math riddle

[dun_noo]

I already know the answer, but my answer was different from my professors.

There exists a horizonal line A, which is bisected by a vertical line B. Two dots are randomly placed on line A. What is the probability that both of these dots lie to the left side of the vertical line B?

Please provide reasoning.

[bunny]

25%
[prob. 1st dot is on lhs] = 0.5
[prob. 2nd dot is on lhs] = 0.5

[prob. both are] = product of these = 0.5 x 0.5 = 0.25

Am I missing something?

[dun_noo]

Wrong. That's what I initially thought, but it's incorrect. The answer is 3/4, try to provide a reasoning for this.

[bunny]

I'm not convinced it's wrong yet. /images/graemlins/tongue.gif

[dun_noo]

Neither am I, which is why I posted this.

[bunny]

*shrug* What I meant was - I think it's 25% so it's difficult for me to try to provide a reason that it's 75%. I'll watch the thread with interest - I can't even see why it would be controversial.

[Sharkey]

How can the left and right sides not be symmetrical?

[Al6Jets18]

The only thing I see possibly wrong in the explanation is the possibility that the number lies on line B, but since there are infinite possibilities, that should be insignificant.

[mrmr]

[ QUOTE ]
Wrong. That's what I initially thought, but it's incorrect. The answer is 3/4, try to provide a reasoning for this.

[/ QUOTE ]

This is not correct. Unless it is a joke of some kind, and not a true math puzzle, the answer is 1/4.

Place dot 1. Whatever side it lands on, call that side X (could be left, or could be right).

So there are two sides, X and anti-X. Place dot 2. Half the time it will be on side X, and half the time it will be on side anti-X. (Because there are only two sides, and we are assuming the probabilities are equal).

Half of the time side X is the left side, and half of the time side X is the right side.

In other words, 1/2 of 1/2 of the time, dot one and two are (i) on the same side, and (ii) that side is the left side.

And of course, 1/2 of 1/2 is 1/4. This should remove all doubt.

[Fly]

[ QUOTE ]
Wrong. That's what I initially thought, but it's incorrect. The answer is 3/4, try to provide a reasoning for this.

[/ QUOTE ]

I assume your professor told you the answer was 3/4? Well let me tell you this, if you believed him for more than an instant you are mentally defective. Just consider the fact that there is absolutely no difference between the left side and right side of the line. So if you rephrased your question to ask, "What is the probablity that both dots are to the right of line B?", by your reasoning that probability must also be 3/4. Obviously we have a problem since they are mutually exclusive events, have a sum greater than 1, and collectively do not even cover the entire sample space! The answer is 1/4 by the way.

[dun_noo]

I didn't say I believed him, I believed it was 1/4. Maybe you should work on comprehension instead of math.

Reviewing it again, it may be a trick question? He could have solved the probability for the right side, which is 1/4. Then, would the other side be 1-1/4, seeing as there is an equal chance for it to be on either side? I'm not sure, maybe it was a trick.

[madnak]

You're ignoring the 1/2 chance that there is one on one side and one on the other.

[soon2bepro]

It all depends on whether or not B has bisected A through it's middle, leaving an equal portion on each side.

If we're assuming that's the case, the chances are 25%. 50% of the time the first dot will be on the left side, and only 50% of those times will the second one also be on the left side.

[Fly]

[ QUOTE ]
It all depends on whether or not B has bisected A through it's middle, leaving an equal portion on each side.


[/ QUOTE ]

Thats the definition.

[madnak]

Hehe, good edit.

With infinity the line bisects the middle by definition.

[HLMencken]

[ QUOTE ]
I didn't say I believed him, I believed it was 1/4. Maybe you should work on comprehension instead of math.

[/ QUOTE ]

You wrote: "Wrong. That's what I initially thought, but it's incorrect." So, yes, you did imply you believed him. Maybe YOU should work on reading comprehension.

[holmansf]

Are you sure the question wasn't, "What is the probability that neither of the dots are on the left side," cuz the answer to that would be 3/4.

[Marko Schmarko]

[ QUOTE ]
Are you sure the question wasn't, "What is the probability that neither of the dots are on the left side," cuz the answer to that would be 3/4.

[/ QUOTE ]
No it wouldn't.
Still 1/4.
The probability that AT LEAST ONE of the dots would be on the left side would be 3/4.
- mark

[holmansf]

Doh. You're right, I meant to say "not both of the dots are on the left."

[HLMencken]

I think you are right and that is the explanation.

Teacher says "What is probability that both dots are not on the left?"

Dun_noo hears "What is the probability that both are on the right?"

Dun_noo--is this possibly the case?

[dun_noo]

It involved a huge huge equation where he somehow arrived at the answer being 3/4. I wasn't listening; maybe I should have though. If I can get the answer on paper, I'll translate it into this thread. I don't see how it could be anything other than 1/4 though.

Oh, and my apologizes before for the comprehension crack, I was in the wrong ^_^

[BigSoonerFan]

[ QUOTE ]
Wrong. That's what I initially thought, but it's incorrect. The answer is 3/4, try to provide a reasoning for this.

