then some possiblities are not mutually exclusive.
or not) on the set. If there doesn't exist a probability distribution, and there
seems to be disagreement on this point, then what you say is simply not the
case.
concerned.
if you prefer, unbounded).
an individual element. The problem with the naturals is a lack of defined
range, which of course has been a sticky issue when I've been trying to put
forth the Inverse Function Rule.
For you, that is equivalent to being unbounded, now, isn't it?
I suppose you could say that, since I'm not really clear on what you seem to
mean by "infinite".
I mean that in standard theory an unbounded set is considered infinite. The
phrase "unboundedly large but finite" doesn't have meaning in the standard
theory, does it? So, when you say the problem is that it's infinite, I think
the problem is the unboundedness of it and the inability to specify any range.
If we consider a specific n, then each individual has probability 1/n, and it
doesn't matter if n is finite or infinite, as long as it's specific.
I disagree. I haven't seen any counterexamples that you've adequately refuted.<snip>Sure, whatever. Do you see that the angle at which your elements intersect the
You're being, at best, very loose with language here: one can't have a pointDo you need this explained? It means that the segment ((x1, y1),(x2, y2)) isThere's really no flaw in proving an equality between expressions in theCould you explain what it means for an infinitesimal (sc. "infinitesimally
infinite case. Inequalities can be invalid in the infinite case, where the
difference causing the inequality has a limit of 0 as n->oo. Here, the flaw was
simply that the staircase never was a diagonal line, even if it appears so
visually as n->oo. Chas mentioned approximating lengths of curves with
infinitesimals, but those are always parallel to the curve, while his are
always at a 45 degree angle, causing the sqrt(2) error by the sine and cosine
of that angle.
long line segment", I suppose) to be "parallel" to a curve?
parallel to the curve if there can be defined a point on the curve
perpendicular to a point on the segment where points (x3, y3) and (x4, y4)
arbitrarily close to that point on either side have the property that (x3-x4)/
(y3-y4)=(x1-x2)/(y1-y2), or at leats the difference is arbitrarily close to 0.
You'll probably have a correction regarding this, but then, why did you ask?
You know what parallel means, and it doesn't mean at a 45 degree angle.
(on the curve or anywhere else) perpendicular to another point (on the segment
or anywhere else). I *think* you mean that one can find a secant line "nearby"
the curve (and yes, I'm aware that I'm using that loosely; I can formalize
it if you'd like) which is parallel to a secant line on the other curve.
diagonal directly accounts for the error of sqrt(2)? If not, I went into
greater detail in an earlier post to Chas.
<snip>I appear to be of a different opinion regarding this matter. You can chalk it
See above. You appear to be misunderstanding the limits of inductive arguments.That is not a violation, as I see it, as long as one proves an equality.Neither is true, because you're using induction on finite quantitiesAnd Han's increasingly large finite sets are still finite, whereas hisNo difference whatsoever, and Chas' conclusion that the infinitely fine
conclusion is drawn about something infinite. What's the difference between
using finite sums of lengths to draw conclusions about an "infinite" (in
some sense) sum of lengths, and using finite sets of equiprobable elements
to draw conclusions about an infinite set of elements?
staircase still requires a line of length 2 is correct, but it's not a diagonal
line. The only mistake is thinking it's a diagonal line. the length is 2, and
the sum of the probabilities is 1, by proper inductive proof of equality
holding for the infinite case. Bith proofs are valid in my opinion.
to assert results about infinite quantities.
up to misunderstanding if that makes you feel better, but my arguments in this
area have all held, as far as I've been able to tell. COunterxamples like this
are easily explained, so my point remains unrefuted.
I just did. Your diagonal is a fractal line, at no point parallel to the line
as a whole. The ratio of error is directly explained by the angle at which your
elements intersect the diagonal. And, the same explanation was published in a
book as cited by Han. So, while you claim the proof is wrong for one reason, I
refuted that claim by showing that it's correct, but fails in the assumption
that the diagonal is the limit of the staircase. You can ignore the obvious
explanation and claim I haven't refuted anything, but this is far from the
first time I've pointed out what the problem is, while others were attributing
it to something unrelated.
The diagonal is the limit of the staircase, according to the standard definition
of a limit (that is, as one increases the number of "steps" in the staircase,
one can find a number of steps beyond which any point is arbitrarily close to
the diagonal. That's what it means to be the limit; and the "assumption" that
the diagonal is the limit of the staircase is not, in fact, an assumption.
Well, it really is. The points of the staircase do indeed become arbitrarily
close to the points of the diagonal, but they are at an angle to the diagonal,
not parallel. It may seem strange to think of points as having direction, but
when those points are line segments in the limit, then they retain the
direction of the elements they're derived from. So, you have a set of
arbitrarily close but non-parallel microsegments you're comparing.
