*
Select a point on a line. Call it A.
Select another point on a line. Call it B.
Select another point on a line. Call it C.
What is the probabiliy that C is between A and B?
earle
*

Earle Jones formulated the question :
> *
> Select a point on a line. Call it A.
> Select another point on a line. Call it B.
> Select another point on a line. Call it C.
> What is the probabiliy that C is between A and B?
> earle
> *

Nearly zero.

On Sat Sep 25 21:51:52 2021 Earle Jones wrote:
> *
> Select a point on a line. Call it A.
> Select another point on a line. Call it B.
> Select another point on a line. Call it C.
> What is the probabiliy that C is between A and B?
> earle
> *

*Note to Archimedes Plutonium and John Gabriel.

Please comment on the question above.

Thanks,

earle
*

FromTheRafters <erratic@nomail.afraid.org> wrote:
> Earle Jones formulated the question :
>> Select a point on a line. Call it A.
>> Select another point on a line. Call it B.
>> Select another point on a line. Call it C.
>> What is the probability that C is between A and B?
> Nearly zero.

How do you see that? It seems like 1/3 is plausible, as follows:
Suppose the points are selected at random in some way, with A,B,C
i.i.d, and distinct with probability 1. Then there are six possible
orderings of A, B, C, each equally likely. Two of those orderings
have C between A and B.

Earle Jones <earle.jones@comcast.net> wrote:
> *
> Select a point on a line. Call it A.
> Select another point on a line. Call it B.
> Select another point on a line. Call it C.
> What is the probabiliy that C is between A and B?

It depends entirely upon what "select" here means, whether the three
lines are parallel or not, or coincide, and what "between" means. For
example, if A, B, and C happen to be at the vertices of an equilateral
triangle, does C count as being between A and B or not?

In short, the question as formulated is silly.

> earle
> *

--
Alan Mackenzie (Nuremberg, Germany).

James Waldby formulated on Sunday :
> FromTheRafters <erratic@nomail.afraid.org> wrote:
>> Earle Jones formulated the question :
>>> Select a point on a line. Call it A.
>>> Select another point on a line. Call it B.
>>> Select another point on a line. Call it C.
>>> What is the probability that C is between A and B?
>
>> Nearly zero.
>
> How do you see that? It seems like 1/3 is plausible, as follows:
> Suppose the points are selected at random in some way, with A,B,C
> i.i.d, and distinct with probability 1. Then there are six possible
> orderings of A, B, C, each equally likely. Two of those orderings
> have C between A and B.

The distance between A and B is finite.

Earle Jones wrote:
> *
> Select a point on a line. Call it A.
> Select another point on a line. Call it B.
> Select another point on a line. Call it C.
> What is the probabiliy that C is between A and B?
> earle
> *
>

The same line in each case? The line being infinite in extent? If yes
and yes, then 0.

--
The world will little note, nor long remember what we say here
Abraham Lincoln at Gettysburg

On 26/09/2021 15:12, Peter wrote:
> Earle Jones wrote:
>> *
>> Select a point on a line.  Call it A.
>> Select another point on a line.  Call it B.
>> Select another point on a line.  Call it C.
>>     What is the probabiliy that C is between A and B?
>> earle
>> *
>>
>
> The same line in each case?  The line being infinite in extent?  If yes
> and yes, then 0.
>

The question as asked doesn't make sense, because "select a point from a
line" needs to be clarified to make any sense at all. If the line is a
finite line segment, then we could at least say we'll make a
"conventional" assumption about the distributions of the points A,B,C
namely that they are each evenly distributed on the line segment, and
are independent of each other. Then the answer would be 1/3.

If the line is infinite in extent, there is no conventional assumption
that applies for the distribution of A,B,C and nothing was said
concerning independence etc.. But if we make an assumption that there
/is/ some fixed distribution, and that A,B,C are independent and each
having that distribution, then we can give the answer 1/3, as someone
else pointed out, using a symmetry argument.

Hmm, actually we need another assumption: that the probability of two
points coinciding is zero. (So e.g. we could say the cumulative
distribution function needs to be continuous, or there is no discreet
component to the distribution or some such wording.)

But are those correct assumptions? I'd guess the OP has no background
to consider such issues.

(I can't instantly think up a natural set of assumptions that would make
the problem properly specified, and give an answer of 0, although I can
see why you said 0.)

Mike.

How does calculus start with slope?

lördag 25 september 2021 kl. 23:51:58 UTC+2 skrev Earle Jones:
> *
> Select a point on a line. Call it A.
> Select another point on a line. Call it B.
> Select another point on a line. Call it C.
> What is the probabiliy that C is between A and B?
> earle
> *
Well, how do you choose A, B and C? What distribution do you use?