[/ QUOTE ]

I think he probably meant 3/8th.

[Colima420]

Yes. the answer is 0.25.

[Colima420]

It is true that all the options add up to 1. But the options are not just left or right. there also middle. by middle I mean one on each side. Basically you have 4 possible outcomes: 1) Both dots on the left. 2)Both dots on the right. 3)One dot on the left and one dot on the right 4)The different dot on the right and the other one on the left.

Therefore the prob. of dot #1 on the left is undoubtely 50%. The probability of dot #2 on the left is also 50%. So if they are both on the left, that is 50% * 50%, which is 25%. Same logic applies to the right side.
Now dot #1 can be on the left, and dot #2 on the right. Also we can have dot #1 on the right and dot #2 on the left. This is therefore (50% * 50%) + (50% * 50%)=.25 +.25= 50%. Prob of both balls on the left + prob of one ball on the left and one ball on the right + prob of two balls on the right = 1. (25% + 50% +25%) =100% = 1.

[Carl_William]

Yep the answer is about .24999999999999999999999999999999999999 where ((.25 minus .24999999999999999999999999999999999999) times 2) is the percentage of time that one of the dots lands on the bisected line B (that's where it intersects line A). Well we all know that dots have zero length and lines have zero thickness -- but just the same there is a probability that a dot could land on line B.

[Carl_William]

Everybodies wrong!

A dot or two could land exactly in the middle.

[Carl_William]

Yes -- evidently to see a dot it must have some thinkness -- there fore it could land in the middle -- half of it on each side of the middle line....

[Nottom]

[ QUOTE ]
Yes -- evidently to see a dot it must have some thinkness -- there fore it could land in the middle -- half of it on each side of the middle line....

[/ QUOTE ]

Even if the dots were the size of New Jersey it wouldn't matter since a line is by definition infinite. Thus the chance of the dot being split down the middle is 0.

[Al6Jets18]

Yes, but by definition a line goes into infinity, so that one spot on infinity is miniscule. I believe that the answer is 24.999999... etc., but in shorthand we just put 25%. I'd love to hear an explanation for a different answer.

[Nottom]

[ QUOTE ]
Yes, but by definition a line goes into infinity, so that one spot on infinity is miniscule. I believe that the answer is 24.999999... etc., but in shorthand we just put 25%. I'd love to hear an explanation for a different answer.

[/ QUOTE ]

Uh oh, lets not let this turn into another "Does 0.9999999999999...=1?" thread.

[AlphaWice]

(1) We can change units so you are choosing 2 random variables: (independently chosen) reals x, y from [-1, 1]. (A random variable is a function taking events to numbers.)

(2) Then we want the probability of event (x<0) AND (y<0).

(3) Since the choice of x and y are independent events, this probability is P(x<0)*P(y<0). (This is the fundamental rule of counting.)

(4) Clearly, P(x<0) = 1/2, similarily for y. (Note: just because the probability P(x=0) = 0, doesn't imply that it is impossible. Certainly impossible events have probability zero, but the converse is not true.)

(5) Then P(x<0)*P(y<0) = 1/4.

If you want clarification at any step, please don't hesitate to ask.

[madnak]

[ QUOTE ]
A dot or two could land exactly in the middle.

[/ QUOTE ]

This is true, but the probability of such an event is 0.

[TomBrooks]

[ QUOTE ]
[ QUOTE ]
A dot or two could land exactly in the middle.

[/ QUOTE ]This is true, but the probability of such an event is 0.

[/ QUOTE ]
0 cannot be correct. Do you see why?

[mosta]

wait is this the same thread about the guy who didn't get into MIT??

[bunny]

[ QUOTE ]
0 cannot be correct. Do you see why?

[/ QUOTE ]
0 is most definitely correct - remember there are an infinite number of choices, not a finite but large number.

[AlphaWice]

[ QUOTE ]
(1) We can change units so you are choosing 2 random variables: (independently chosen) reals x, y from [-1, 1]. (A random variable is a function taking events to numbers.)

(2) Then we want the probability of event (x<0) AND (y<0).

(3) Since the choice of x and y are independent events, this probability is P(x<0)*P(y<0). (This is the fundamental rule of counting.)

(4) Clearly, P(x&lt;0) = 1/2, similarily for y. <u>(Note: just because the probability P(x=0) = 0, doesn't imply that it is impossible. Certainly impossible events have probability zero, but the converse is not true.)</u>

(5) Then P(x&lt;0)*P(y&lt;0) = 1/4.

If you want clarification at any step, please don't hesitate to ask.

[/ QUOTE ]

[Sharkey]

[ QUOTE ]
(Note: just because the probability P(x=0) = 0, doesn't imply that it is impossible. Certainly impossible events have probability zero, but the converse is not true.)

[/ QUOTE ]

Likewise, P(x) = 0 for any x, so the dots can’t fall anywhere, ‘twould seem.