The set of points in this "limiting staircase" is precisely the diagonal.But, the flaw in that argument is not what makes it similar to han's argument.That's exactly the point. That argument was intentionally posed as a flawedLike your ill defined diagonal as the limit of the staircase as n->oo.That's how reasoning works. One may arrive at a true conclusion usingWaaahhh...Again, you haven't defined infinitesimals rigorously and non-circularly,I claim that if you can figure out why the /argument/ is wrong (notBut it's not. Archimedean principle is mitigated at the infinitesimal level by
just the conclusion), then you will also see why Han's argument is
wrong (above and beyond the fact that his conclusion is also
independently false, although perhaps not as obviously, by appeal to
Archimedes).
the fact that neighboring infinitesimals do not constitute distinct standard
reals, and so on the finite level, any discreteness of the infinitesimals, if
they are taken to be discrete, does not violate Archimedean principle on that
level. Further, on the infintiesimal level, Archimedean principle CAN be
preserved by defining midpoints in the infinitesimal intervals, because at that
level, the endpoint ARE distinct values in their nonstandard sense.
so referring to them in an argument can't lead to valid conclusions.
ill-defined terms, but the argument leading to the conclusion will be an
invalid one.
argument, to point out the flaws in Han's argument.
The only flaw is in equating the fractal diagonal with the normal self-parallel
diagonal. They're different animals, despite the fact that your microscope is
too feeble to see the difference.
Charles has in fact proved this.
No, he proved the points are "indistiguishable" from those int he diagonal 1D
line. That's a crucial point.
If the elements of a set (e.g. a set of points) A are indistinguishable from
the elements of a set B, then the elements are the same, and thus the sets are
identical.
If they are indistinguishable, then they are the same. But if there is a
distinction that is not being made, then they may only appear to be the same.
In this case, the direction of the elements is being ignored, but is the
crucial difference in whether you get a reasonable answer or not.
I honestly don't know what this means. This may be why you "get complaints aboutI haven't been doing that all over the newsgroup. Just getting basic notionsI believe I did bring up objections regarding your usage of infinity; we haveFirst of all, if you have an objection regarding the definition of infinity,Smaller than any finite but non zero. The multiplicative inverse of anThat's circular, because you haven't come up with a coherent definition of
infinity.
infinity, nor any reason to believe that the inverse of "an infinity"
will be non-zero.
you should have brought that up weeks ago. Secondly, the argument regarding the
chance of one of an infinite set being chosen being smaller than any finite
value, and yet nonzero, has been exhaustively discussed without any valid
objection. Each has a probability of 0% in the standard world, but is not
without a chance in reality. Been there, done that. Each of n equally likely
events has probability 1/n, and if n is infinite, as put forth to begin with,
this probability is infinitesimal.
discussed elsewhere the need for you to formalize and axiomatize your notions
of infinity, and you agreed that it was necessary and important. I haven't seen
any development in this area.
like the fractal natural of Chas' line is difficult enough. When I state a rule
clearly, I get complaints about every term. I think the theory ends up putting
forth infinite sets in the context of order, such that:
x<y<z -> x<z Order
x<z -> x<y<z Internal Infinity
y -> x<y<z External Infinity
every term". You abbreviate so much that it's difficult to tell what you mean,
and when you don't abbreviate, you use terms in senses that you don't always
make clear. Until you can do this, I can't accept any arguments you make
based on your terminology.
Honestly? (sigh)
for x,y x e R
x<y<z -> x<z:
In an ordered set, if x is less than y and y is less than z, then x is less
than z.
Yes, that's part of the definition of a total ordering.
Right, it's the very basics of order.
x<z -> x<y<z:
In an internally infinite (continuous) set, given any two distinct elements,
there exists an element between them.
Okay...you're talking about what might be called a "dense" set.
Right, a set with internal infinity, infinitely subdividable.
y -> x<y<z:
In an externally infinite (unbounded) set, given any element, there is an
element less than it and an element greater than it.
Okay...
:)
Did that help?
Somewhat. Part of my problem is that it doesn't make sense to me that
the number of elements in a set seems to be dependent on its order.
When dealing with infinite sets, the notion of a bijection is part of the
solution, as long as the mapping function is accounted for, and establishing
bijections between infinite sets generally depends on there being an order on
each. Can you think of a bijection between infinite sets that doesn't involve
the notion of comparison and order?
Part of the problem is that one can have a least or greatest element in
an unbounded set; think of the naturals (or their additive inverses in
the integers), for example.
You mean the set can have a bound on one end and not the other? True. But that
doesn't cover the entire real line. We CAN restate the External oo rule as y ->
y<z and just follow Peano, but why not kill two birds with one stone?
And another part is that (for example) one
can't derive from these three descriptions the idea that "an infinite
number is greater than any finite number."