On Sun Sep 26 06:21:06 2021 FromTheRafters wrote:
> James Waldby formulated on Sunday :
> > FromTheRafters <erratic@nomail.afraid.org> wrote:
> >> Earle Jones formulated the question :
> >>> Select a point on a line. Call it A.
> >>> Select another point on a line. Call it B.
> >>> Select another point on a line. Call it C.
> >>> What is the probability that C is between A and B?
> >
> >> Nearly zero.
> >
> > How do you see that? It seems like 1/3 is plausible, as follows:
> > Suppose the points are selected at random in some way, with A,B,C
> > i.i.d, and distinct with probability 1. Then there are six possible
> > orderings of A, B, C, each equally likely. Two of those orderings
> > have C between A and B.
>
> The distance between A and B is finite.
**
I have seen attempted justificatiopns for several answers to the question.
Answers of "0", "1/2", and "1/3" have all been offered.

My favorite answer is 1/3 for this reason:

For the three points A, B, and C: One of them must lie between the other two.
Since they all have equal probability, the probability that it is "C" is 1/3.

earle
*

After serious thinking Earle Jones wrote :
> On Sun Sep 26 06:21:06 2021 FromTheRafters wrote:
>> James Waldby formulated on Sunday :
>>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>>> Earle Jones formulated the question :
>>>>> Select a point on a line. Call it A.
>>>>> Select another point on a line. Call it B.
>>>>> Select another point on a line. Call it C.
>>>>> What is the probability that C is between A and B?
>>>> Nearly zero.
>>>
>>> How do you see that? It seems like 1/3 is plausible, as follows:
>>> Suppose the points are selected at random in some way, with A,B,C
>>> i.i.d, and distinct with probability 1. Then there are six possible
>>> orderings of A, B, C, each equally likely. Two of those orderings
>>> have C between A and B.
>>
>> The distance between A and B is finite.
> **
> I have seen attempted justificatiopns for several answers to the question.
> Answers of "0", "1/2", and "1/3" have all been offered.
>
> My favorite answer is 1/3 for this reason:
>
> For the three points A, B, and C: One of them must lie between the other
> two. Since they all have equal probability, the probability that it is "C"
> is 1/3.
>
> earle
> *

I tend to think about it visually. A straight curve (a line) is
infinite both to the 'left' and 'right'. I'm shooting random missiles
at an infinite probability space with a target marked off by two points
on the line -- A and B. My probability space is infinite and I can
"zoom out" to get the bigger picture and watch the target segment
marked off by A and B shrink to non-existence -- almost.

It doesn't make sense to me to use the density (which is the same
everywhere in this scenario) when an easy measure is to ignore the
field aspect and use integers -- rounding up or down as needed.

Then, the "line" is only countably infinite and the two rays retain
that property while the segment between them is finite. Now we know of
course that "Almost All" of the probability space is outside the AB
segment.

That is why I prefer the _nearly_ zero answer.

markus...@gmail.com wrote on 9/27/2021 :
> lördag 25 september 2021 kl. 23:51:58 UTC+2 skrev Earle Jones:
>> *
>> Select a point on a line. Call it A.
>> Select another point on a line. Call it B.
>> Select another point on a line. Call it C.
>> What is the probabiliy that C is between A and B?
>> earle
>> *
> Well, how do you choose A, B and C? What distribution do you use?

As badly written as it was, I assumed one infinite line (because they
are by definition) and three points on that same line chosen at random.
The first two A and B set the target and the last C is the missile.

Sometimes the hardest thing about writing problem excercises is to make
sure it is not ambiguous or otherwise easily misconstrued.

On 9/26/2021 6:21 AM, FromTheRafters wrote:
> James Waldby formulated on Sunday :
>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>> Earle Jones formulated the question :

>>>> Select a point on a line.  Call it A.
>>>> Select another point on a line.  Call it B.
>>>> Select another point on a line.  Call it C.
>>>> What is the probability that C is between A and B?
>>
>>> Nearly zero.
>>
>> How do you see that?  It seems like 1/3 is plausible, as follows:
>> Suppose the points are selected at random in some way, with A,B,C
>> i.i.d, and distinct with probability 1.  Then there are six possible
>> orderings of A, B, C, each equally likely.  Two of those orderings
>> have C between A and B.
>
> The distance between A and B is finite.

On average, the distance between A and B is infinite.

I suggest that we fill in the blank places in the question
by tipping a delta function over on its side.
Define the distribution of the points to be flat between
-M/2 and +M/2, zero outside, answer the question for that,
and then let M --> +inf

We get the probability of C between A and B to be 1/3.
Also, we get the average distance between A and B
to be M/3.

----
There are other ways to fill in the blank places.
My first thoughts ran to an exponential "decay" or a power law.
However, that symmetry argument looks very good.
I haven't worked out any other answer. I bet they'll be 1/3 also.