[TomCollins]

[ QUOTE ]
[ QUOTE ]
(Note: just because the probability P(x=0) = 0, doesn't imply that it is impossible. Certainly impossible events have probability zero, but the converse is not true.)

[/ QUOTE ]

Likewise, P(x) = 0 for any x, so the dots can’t fall anywhere, ‘twould seem.

[/ QUOTE ]

Incorrect. It seems that way, but it is not that way.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
(Note: just because the probability P(x=0) = 0, doesn't imply that it is impossible. Certainly impossible events have probability zero, but the converse is not true.)

[/ QUOTE ]

Likewise, P(x) = 0 for any x, so the dots can’t fall anywhere, ‘twould seem.

[/ QUOTE ]

Incorrect. It seems that way, but it is not that way.

[/ QUOTE ]

No one ever threw a dart and hit the dartboard, apparently.

[jason_t]

[ QUOTE ]
Yep the answer is about .24999999999999999999999999999999999999 where ((.25 minus .24999999999999999999999999999999999999) times 2) is the percentage of time that one of the dots lands on the bisected line B (that's where it intersects line A). Well we all know that dots have zero length and lines have zero thickness -- but just the same there is a probability that a dot could land on line B.

[/ QUOTE ]

That probability is zero.

[jason_t]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
A dot or two could land exactly in the middle.

[/ QUOTE ]This is true, but the probability of such an event is 0.

[/ QUOTE ]
0 cannot be correct. Do you see why?

[/ QUOTE ]

You certainly don't. Zero is correct.

[mosta]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
(Note: just because the probability P(x=0) = 0, doesn't imply that it is impossible. Certainly impossible events have probability zero, but the converse is not true.)

[/ QUOTE ]

Likewise, P(x) = 0 for any x, so the dots can’t fall anywhere, ‘twould seem.

[/ QUOTE ]

Incorrect. It seems that way, but it is not that way.

[/ QUOTE ]

No one ever threw a dart and hit the dartboard, apparently.

[/ QUOTE ]

pretty counter-intuitive isn't it? infinities are confusing.

dealing with a continuumm you can through out any single point and the set is no smaller. you can through out any finite subset of points. you can throw out any infinite countable set of points. I would go on but I'm out of my depth. but that point is that all of those things = a P of zero.

put in a slightly less galling wording: the P of hitting any particular point is so small as to be indistinguishable from zero. well, in math indistinguishable from zero IS zero. the concept of the limit. if you don't accept that, you have to throw out calculus and several centuries worth of work.

[madnak]

0 probability doesn't mean impossible. Trash that assumption.

[DrunkHamster]

http://en.wikipedia.org/wiki/Almost_surely

[Matt R.]

[ QUOTE ]
0 probability doesn't mean impossible. Trash that assumption.

[/ QUOTE ]

But what if the temperature passes through 72 degrees? /images/graemlins/grin.gif

[Sharkey]

Oddly, the probability of hitting a specific point is indistinguishable form zero, yet it can’t be zero.

The definition “the probability of an event being 0 means that as the number of trials tends to infinity, the limit superior of the ratio of successes to trials is zero” seems like a bit of a dodge to me, since the interval between hits will always have some sort of a variance.

Maybe the way to look at it is the following: Before the event, the probability of a given point being hit is zero. After the event, the probability of a given point being hit is undefined.

[Sharkey]

[ QUOTE ]
the P of hitting any particular point is so small as to be indistinguishable from zero. well, in math indistinguishable from zero IS zero.

[/ QUOTE ]

That’s why I think there’s a problem with definitions.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
the P of hitting any particular point is so small as to be indistinguishable from zero. well, in math indistinguishable from zero IS zero.

[/ QUOTE ]

That’s why I think there’s a problem with definitions.

[/ QUOTE ]
The problem is not with the definition but a lack of understanding. If the prob' of each point was non-zero then you wouldn't have a probability distribution at all.

chez

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
the P of hitting any particular point is so small as to be indistinguishable from zero. well, in math indistinguishable from zero IS zero.

[/ QUOTE ]

That’s why I think there’s a problem with definitions.

[/ QUOTE ]
The problem is not with the definition but a lack of understanding. If the prob' of each point was non-zero then you wouldn't have a probability distribution at all.

chez

[/ QUOTE ]

Then maybe you can contribute some understanding.

How can an event with zero probability of occurrence occur? And, if you’re going to the above cited definition of the probability of an event being 0, how do you account for variance?

[chezlaw]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
the P of hitting any particular point is so small as to be indistinguishable from zero. well, in math indistinguishable from zero IS zero.

[/ QUOTE ]

That’s why I think there’s a problem with definitions.

[/ QUOTE ]
The problem is not with the definition but a lack of understanding. If the prob' of each point was non-zero then you wouldn't have a probability distribution at all.

chez

[/ QUOTE ]

Then maybe you can contribute some understanding.

How can an event with zero probability of occurrence occur? And, if you’re going to the above cited definition of the probability of an event being 0, how do you account for variance?