Well, there is an argument that goes with those statements that says that since
one element leads to at least one other, if this process is performed a finite
number of times it produces a finite set, but if it is performed more than any
finite number of times, it produces a set with more than any finite number of
elements. Perhaps the definition might be something like an infinite number is
how many iterations must be performed to generate every real in any finite
interval.
(ooh almost time to leave, and won't be back until Monday - have a nice
weekend)
I gave a bunch of this to Tribble a couple days ago, and went into discussionWhat do you mean, "an infinite upper bound"?
of how both use the same formula for generating elements in opposite
directions, at least in one sense. That needs to be expanded to include another
interpretation of successor that produces the H-riffics. In any case, can you
think of a sense of infinity where a nonzero reciprocal could be finite, or
anything but infinitesimal?
The fact that you have seen no valid objection to an argument is notWhy not? It's impossible to define a uniform probability distribution on the
necessarily an indicator that there is no valid objection; it just means that
you haven't seen one. As I said above, the assumption you're making here (one
of the assumptions, at any rate) is that it makes sense, when talking about
infinite sets, to consider the phrase "equally likely" as meaningful. It's not.
naturals due to its unboundedness, but given an infinite upper bound of n, it's
not impossible to define the likelihood of individual elements. It's just not
part of standard mathematics.
I mean a particular infinite maximum value, such as the index of the real
infinitesimal at x=1, which is 1:000...000. :D
I really can't evaluate the truth of that without knowing exactly what your set
of rules for dealing with them is.
Er, okay.
But is there an infinite n? I think that's part of the problem.It works for any set from 1 through n, whether n is finite or infinite.That works when n is finite; that is, it works for any set 1 through n.Sum(x=1->n: 1/n)=1. We already DEFINED the probability distribution to beBut how do you know that your probabilities add? Without a probabilityBecause they add linearly, without the need to be parallel to what theyWhy is your conclusion wrong? It's not, if you're talking about a line brokenWhat gives you the idea that one can "sum" the probabilities of an infinite
into vertical and horizontal elements. It will always be equal to the sum of
the vertical and horizontal distances traveled, because it's NOT the diagonal
between the starting and ending points.
number of points to get the whole (1) any more than one can add up infinite
numbers of vertical and horizontal lengths to get a diagonal length?
"approximate", since the values are scalar and cannot be anything BUT parallel,
as opposed to the vectors that the treads and risers represent. If all your
infinitesimal segments were PARALLEL to the diagonal, then indeed they would
sum to, guess what, sqrt(2)!
distribution (known or unknown), they don't.
uniform, not over the naturals, but from 1 through n.
Indeed, there is Big'un, and all formulaic expressions on Big'un. Of course,
that's not in standard theory, but we're discussing extensions to standard
theory, no?
We're discussing reasonably consistent extensions, okay. I'm not quite sure
you have one.
That's understandable. I can't gurarantee that everything will be perfect, but
as it fits together, I'm surprised how seamless it seems. I know it sounds
foreign and it doesn't play well with standard theory, so it's hard to even
consider, but I gotta do what I gotta do, even if it sounds wrong at first.
Like I said, I am sticking to the basic tenet of "plus means more".
The problem is that the consequences of the idea lead to contradictions whichBut none of these sets is the natural numbers.Good. To hell with the naturals. They make a crappy domain.
If you're not talking about an infiniteThat's opinion, not logic. If you throw a dart at a board randomly, every point
set, we have no argument. But if you are, then what infinite set are you
intending to discuss? The same objection holds for any infinite set: the
concept of "probability of an individual event occurring" is meaningless in
such a context.
has a chance of being hit, but there are an infinity of points, so that chance
is effectively 0, but not really 0. Is it meaningless to talk about the chance
of a single point being hit? That idea has meaning for me.
can only be resolved by dealing much more rigorously than you have with concepts
of infinity and infinitesimals. The idea of something being "effectively 0, but
not really 0" has no meaning for me.
Hmmm. How would you describe the chance of a particular one of an infinite set
being chosen at random? Is it really 0, if one of them is being picked?
No, it's not really 0. It's undefined.
Is it undefinable?
Flip a coin aleph_1 times and generate bits to find a point in [0,1). Is thereNo, nor is it "as likely as any other" - the concept of "likelihood" needn't
not a uniform probability distribution among the points? Throw your balls all
in a vase by noon, shake vigorously an infinite number of times, and pick. Is
one ball more likely than any other?
apply here.
"Needn't"?
Can it? I'll refer you to Bob the Builder on that one. :|
???
"Can we do it? Yes we can!!" Bob the Builder, an animated show for little kids.
I am saying that if we can apply some reasonable notion of likelihood to such
an event, then we should, rather than just say it's "undefined". I see that as
a temporary condition, this undefinedness.
Matt
--
Smiles,
Tony
.
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