Jim Burns formulated the question :
> On 9/26/2021 6:21 AM, FromTheRafters wrote:
>> James Waldby formulated on Sunday :
>>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>>> Earle Jones formulated the question :
>
>>>>> Select a point on a line.  Call it A.
>>>>> Select another point on a line.  Call it B.
>>>>> Select another point on a line.  Call it C.
>>>>> What is the probability that C is between A and B?
>>>
>>>> Nearly zero.
>>>
>>> How do you see that?  It seems like 1/3 is plausible, as follows:
>>> Suppose the points are selected at random in some way, with A,B,C
>>> i.i.d, and distinct with probability 1.  Then there are six possible
>>> orderings of A, B, C, each equally likely.  Two of those orderings
>>> have C between A and B.
>>
>> The distance between A and B is finite.
>
> On average, the distance between A and B is infinite.
>
> I suggest that we fill in the blank places in the question
> by tipping a delta function over on its side.
> Define the distribution of the points to be flat between
> -M/2 and +M/2, zero outside, answer the question for that,
> and then let M --> +inf
>
> We get the probability of C between A and B to be 1/3.
> Also, we get the average distance between A and B
> to be M/3.
>
> ----
> There are other ways to fill in the blank places.
> My first thoughts ran to an exponential "decay" or a power law.
> However, that symmetry argument looks very good.
> I haven't worked out any other answer. I bet they'll be 1/3 also.

I suggest, naively, that there are just as many points within the
segment as there are without, so one half could be a (useless IMO)
answer too.

måndag 27 september 2021 kl. 20:51:49 UTC+2 skrev Jim Burns:
> On 9/26/2021 6:21 AM, FromTheRafters wrote:
> > James Waldby formulated on Sunday :
> >> FromTheRafters <err...@nomail.afraid.org> wrote:
> >>> Earle Jones formulated the question :
>
> >>>> Select a point on a line. Call it A.
> >>>> Select another point on a line. Call it B.
> >>>> Select another point on a line. Call it C.
> >>>> What is the probability that C is between A and B?
> >>
> >>> Nearly zero.
> >>
> >> How do you see that? It seems like 1/3 is plausible, as follows:
> >> Suppose the points are selected at random in some way, with A,B,C
> >> i.i.d, and distinct with probability 1. Then there are six possible
> >> orderings of A, B, C, each equally likely. Two of those orderings
> >> have C between A and B.
> >
> > The distance between A and B is finite.
> On average, the distance between A and B is infinite.
>
> I suggest that we fill in the blank places in the question
> by tipping a delta function over on its side.
> Define the distribution of the points to be flat between
> -M/2 and +M/2, zero outside, answer the question for that,
> and then let M --> +inf
>
> We get the probability of C between A and B to be 1/3.
> Also, we get the average distance between A and B
> to be M/3.
>
> ----
> There are other ways to fill in the blank places.
> My first thoughts ran to an exponential "decay" or a power law.
> However, that symmetry argument looks very good.
> I haven't worked out any other answer. I bet they'll be 1/3 also.
Well, that depends on the distribution.

On 27/09/2021 20:24, FromTheRafters wrote:
> Jim Burns formulated the question :
>> On 9/26/2021 6:21 AM, FromTheRafters wrote:
>>> James Waldby formulated on Sunday :
>>>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>>>> Earle Jones formulated the question :
>>
>>>>>> Select a point on a line.  Call it A.
>>>>>> Select another point on a line.  Call it B.
>>>>>> Select another point on a line.  Call it C.
>>>>>> What is the probability that C is between A and B?
>>>>
>>>>> Nearly zero.
>>>>
>>>> How do you see that?  It seems like 1/3 is plausible, as follows:
>>>> Suppose the points are selected at random in some way, with A,B,C
>>>> i.i.d, and distinct with probability 1.  Then there are six possible
>>>> orderings of A, B, C, each equally likely.  Two of those orderings
>>>> have C between A and B.
>>>
>>> The distance between A and B is finite.
>>
>> On average, the distance between A and B is infinite.

Well, that depends entirely on the distribution for A and B. Some
distributions give infinite expectation, some finite.

>>
>> I suggest that we fill in the blank places in the question
>> by tipping a delta function over on its side.
>> Define the distribution of the points to be flat between
>> -M/2 and +M/2, zero outside, answer the question for that,
>> and then let M --> +inf
>>
>> We get the probability of C between A and B to be 1/3.
>> Also, we get the average distance between A and B
>> to be M/3.
>>
>> ----
>> There are other ways to fill in the blank places.
>> My first thoughts ran to an exponential "decay" or a power law.
>> However, that symmetry argument looks very good.
>> I haven't worked out any other answer. I bet they'll be 1/3 also.