[/ QUOTE ]
because zero probability doesn't mean impossible.

Whats your problem with variance?

chez

[Copernicus]

If p=0 (not so close to 0 as to be indistinguishable from 0), what is the difference between that an "impossible"?

Give me an example of something that is possible but has p=0.

[chezlaw]

[ QUOTE ]
If p=0 (not so close to 0 as to be indistinguishable from 0), what is the difference between that an "impossible"?

Give me an example of something that is possible but has p=0.

[/ QUOTE ]
Me being born.

chez

[Sharkey]

[ QUOTE ]
because zero probability doesn't mean impossible.

[/ QUOTE ]

What is the probability of the non-occurrence of an event whose probability of occurrence is equal to zero?

[chezlaw]

[ QUOTE ]
[ QUOTE ]
because zero probability doesn't mean impossible.

[/ QUOTE ]

What is the probability of the non-occurrence of an event whose probability of occurrence is equal to zero?

[/ QUOTE ]
1 but prob = 1 doesn't mean inevitable.

chez

[Sharkey]

[ QUOTE ]
1 but prob = 1 doesn't mean inevitable.

[/ QUOTE ]

Give me an example of an event that had a 100% probability of occurrence, yet did not occur.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
1 but prob = 1 doesn't mean inevitable.

[/ QUOTE ]

Give me an example of an event that had a 100% probability of occurrence, yet did not occurred.

[/ QUOTE ]
me not being born

chez

[Sharkey]

On what basis do you conclude there was a 100% probability that you would not be born?

[chezlaw]

[ QUOTE ]
On what basis do you conclude there was a 100% probability that you would not be born?

[/ QUOTE ]
on the basis that there's a zero probability that I would be born and that p(something happing) = 1, p(a particular something happening) = 0.

chez

[Sharkey]

[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[/ QUOTE ]
You just quoted the basis.

chez

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[/ QUOTE ]
You just quoted the basis.

chez

[/ QUOTE ]

Then the next question would have to be why your being born belongs to such a distribution when, for example, winning the lottery does not.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[/ QUOTE ]
You just quoted the basis.

chez

[/ QUOTE ]

Then the next question would have to be why your being born belongs to such a distribution when, for example, winning the lottery does not.

[/ QUOTE ]
One's finite the other isn't.

I'm assuming the universe has an infinite number of states to keep the analogy with point on a line.

chez

[madnak]

Winning the lottery is based on a finite, rather than an infinite, mechanic.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[/ QUOTE ]
You just quoted the basis.

chez

[/ QUOTE ]

Then the next question would have to be why your being born belongs to such a distribution when, for example, winning the lottery does not.

[/ QUOTE ]
One's finite the other isn't.

I'm assuming the universe has an infinite number of states to keep the analogy with point on a line.

chez

[/ QUOTE ]

Yes, of course, but that’s quite a claim.

How do you figure your biology to belong to an infinite set of possible outcomes?

[chezlaw]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[/ QUOTE ]
You just quoted the basis.

chez

[/ QUOTE ]

Then the next question would have to be why your being born belongs to such a distribution when, for example, winning the lottery does not.

[/ QUOTE ]
One's finite the other isn't.

I'm assuming the universe has an infinite number of states to keep the analogy with point on a line.

chez

[/ QUOTE ]

Yes, of course, but that’s quite a claim.

How do your figure your biology to belong to an infinite set of possible outcomes?

[/ QUOTE ]
Doesn't seem much of a claim but its moving of the topic. If the probability space of the universe is finite then P(me being born) isn't zero but then its not like points on a line.

chez

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[/ QUOTE ]
You just quoted the basis.

chez

[/ QUOTE ]

Then the next question would have to be why your being born belongs to such a distribution when, for example, winning the lottery does not.

[/ QUOTE ]
One's finite the other isn't.

I'm assuming the universe has an infinite number of states to keep the analogy with point on a line.

chez

[/ QUOTE ]

Yes, of course, but that’s quite a claim.

How do your figure your biology to belong to an infinite set of possible outcomes?

[/ QUOTE ]
Doesn't seem much of a claim but its moving of the topic. If the probability space of the universe is finite then P(me being born) isn't zero but then its not like points on a line.

chez

[/ QUOTE ]

Since the probability space of the physical universe hasn’t been shown to be infinite, that leaves us still waiting for an example of an event with zero probability of occurrence occurring.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
p(something happing) = 1, p(a particular something happening) = 0.

[/ QUOTE ]

Maybe we’re getting somewhere, yet I have to ask on what basis you conclude there was a zero probability that you would be born.

[/ QUOTE ]
You just quoted the basis.

chez

[/ QUOTE ]

Then the next question would have to be why your being born belongs to such a distribution when, for example, winning the lottery does not.

[/ QUOTE ]
One's finite the other isn't.

I'm assuming the universe has an infinite number of states to keep the analogy with point on a line.

chez

[/ QUOTE ]

Yes, of course, but that’s quite a claim.