Yes, with a couple of extra assumptions:
a) one fixed distibution (the same one) for each of A, B, C
b) A, B, C distributions are independent
c) no "discrete" component in the distibutions, so the probability
of and of A, B, C coinciding is zero.

I'd say this is the most reasonable way to go, because in practice
someone suggesting this experiment can be challenged to say what they
mean by "select a point on a line", and if we press them, hopefully they
would come up with some pseudo-mechanical process for generating the
point, which will hopefully have some well defined distribution we could
apply, meeting the criteria (a)(b)(c). Then the symmetry argument works.

E.g. "put a spinner 1m from the line and spin it around. [AKA let the
angle the spinner points be evenly distributed on [0, 2Pi)]. Extend a
line through where the spinner stops, and the selected point on the
original line is where that line intersects the spinner line.

>
> I suggest, naively, that there are just as many points within the
> segment as there are without, so one half could be a (useless IMO)
> answer too.

Probability is not based on "the number of points", so that would indeed
be a useless answer. :) (..and one not justified by probability
arguments..)

Mike.

On 27/09/2021 19:49, FromTheRafters wrote:
> markus...@gmail.com wrote on 9/27/2021 :
>> lördag 25 september 2021 kl. 23:51:58 UTC+2 skrev Earle Jones:
>>> * Select a point on a line. Call it A. Select another point on a
>>> line. Call it B. Select another point on a line. Call it C. What is
>>> the probabiliy that C is between A and B? earle *
>> Well, how do you choose A, B and C? What distribution do you use?
>
> As badly written as it was, I assumed one infinite line (because they
> are by definition) and three points on that same line chosen at random.

But FromTheRafters was after the probability distibution - just saying
"chosen at random" doesn't answer that.

If you have a background where you've studied probability a little,
you'll probably know that you can't say "just choose a point randomly so
that each point is equally likely". That just doesn't work for an
infinite line, or even for a finite line segment although the latter can
at least be sorted out to make sense with enough care.

If you don't have that kind of background, a way forward would be to
specify the experiment you'll perform to select a point. E.g. "I'll
stand 10m away from a wall, blindfolded, then I'll spin round and round
until I'm really dizzy and disorientated, then I'll stop and fire.
Where I hit the wall will be my point A. [Yeah, 50% of shots will be
wasted and you'll have to try again until you hit. Or better, use a
double ended gun that fires both forwards and backwards so one bullet
will hit. Yeah, in practice not so, but you get what I mean...]

Or if you want to stay maths-based, perhaps give the cumulative
distribution function F(x) which gives

F(x) = Prob( A <= x )

(That assumes we've identified our line with the real numbers, so we can
phrase everything in terms of real numbers. F(x) is the probability A
is to the left of point x.)

Or (needing more background), specify the distribution, e.g. the
distribution for A will be Normal, centred at 0 with standard deviation
1. Or A will be evenly distributed on the interval [-M/2, M/2], or
whatever.

> The first two A and B set the target and the last C is the missile.
>
> Sometimes the hardest thing about writing problem excercises is to make
> sure it is not ambiguous or otherwise easily misconstrued.

I'm with you there! (And especially so with probability questions...)

Mike.

Mike Terry explained :
> On 27/09/2021 20:24, FromTheRafters wrote:
>> Jim Burns formulated the question :
>>> On 9/26/2021 6:21 AM, FromTheRafters wrote:
>>>> James Waldby formulated on Sunday :
>>>>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>>>>> Earle Jones formulated the question :
>>>
>>>>>>> Select a point on a line.  Call it A.
>>>>>>> Select another point on a line.  Call it B.
>>>>>>> Select another point on a line.  Call it C.
>>>>>>> What is the probability that C is between A and B?
>>>>>
>>>>>> Nearly zero.
>>>>>
>>>>> How do you see that?  It seems like 1/3 is plausible, as follows:
>>>>> Suppose the points are selected at random in some way, with A,B,C
>>>>> i.i.d, and distinct with probability 1.  Then there are six possible
>>>>> orderings of A, B, C, each equally likely.  Two of those orderings
>>>>> have C between A and B.
>>>>
>>>> The distance between A and B is finite.
>>>
>>> On average, the distance between A and B is infinite.
>
> Well, that depends entirely on the distribution for A and B. Some
> distributions give infinite expectation, some finite.
>
>>>
>>> I suggest that we fill in the blank places in the question
>>> by tipping a delta function over on its side.
>>> Define the distribution of the points to be flat between
>>> -M/2 and +M/2, zero outside, answer the question for that,
>>> and then let M --> +inf
>>>
>>> We get the probability of C between A and B to be 1/3.
>>> Also, we get the average distance between A and B
>>> to be M/3.
>>>
>>> ----
>>> There are other ways to fill in the blank places.
>>> My first thoughts ran to an exponential "decay" or a power law.
>>> However, that symmetry argument looks very good.
>>> I haven't worked out any other answer. I bet they'll be 1/3 also.
>
> Yes, with a couple of extra assumptions:
> a) one fixed distibution (the same one) for each of A, B, C
> b) A, B, C distributions are independent
> c) no "discrete" component in the distibutions, so the probability
> of and of A, B, C coinciding is zero.
>
> I'd say this is the most reasonable way to go, because in practice someone
> suggesting this experiment can be challenged to say what they mean by "select
> a point on a line", and if we press them, hopefully they would come up with
> some pseudo-mechanical process for generating the point, which will hopefully
> have some well defined distribution we could apply, meeting the criteria
> (a)(b)(c). Then the symmetry argument works.
>
> E.g. "put a spinner 1m from the line and spin it around. [AKA let the angle
> the spinner points be evenly distributed on [0, 2Pi)]. Extend a line through
> where the spinner stops, and the selected point on the original line is where
> that line intersects the spinner line.
>
>>
>> I suggest, naively, that there are just as many points within the segment
>> as there are without, so one half could be a (useless IMO) answer too.
>
> Probability is not based on "the number of points", so that would indeed be a
> useless answer. :) (..and one not justified by probability arguments..)