How do your figure your biology to belong to an infinite set of possible outcomes?

[/ QUOTE ]
Doesn't seem much of a claim but its moving of the topic. If the probability space of the universe is finite then P(me being born) isn't zero but then its not like points on a line.

chez

[/ QUOTE ]

Since the probability space of the physical universe hasn’t been shown to be infinite, that leaves us still waiting for an example of an event with zero probability of occurrence occurring.

[/ QUOTE ]
Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[Masquerade]

Actually the answer is 0.5 not 0.25

Suppose we choose the dots first then add the line. Let the dots be [a, b]. Then the probability that the vertical line is to the left of a is 0.5 and the probability that it's to the right of b is also 0.5. The probability of it landing in [a,b] or any finite interval is zero.

[Sharkey]

[ QUOTE ]
Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[felson]

[ QUOTE ]
I already know the answer, but my answer was different from my professors.

There exists a horizonal line A, which is bisected by a vertical line B. Two dots are randomly placed on line A. What is the probability that both of these dots lie to the left side of the vertical line B?

Please provide reasoning.

[/ QUOTE ]

This question cannot be answered without specifying the joint distribution of the random placement of the dots. If line A were a finite line segment, and if we assumed the uniform distribution, and that the placements are independent... then, since B bisects A, then you would have 0.25. But those assumptions would have to be specified.

When line A is infinite, then the question is even more vague, because there's no such thing as a uniform distribution over the infinite real set (as far as I know). If you don't specify the distributions, you get all kinds of possibilities, including perhaps the 0.5 value which Masquerade suggested.

I don't think the question can be answered as phrased. Perhaps the professor phrased it differently.

[Sharkey]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[chezlaw]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[Sharkey]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[/ QUOTE ]

It’s like the difference between anyone winning the lottery and someone in particular.

[felson]

Here's an illustration of why the question is poorly phrased.

1. "A real number X is chosen at random. What is the probability that X is less than 0?"

The natural answer is 0.5. That actually depends on your assumption of the distribution of the number and may or may not be correct. But okay, let's assume symmetry about zero. Seems reasonable, right?

2. "What is the probability that X is less than 2?"

Uh, I guess the answer has to be &gt;= 0.5. But what, exactly?

If X has the zero-mean Gaussian distribution with unit variance, then P(X&lt;2) is ~.975. If X is uniformly distributed on [0,1], then P(X&lt;2) = 1 exactly. Of course, with this distribution, we also have P(X&lt;0) = 0, which meant that we guessed wrong on question #1. Oh well.

So the question is ill-posed.

[chezlaw]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[/ QUOTE ]

It’s like the difference between anyone winning the lottery and someone in particular.

[/ QUOTE ]
No it isn't. Assume the list of enumerated points is made independantly of the point picked from the line - how can that list change the probability value of the point picked?

chez

[Sharkey]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[/ QUOTE ]

It’s like the difference between anyone winning the lottery and someone in particular.

[/ QUOTE ]
No it isn't. Assume the list of enumerated points is made independantly of the point picked from the line - how can that list change the probability value of the point picked?

chez

[/ QUOTE ]

In that case, the set consisting of the sum of the list of enumerated points and x = 0 is another enumerated set, all the elements of which are impossible future outcomes.

[chezlaw]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[/ QUOTE ]

It’s like the difference between anyone winning the lottery and someone in particular.

[/ QUOTE ]
No it isn't. Assume the list of enumerated points is made independantly of the point picked from the line - how can that list change the probability value of the point picked?

chez

[/ QUOTE ]

In that case, the set consisting of the sum of the list of enumerated points and x = 0 is another enumerated set, all the elements of which are impossible future outcomes.

[/ QUOTE ]
This doesn't respond to the previous post.

chez

[Sharkey]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[/ QUOTE ]

It’s like the difference between anyone winning the lottery and someone in particular.

[/ QUOTE ]
No it isn't. Assume the list of enumerated points is made independantly of the point picked from the line - how can that list change the probability value of the point picked?

chez

[/ QUOTE ]

In that case, the set consisting of the sum of the list of enumerated points and x = 0 is another enumerated set, all the elements of which are impossible future outcomes.

[/ QUOTE ]
This doesn't respond to the previous post.

chez

[/ QUOTE ]

Well, then clarify what you meant.

[chezlaw]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[/ QUOTE ]

It’s like the difference between anyone winning the lottery and someone in particular.

[/ QUOTE ]
No it isn't. Assume the list of enumerated points is made independantly of the point picked from the line - how can that list change the probability value of the point picked?

chez

[/ QUOTE ]

In that case, the set consisting of the sum of the list of enumerated points and x = 0 is another enumerated set, all the elements of which are impossible future outcomes.

[/ QUOTE ]
This doesn't respond to the previous post.

chez

[/ QUOTE ]

Well, then clarify what you meant.