That's what I mean, there is no measure. With my interpretation we had
a line, a point on that line, so now two opposing rays can be
envisioned. Another 'different' point added and we have the opposing
rays with a line segment between them - and line segments have two
endpoints so they can't be infinite by measure and they are additive.

On Tuesday, September 28, 2021 at 2:27:40 AM UTC-7, FromTheRafters wrote:
> Mike Terry explained :
> > On 27/09/2021 20:24, FromTheRafters wrote:
> >> Jim Burns formulated the question :
> >>> On 9/26/2021 6:21 AM, FromTheRafters wrote:
> >>>> James Waldby formulated on Sunday :
> >>>>> FromTheRafters <err...@nomail.afraid.org> wrote:
> >>>>>> Earle Jones formulated the question :
> >>>
> >>>>>>> Select a point on a line. Call it A.
> >>>>>>> Select another point on a line. Call it B.
> >>>>>>> Select another point on a line. Call it C.
> >>>>>>> What is the probability that C is between A and B?
> >>>>>
> >>>>>> Nearly zero.
> >>>>>
> >>>>> How do you see that? It seems like 1/3 is plausible, as follows:
> >>>>> Suppose the points are selected at random in some way, with A,B,C
> >>>>> i.i.d, and distinct with probability 1. Then there are six possible
> >>>>> orderings of A, B, C, each equally likely. Two of those orderings
> >>>>> have C between A and B.
> >>>>
> >>>> The distance between A and B is finite.
> >>>
> >>> On average, the distance between A and B is infinite.
> >
> > Well, that depends entirely on the distribution for A and B. Some
> > distributions give infinite expectation, some finite.
> >
> >>>
> >>> I suggest that we fill in the blank places in the question
> >>> by tipping a delta function over on its side.
> >>> Define the distribution of the points to be flat between
> >>> -M/2 and +M/2, zero outside, answer the question for that,
> >>> and then let M --> +inf
> >>>
> >>> We get the probability of C between A and B to be 1/3.
> >>> Also, we get the average distance between A and B
> >>> to be M/3.
> >>>
> >>> ----
> >>> There are other ways to fill in the blank places.
> >>> My first thoughts ran to an exponential "decay" or a power law.
> >>> However, that symmetry argument looks very good.
> >>> I haven't worked out any other answer. I bet they'll be 1/3 also.
> >
> > Yes, with a couple of extra assumptions:
> > a) one fixed distibution (the same one) for each of A, B, C
> > b) A, B, C distributions are independent
> > c) no "discrete" component in the distibutions, so the probability
> > of and of A, B, C coinciding is zero.
> >
> > I'd say this is the most reasonable way to go, because in practice someone
> > suggesting this experiment can be challenged to say what they mean by "select
> > a point on a line", and if we press them, hopefully they would come up with
> > some pseudo-mechanical process for generating the point, which will hopefully
> > have some well defined distribution we could apply, meeting the criteria
> > (a)(b)(c). Then the symmetry argument works.
> >
> > E.g. "put a spinner 1m from the line and spin it around. [AKA let the angle
> > the spinner points be evenly distributed on [0, 2Pi)]. Extend a line through
> > where the spinner stops, and the selected point on the original line is where
> > that line intersects the spinner line.
> >
> >>
> >> I suggest, naively, that there are just as many points within the segment
> >> as there are without, so one half could be a (useless IMO) answer too.
> >
> > Probability is not based on "the number of points", so that would indeed be a
> > useless answer. :) (..and one not justified by probability arguments..)
> That's what I mean, there is no measure. With my interpretation we had
> a line, a point on that line, so now two opposing rays can be
> envisioned. Another 'different' point added and we have the opposing
> rays with a line segment between them - and line segments have two
> endpoints so they can't be infinite by measure and they are additive.