[/ QUOTE ]
It looks clear to me. I don't think I can do better.

chez

[Sharkey]

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Nothing to do with your original objection to the p(point) being 0 on a line. Do I take it you now see why that is so?

chez

[/ QUOTE ]

Not at all, and I’ll raise ya:

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
You always want to complicate things before understanding the relatively simple. I'm not clever enough to do that, so I have to stick to the relatively simple bit first.

Can you define a probability distribution that would allow each point on a line to have a non-zero probability of being randomly selected?

chez

[/ QUOTE ]

That’s impossible, but this problem is about the specific outcome x = 0.

[/ QUOTE ]
Okay, but what special about x=0? if P(X=0)&gt;0 then its also true for all other points. This, as you say, is impossible.

chez

[/ QUOTE ]

The value x = 0 was made explicit, so the treatment is different than when dealing with an infinite set.

[/ QUOTE ]
how did making it explicit change its probability value?

chez

[/ QUOTE ]

No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]
So just saying the name of a point changes its probability from zero to non-zero. What if you whisper?

chez

[/ QUOTE ]

It’s like the difference between anyone winning the lottery and someone in particular.

[/ QUOTE ]
No it isn't. Assume the list of enumerated points is made independantly of the point picked from the line - how can that list change the probability value of the point picked?

chez

[/ QUOTE ]

In that case, the set consisting of the sum of the list of enumerated points and x = 0 is another enumerated set, all the elements of which are impossible future outcomes.

[/ QUOTE ]
This doesn't respond to the previous post.

chez

[/ QUOTE ]

Well, then clarify what you meant.

[/ QUOTE ]
It looks clear to me. I don't think I can do better.

chez

[/ QUOTE ]

Okay.

[TomCollins]

Simple question to Sharkey.

Does each point have an equal likelyhood to get picked?

[Sharkey]

On a point-by-point basis, P(x) = 0 for all x.

[bunny]

[ QUOTE ]
On a point-by-point basis, P(x) = 0 for all x.

[/ QUOTE ]
Doesnt this contradict the position you were taking earlier when arguing with chezlaw...?

[ QUOTE ]
No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]

[Sharkey]

No. Where?

[chezlaw]

[ QUOTE ]
[ QUOTE ]
On a point-by-point basis, P(x) = 0 for all x.

[/ QUOTE ]
Doesnt this contradict the position you were taking earlier when arguing with chezlaw...?

[ QUOTE ]
No enumerated set of possible outcomes where the probability of each is zero will eventually contain a positive result.

[/ QUOTE ]

[/ QUOTE ]
Its actually agreeing with chezlaw. Then it disappears into Sharkyland.

chez

[bunny]

I thought you were arguing that the P(x) couldnt be zero for all x because if you then summed them all you wouldnt get a total probability of 1. Then it seemed you claimed that P(x=0) was equal to zero (and presumably the probability of any specific x).

[Sharkey]

Sorry for any confusion.

My argument was intended to be that the sum of any enumerated set of P(x) is equal to zero.

[UnitBubble]

OK, Sharkey, try this...

Throw a dart at a dart board. The dart HITS the board. P(dart hits some point on board) = 1
this is just given in the problem

now i claim that P(dart hitting point x) = 0
why? choose a different real for that probability...lets say P(dart hitting point x) = r (r is a real number, non-zero)
this raises 2 contradictions:
(1) if i sum infinite values with positive non-zero values, the result is infinity (not 1 as it should be)
(2) there exists infinite reals in (0, r) (this is 0-r non-inclusive), wouldn't is make more sense to assign P(dart hitting point x) = r0 (r0 is element of (0,r)), then after that wouldnt it make more sense to assign P(dart hitting point x) = r1 (r1 is element of (0, r0)), etc etc...until, wouldnt it make more sense to assign P(dart hitting point x) = lim x -&gt; inf of 1/x?
this is correct answer...P(dart hitting point x) = lim x -&gt; inf of 1/x

sum, from 0 to inf, lim x -&gt; inf of 1/x...this is 1

[UnitBubble]

One more comment Sharkey,

Perhaps your arguments lie in the definitions of the words that we are using. These words are predefined for our purposes. If you would liek to invent your own system, go ahead, but arguing against the definition of a word does not make any sense.

[Sharkey]

[ QUOTE ]
this is correct answer...P(dart hitting point x) = lim x -&gt; inf of 1/x

sum, from 0 to inf, lim x -&gt; inf of 1/x...this is 1

[/ QUOTE ]

Looks good.

Note the word “enumerated” in my post. All the elements of an infinite set, like from 0 to infinity, are never enumerated, for obvious reasons.

[chezlaw]

[ QUOTE ]
[ QUOTE ]
this is correct answer...P(dart hitting point x) = lim x -&gt; inf of 1/x

sum, from 0 to inf, lim x -&gt; inf of 1/x...this is 1

[/ QUOTE ]

Looks good.

Note the word “enumerated” in my post. All the elements of an infinite set, like from 0 to infinity, are never enumerated, for obvious reasons.