I supposed the expected values AB = BC = AC.

Most people are familiar with two shapes of distributions,
those with a mean like the normal and those without a mean
like the uniform.

Many results are built in probability about the Central Limit
Theorem after various laws of large numbers but there's space
for some Uniformization Limit Theorem in laws of large numbers.

Lots of people might write "given events A, B, C, assume A = 0
'without loss of generality' then B > A because otherwise instead relabel
A and B for A = 0 and A < B, then, similarly if C would be < 0 that
what results are 3 events that have an order A <= B and A <= C",
that there is basically "sampling with relabeling", where mostly people
are familiar with "sampling w/w-o replacement".

There are 6 permutations of {A, B, C}, in their lexicographic ordering,
which is a total ordering.

ABC
ACB
BAC
BCA
CAB
CBA

From these P(A<C<B | B<C<A) = 1/3.

About a uniform natural integer or random point on the semi-infinite line,
if A = 0 then P(B>C) = P(C > B) makes for P(B>C) = P(C>B) = 1/2.

I.e. whether one point is fixed makes for perspective of a sort.

Why choose a point and then use other points?
If you need to calculate slope why not just start at the point you know?
How could you be any more accurate at the surrounding slopes?

FromTheRafters <erratic@nomail.afraid.org> wrote:
> Mike Terry explained :
>> On 27/09/2021 20:24, FromTheRafters wrote:
>>> Jim Burns formulated the question :
>>>> On 9/26/2021 6:21 AM, FromTheRafters wrote:
>>>>> James Waldby formulated on Sunday :
>>>>>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>>>>>> Earle Jones formulated the question :
>>>>
>>>>>>>> Select a point on a line.  Call it A.
>>>>>>>> Select another point on a line.  Call it B.
>>>>>>>> Select another point on a line.  Call it C.
>>>>>>>> What is the probability that C is between A and B?
>>>>>>
>>>>>>> Nearly zero.
>>>>>>
>>>>>> How do you see that?  It seems like 1/3 is plausible, as follows:
>>>>>> Suppose the points are selected at random in some way, with A,B,C
>>>>>> i.i.d, and distinct with probability 1.  Then there are six possible
>>>>>> orderings of A, B, C, each equally likely.  Two of those orderings
>>>>>> have C between A and B.
>>>>>
>>>>> The distance between A and B is finite.
>>>>
>>>> On average, the distance between A and B is infinite.
>>
>> Well, that depends entirely on the distribution for A and B. Some
>> distributions give infinite expectation, some finite.
>>>>
>>>> I suggest that we fill in the blank places in the question
>>>> by tipping a delta function over on its side.
>>>> Define the distribution of the points to be flat between
>>>> -M/2 and +M/2, zero outside, answer the question for that,
>>>> and then let M --> +inf
>>>>
>>>> We get the probability of C between A and B to be 1/3.
>>>> Also, we get the average distance between A and B
>>>> to be M/3.
>>>>
>>>> ----
>>>> There are other ways to fill in the blank places.
>>>> My first thoughts ran to an exponential "decay" or a power law.
>>>> However, that symmetry argument looks very good.
>>>> I haven't worked out any other answer. I bet they'll be 1/3 also.
>>
>> Yes, with a couple of extra assumptions:
>> a) one fixed distibution (the same one) for each of A, B, C
>> b) A, B, C distributions are independent
>> c) no "discrete" component in the distibutions, so the probability
>> of and of A, B, C coinciding is zero.
>>
>> I'd say this is the most reasonable way to go, because in practice someone
>> suggesting this experiment can be challenged to say what they mean by "select
>> a point on a line", and if we press them, hopefully they would come up with
>> some pseudo-mechanical process for generating the point, which will hopefully
>> have some well defined distribution we could apply, meeting the criteria
>> (a)(b)(c). Then the symmetry argument works.
>>
>> E.g. "put a spinner 1m from the line and spin it around. [AKA let the angle
>> the spinner points be evenly distributed on [0, 2Pi)]. Extend a line through
>> where the spinner stops, and the selected point on the original line is where
>> that line intersects the spinner line.
>>
>>> I suggest, naively, that there are just as many points within the segment
>>> as there are without, so one half could be a (useless IMO) answer too.
>>
>> Probability is not based on "the number of points", so that would indeed be a
>> useless answer. :) (..and one not justified by probability arguments..)
>
> That's what I mean, there is no measure. With my interpretation we had
> a line, a point on that line, so now two opposing rays can be
> envisioned. Another 'different' point added and we have the opposing
> rays with a line segment between them - and line segments have two
> endpoints so they can't be infinite by measure and they are additive.