[/ QUOTE ]
I guess we all noted the word. The question is why you think it makes any difference introducing it?

chez

[Sharkey]

Spare me the “we all” nonsense.

That word was used because an enumerated subset, i.e. x = 0, was the issue.

[chezlaw]

[ QUOTE ]
Spare me the “we all” nonsense.

That word was used because an enumerated subset, i.e. x = 0, was the issue.

[/ QUOTE ]
x=0 is an arbitrary point. Still unexplained why you think enumerating it makes any difference to the issue at hand.

chez

[Sharkey]

To review:

The possibility was brought up that one of the dots might fall on the x-axis. My response was that no enumerated subset of a continuum where P(x) = 0 will return a future positive result.

[chezlaw]

[ QUOTE ]
To review:

The possibility was brought up that one of the dots might fall on the x-axis. My response was that no enumerated subset of a continuum where P(x) = 0 will return a future positive result.

[/ QUOTE ]
yes but you agree that the p(x) = 0 for all x. Still no explanation as to why mentioning a value makes any difference to that.

chez

[Sharkey]

Correction: It was the y-axis (x = 0) that was mentioned.

The dots will have to fall only on P(x) = 0, because there’s no other option. However, the probability that a dot will fall within the finite enumerated subset rather the infinite unenumerated subset of the entire line is zero.

[chezlaw]

[ QUOTE ]
Correction: It was the y-axis (x = 0) that was mentioned.

The dots will have to fall only on P(x) = 0, because there’s no other option. However, the probability that a dot will fall within the finite enumerated subset rather the infinite unenumerated subset of the entire line is zero.

[/ QUOTE ]
there's still no explanation as to how introducing this enumerated subset changes the probability of any point being the chosen one.

By best guest is that you're just saying the chance of predicting which finite subset of points will be chosen is zero but that's obviously true and not relevent.

It would seem easiest if you just explained what the relevence of the enumerated set is.

chez

[jason_t]

Most of you clearly don't even know what probability means. You have an intuitive understanding of it, but no formal understanding. This is leading you to not understand what things like probability zero and probability one mean. I would suggest reading this article (http://en.wikipedia.org/wiki/Probability) in Wikipedia before trying to discuss this further. In particular, note that the article says

[ QUOTE ]
The probability of an event is generally represented as a real number between 0 and 1, inclusive. An impossible event has a probability of exactly 0, and a certain event has a probability of 1, but the converses are not always true: probability 0 events are not always impossible, nor probability 1 events certain. The rather subtle distinction between "certain" and "probability 1" is treated at greater length in the article on "almost surely".

[/ QUOTE ]

If you have questions AFTER reading the article there are several capable posters in this forum that can answer them.

[SNOWBALL]

there are four possible outcomes.

left, left
right, right
right, left
left, right

they each have a probability of 1/4. The answer is 1/4th

[sweetjazz]

It is possible that the probability distrubtion on the horizontal line is not uniform. The OP suggested that his professor gave a long calculation to justify his solution, suggesting that the OP missed some aspect of the problem that makes it non-trivial.

[jason_t]

[ QUOTE ]
It is possible that the probability distrubtion on the horizontal line is not uniform. The OP suggested that his professor gave a long calculation to justify his solution, suggesting that the OP missed some aspect of the problem that makes it non-trivial.

[/ QUOTE ]

I agree.

[HLMencken]

Okay, ad hominem attack said to satisfy Sharkey...

Seriously, why are you guys debating this issue with him? It is a very simple concept that anybody who has studied any probability and statistics knows. The probability of a particular outcome is zero if there are infinite possible outcomes. However, the definition of probability is zero is not synonymous with "impossible", as Sharkey keeps alluding to. If he really wanted to LEARN why his statements are wrong, rather than continue to weave complicated threads to obscure the point in question in order to protect the fallacy his position rests upon, then there is plenty in this thread he could read or even a 10-second google would explain this "paradox". But it is once again clearly evident that Sharkey doesn't know what he's talking about and doesn't want to consider that as a possibility. This is a ridiculous thread--ANY student in a basic probability course can explain this paradox. It really just comes down to 1/infinity = 0, but this is not the same as 0/infinity! If Sharkley intentionally chooses to ignore this difference, so be it and let this thread die. He clearly is incapable of recognizing these very basic concepts because he chooses to be oblivious to them.

Let him bask in his self-righteous ignorance in yet another thread.

[chief444]

X/N = 0 as N approaches infinity. This is true for any finite X...a numerical value (point) or a finite subset. So sure. But what is your point relative to the question at hand?

[Sharkey]

[ QUOTE ]
X/N = 0 as N approaches infinity. This is true for any finite X...a numerical value (point) or a finite subset. So sure. But what is your point relative to the question at hand?

[/ QUOTE ]

Basically, that is my point, as applied to x = 0. Or, to quote myself from 8 posts up in this thread:

The probability that a dot will fall within the finite enumerated subset rather the infinite unenumerated subset of the entire line is zero.

[chief444]

Yes. And I'm agreeing with that. But as with the point P(x)=0 does not mean it's an impossibility.