It looks like your interpretation may be based on some unstated or
unsupported assumptions, such as particular probability distributions
or choice of C depending on choices of A, B.

In my post, the essential phrasing is this: "Suppose the points are
selected at random in some way, with A,B,C i.i.d, and distinct with
probability 1". That sentence states the following assumptions:

1. The points are selected at random in some way.
2. The random variables A, B, C are independent. (The first i in i.i.d)
3. The random variables A, B, C are identically distributed. (The i.d part)
4. A, B, C are distinct with probability 1.

Items 1-3 address some vague parts of the problem statement, and state
a fairly general scenario. It doesn't matter what the probability
distribution is; it doesn't matter if the distribution is discrete or
not, continuous or not; it merely matters that there is a way of
selecting points all the same way from some line and that the choice
of one does not influence the choice of another. Under these
assumptions, all six possible orders of A, B, C are equally likely.

Item 4 means that the probability is 0 that two points coincide. As a
consequence, points coinciding doesn't change the expected outcome.

James Waldby wrote on 9/29/2021 :
> FromTheRafters <erratic@nomail.afraid.org> wrote:
>> Mike Terry explained :
>>> On 27/09/2021 20:24, FromTheRafters wrote:
>>>> Jim Burns formulated the question :
>>>>> On 9/26/2021 6:21 AM, FromTheRafters wrote:
>>>>>> James Waldby formulated on Sunday :
>>>>>>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>>>>>>> Earle Jones formulated the question :
>>>>>>>>> Select a point on a line.  Call it A.
>>>>>>>>> Select another point on a line.  Call it B.
>>>>>>>>> Select another point on a line.  Call it C.
>>>>>>>>> What is the probability that C is between A and B?
>>>>>>>> Nearly zero.
>>>>>>>
>>>>>>> How do you see that?  It seems like 1/3 is plausible, as follows:
>>>>>>> Suppose the points are selected at random in some way, with A,B,C
>>>>>>> i.i.d, and distinct with probability 1.  Then there are six possible
>>>>>>> orderings of A, B, C, each equally likely.  Two of those orderings
>>>>>>> have C between A and B.
>>>>>>
>>>>>> The distance between A and B is finite.
>>>>>
>>>>> On average, the distance between A and B is infinite.
>>>
>>> Well, that depends entirely on the distribution for A and B. Some
>>> distributions give infinite expectation, some finite.
>>>>>
>>>>> I suggest that we fill in the blank places in the question
>>>>> by tipping a delta function over on its side.
>>>>> Define the distribution of the points to be flat between
>>>>> -M/2 and +M/2, zero outside, answer the question for that,
>>>>> and then let M --> +inf
>>>>>
>>>>> We get the probability of C between A and B to be 1/3.
>>>>> Also, we get the average distance between A and B
>>>>> to be M/3.
>>>>>
>>>>> ----
>>>>> There are other ways to fill in the blank places.
>>>>> My first thoughts ran to an exponential "decay" or a power law.
>>>>> However, that symmetry argument looks very good.
>>>>> I haven't worked out any other answer. I bet they'll be 1/3 also.
>>>
>>> Yes, with a couple of extra assumptions:
>>> a) one fixed distibution (the same one) for each of A, B, C
>>> b) A, B, C distributions are independent
>>> c) no "discrete" component in the distibutions, so the probability
>>> of and of A, B, C coinciding is zero.
>>>
>>> I'd say this is the most reasonable way to go, because in practice someone
>>> suggesting this experiment can be challenged to say what they mean by
>>> "select a point on a line", and if we press them, hopefully they would
>>> come up with some pseudo-mechanical process for generating the point,
>>> which will hopefully have some well defined distribution we could apply,
>>> meeting the criteria (a)(b)(c). Then the symmetry argument works.
>>>
>>> E.g. "put a spinner 1m from the line and spin it around. [AKA let the
>>> angle the spinner points be evenly distributed on [0, 2Pi)]. Extend a
>>> line through where the spinner stops, and the selected point on the
>>> original line is where that line intersects the spinner line.
>>>
>>>> I suggest, naively, that there are just as many points within the segment
>>>> as there are without, so one half could be a (useless IMO) answer too.
>>>
>>> Probability is not based on "the number of points", so that would indeed be
>>> a useless answer. :) (..and one not justified by probability
>>> arguments..)
>>
>> That's what I mean, there is no measure. With my interpretation we had
>> a line, a point on that line, so now two opposing rays can be
>> envisioned. Another 'different' point added and we have the opposing
>> rays with a line segment between them - and line segments have two
>> endpoints so they can't be infinite by measure and they are additive.
>
> It looks like your interpretation may be based on some unstated or
> unsupported assumptions, such as particular probability distributions
> or choice of C depending on choices of A, B.
>
> In my post, the essential phrasing is this: "Suppose the points are
> selected at random in some way, with A,B,C i.i.d, and distinct with
> probability 1". That sentence states the following assumptions:
>
> 1. The points are selected at random in some way.
> 2. The random variables A, B, C are independent. (The first i in i.i.d)
> 3. The random variables A, B, C are identically distributed. (The i.d part)
> 4. A, B, C are distinct with probability 1.
>
> Items 1-3 address some vague parts of the problem statement, and state
> a fairly general scenario. It doesn't matter what the probability
> distribution is; it doesn't matter if the distribution is discrete or
> not, continuous or not; it merely matters that there is a way of
> selecting points all the same way from some line and that the choice
> of one does not influence the choice of another. Under these
> assumptions, all six possible orders of A, B, C are equally likely.
>
> Item 4 means that the probability is 0 that two points coincide. As a
> consequence, points coinciding doesn't change the expected outcome.