[Sharkey]

[ QUOTE ]
Yes. And I'm agreeing with that. But as with the point P(x)=0 does not mean it's an impossibility.

[/ QUOTE ]

Of course. In fact, in this case it’s an inevitability.

[TomBrooks]

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A dot or two could land exactly in the middle.

[/ QUOTE ]This is true, but the probability of such an event is 0.

[/ QUOTE ]
Tom sed: 0 cannot be correct. Do you see why?

[/ QUOTE ]jason sed: You certainly don't. Zero is correct.

[/ QUOTE ]
I studied the information in Wikipedia that you linked to in another post, jason, and I see that the way mathematitions and statisticians have defined zero, it means the chances are practically zero, but does not exclude the possibility of the event happening.

So by conventional mathamatical terminology, the chance of the dot landing directly in the middle is 0, but it could happen.

I was thinking the chances were infintisimally small, but not impossible. It seems to me a very small number, even an infantisimally small number that keeps getting smaller (as the line gets longer) just never reaches zero. For all practical purposes it might as well be zero, but really it never is quite zero.

[speakerfreak]

No that's not right, the probability of the dot landing on the line is 0 (with probability 1). In fact the probability of it ending up at any one point is 0, the question does not make sense as the line is continuous.

It has an infinite number of points it could be at, for example 0.0000000000000004312, close to zero but EXACTLY zero.

[Central Limit]

[ QUOTE ]
I already know the answer, but my answer was different from my professors.

There exists a horizonal line A, which is bisected by a vertical line B. Two dots are randomly placed on line A. What is the probability that both of these dots lie to the left side of the vertical line B?

Please provide reasoning.

[/ QUOTE ]

I get the probability is 1/3 by the following reasoning:

Line A is infinite and therefore has no center. So, verticle line B can cross it anywhere. So, the intersection of A&amp;B is simply a random point on A.

Now, what is the probability that two other random points (call them x and y) are to the left of B?

Well, along line A, the points and the vertical line can arrange themselves in 6 patterns, each equally likely because they are all three random:

x y B
y x B
x B y
y B x
B x y
B y x.

Two of these six have both x and y to the left of B. Therefore, probability = 2/6 = 1/3.

[holmansf]

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The probability that a dot will fall within the finite enumerated subset rather the infinite unenumerated subset of the entire line is zero.

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You're right, and in fact you can do better: the probability that the dot will fall on a rational number is 0.

[sweetjazz]

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The probability that a dot will fall within the finite enumerated subset rather the infinite unenumerated subset of the entire line is zero.

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You're right, and in fact you can do better: the probability that the dot will fall on a rational number is 0.

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You can do even better: the probability that the dot will fall on an algebraic number is 0.

[bunny]

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I already know the answer, but my answer was different from my professors.

There exists a horizonal line A, which is bisected by a vertical line B. Two dots are randomly placed on line A. What is the probability that both of these dots lie to the left side of the vertical line B?

Please provide reasoning.

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I get the probability is 1/3 by the following reasoning:

Line A is infinite and therefore has no center. So, verticle line B can cross it anywhere. So, the intersection of A&amp;B is simply a random point on A.

Now, what is the probability that two other random points (call them x and y) are to the left of B?

Well, along line A, the points and the vertical line can arrange themselves in 6 patterns, each equally likely because they are all three random:

x y B
y x B
x B y
y B x
B x y
B y x.

Two of these six have both x and y to the left of B. Therefore, probability = 2/6 = 1/3.

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This is an awesome false proof - I love things like this!

Two points I would make- first the original problem states that line A is bisected by line B which implies A is finite. Second is that you have assumed that all 6 distributions have an equal probability. I think the actual probabilities are:

xyB - 12.5%
yxB - 12.5%
xBy - 25%
yBx - 25%
Bxy - 12.5%
Byx - 12.5%

[PancakeBoy]

Sharkey, suppose that the probability of landing on any particular point on the line is e, where e &gt; 0.

There are an infinite # of points on the line. Calculate then the probability of landing on ANY point. It would be infinity (add e an infinite # of times). This is contradiction as probabilities are defined between 0 and 1 inclusive.



Another way to think about it. One can calculate the probability of a particular random event occuring by summing the area under the curve of its asoociated probability distribution function. If we sum the area under this curve below just one point, the value will be 0, i.e. probability = 0 of an event that certainly can occur.

[Sharkey]

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One can calculate the probability of a particular random event occuring by summing the area under the curve of its asoociated probability distribution function. If we sum the area under this curve below just one point, the value will be 0, i.e. probability = 0 of an event that certainly can occur.

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This brings back what I was getting at earlier, i.e. the possibility of a function outputting continuous values in real time. Or, put another way, can an operational algorithm uniquely identify every point in a line segment? If not, then the entire P(x) = 0 issue disappears, maybe.

[jason_t]

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Or, put another way, can an operational algorithm uniquely identify every point in a line segment?

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No. The so-called computable numbers are countable.