It seems analogous to the shell game or three card monte shuffle. The
probability of the pea under the walnut shell being in the center
position after stopping the juggling at 'random' times is 1/3 as one
would expect. My thought was that stating it as being on a line gives
us a notion of size (length in this case) where the line segment "set"
is "smaller" (a proper subset of the line) than the the infinite ray
"sets" are. The segment being the target.

It might have been offered as an area also if one considers a first
point on an infinite plane being a center, the second different point
determines a radius for the circle at that center and a third either
hits the associated disk or it misses. As big as you want the target to
be, you can still zoom out to see how "small" it is in comparison to
the vast surrounding infinite plane.

On 9/29/2021 4:23 AM, FromTheRafters wrote:
> James Waldby wrote on 9/29/2021 :
>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>> Mike Terry explained :
>>>> On 27/09/2021 20:24, FromTheRafters wrote:
>>>>> Jim Burns formulated the question :
>>>>>> On 9/26/2021 6:21 AM, FromTheRafters wrote:
>>>>>>> James Waldby formulated on Sunday :
>>>>>>>> FromTheRafters <erratic@nomail.afraid.org> wrote:
>>>>>>>>> Earle Jones formulated the question :

>>>>>>>>>> Select a point on a line.  Call it A.
>>>>>>>>>> Select another point on a line.  Call it B.
>>>>>>>>>> Select another point on a line.  Call it C.
>>>>>>>>>> What is the probability that C is between A and B?

> It might have been offered as an area also if one considers
> the second different point a first point on an infinite
> plane being a center, determines a radius for the circle at
> that center and a third either hits the associated disk or
> it misses. As big as you want the target to be, you can
> still zoom out to see how "small" it is in comparison to
> the vast surrounding infinite plane.

You have reminded me of Joseph Bertrand's paradox (1889).
https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)
| | The Bertrand paradox is generally presented as follows:[3]
| Consider an equilateral triangle inscribed in a circle.
| Suppose a chord of the circle is chosen at random.
| What is the probability that the chord is longer than
| a side of the triangle?
| | Bertrand gave three arguments (each using
| the principle of indifference), all apparently valid,
| yet yielding different results:

1.
Pick two points on the circle. Let the chord be between
the two. Probability 1/3.

2.
Pick a radius of the circle. Pick a point on that radius.
Let the chord be perpendicular to the radius at the point.
Probability 1/2.

3.
Pick a point in the circle. Let the chord have that point
as the chord's midpoint. Probability 1/4.

The devil is in the details.

I remember thinking at the time (a little while back)
(no, not 1889), Bertrand's paradox could be resolved by being
indifferent to _enough_ If there are different answers,
a bias must have been introduced somehow. Eliminate the bias.

In this particular case, a random chord would be indifferent to
the existence of the circle. So, the distribution _in the plane_
of lines that may or may not intersect the circle should not vary
under translations and rotations in the plane. This gives a
single answer, for what it's worth. (I only vaguely remember
what it is.)

Today, I am less confident that I defeated the paradox.
Yes, I have an answer. Is it THE answer?

----
It seems to me that "being indifferent to enough" is a well-
-regarded technique in physics. A description of a physical
system that is indifferent to what time it is has time symmetry.
Time symmetry implies energy conservation. And so on.

https://en.wikipedia.org/wiki/Noether%27s_theorem

It is perhaps too tenuous a connection between indifference in
a-priori probability and indifference is conservation laws.
Perhaps. But I wonder.