Peak Oil
Donate Bitcoin

Donate Paypal


PeakOil is You

PeakOil is You

Statistics Question

What's on your mind?
General interest discussions, not necessarily related to depletion.
Post a reply

Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 03:19:24

$this->bbcode_second_pass_quote('jbeckton', '
')AGAIN, I will continue to debate your statistical ignorance if you start your own thread.
link

Here's the question:

$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of BOTH occurring must be even lower.


True or False?
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby ZombieMalthus » Tue 06 Jan 2009, 04:17:48

$this->bbcode_second_pass_quote('Koyaanisqatsi', '')$this->bbcode_second_pass_quote('jbeckton', '
')AGAIN, I will continue to debate your statistical ignorance if you start your own thread.
link

Here's the question:

$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of BOTH occurring must be even lower.


True or False?


False, q (falling down the stairs) and p (breaking your leg) are not independent events.

Flipping heads four times in a row with a quarter, q1, q2, q3, q4 (each with probability of 1/2) ARE independent events so the probability of is q1 * q2 * q3 * q4 IS lower than q1 or q2 or q3 or q4 happening by themselves.
User avatar
ZombieMalthus
Wood
Wood
 
Posts: 32
Joined: Wed 14 May 2008, 03:00:00
Top

Re: Statistics Question

Unread postby smallpoxgirl » Tue 06 Jan 2009, 04:43:20

$this->bbcode_second_pass_quote('Koyaanisqatsi', '')$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of BOTH occurring must be even lower.


True or False?


You example isn't quite precise enough, IMHO.

If the statements are:
"The chances of me falling down the stairs are low.
If I fall down the stairs, the chances of me breaking my leg are low.
Therefore the chances of me falling down the stairs and breaking my leg are even lower." That's fine.

OTOH "The chances of me falling down the stairs are low.
The chances of me breaking my leg are low.
Therefore the chances of me falling down the stairs are even lower."

That's incorrect logic, because it doesn't exclude the possibility that "If and only if, I fall down the stairs, I will break my leg" If everyone who falls down the stairs breaks their leg, then the chances of falling down the stairs and breaking your leg are exactly the same as the chances of falling down the stairs.
"We were standing on the edges
Of a thousand burning bridges
Sifting through the ashes every day
What we thought would never end
Now is nothing more than a memory
The way things were before
I lost my way" - OCMS
User avatar
smallpoxgirl
Expert
Expert
 
Posts: 7258
Joined: Mon 08 Nov 2004, 04:00:00
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 04:51:06

$this->bbcode_second_pass_quote('ZombieMalthus', '')$this->bbcode_second_pass_quote('Koyaanisqatsi', '')$this->bbcode_second_pass_quote('jbeckton', '
')AGAIN, I will continue to debate your statistical ignorance if you start your own thread.
link

Here's the question:

$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of BOTH occurring must be even lower.


True or False?


False, q (falling down the stairs) and p (breaking your leg) are not independent events.

Flipping heads four times in a row with a quarter, q1, q2, q3, q4 (each with probability of 1/2) ARE independent events so the probability of is q1 * q2 * q3 * q4 IS lower than q1 or q2 or q3 or q4 happening by themselves.


You are saying that if the events were independent then the statement would be true. However, just because they are not independent doesn't mean that the statement isn't true.
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 05:03:21

$this->bbcode_second_pass_quote('smallpoxgirl', '')$this->bbcode_second_pass_quote('Koyaanisqatsi', '')$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of BOTH occurring must be even lower.


True or False?


You example isn't quite precise enough, IMHO.

If the statements are:
"The chances of me falling down the stairs are low.
If I fall down the stairs, the chances of me breaking my leg are low.
Therefore the chances of me falling down the stairs and breaking my leg are even lower." That's fine.

OTOH "The chances of me falling down the stairs are low.
The chances of me breaking my leg are low.
Therefore the chances of me falling down the stairs are even lower."

That's incorrect logic, because it doesn't exclude the possibility that "If and only if, I fall down the stairs, I will break my leg" If everyone who falls down the stairs breaks their leg, then the chances of falling down the stairs and breaking your leg are exactly the same as the chances of falling down the stairs.


I read it as:

The chances of me falling down the stairs are low.
The chances of me breaking my leg are low.
Therefore the chances of me falling down the stairs and breaking my leg are even lower.

This frames the events as being independent. As ZombieMalthus noted, the events are actually dependent. I contend that the probability of both occurring together is lower irrespective of whether the events are conceived as dependent or independent.
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 05:18:20

$this->bbcode_second_pass_quote('', 'W')hen events are dependent, each possible outcome is related to the other. Given two events A and B, the probability of obtaining both A and B is the product of the probability of obtaining one of the events times the conditional probability of obtaining the other event, given the first event has occurred.

P(A and B) = P(A) . P(B|A)

This rule says that for both of the two events to occur, the first one must occur ( (P(A) ) and then, given the first event has occurred, the second event occurs ( (P(B|A) ).
link

P(A) = Probability of falling down stairs
P(B) = Probability of breaking leg.
P(B|A) = Probability of breaking leg given falling down stairs.

According to the question, both probabilities are low, so:
P(A) ~ P(B) << 1

Further,
P(B|A) < 1
P(A and B) = P(A)*P(B|A)
Therefore P(A and B) < P(A)

Note that in the limit if P(A) = 1 then P(B|A) = P(B) and P(A and B) = P(B). In other words, P(A and B) cannot be greater than P(B). But in fact P(A) < 1, therefore P(A and B) < P(B).

The statement is true. The probability of you falling down stairs may be small, the probability of you breaking your leg may be small, and the probability of you breaking your leg GIVEN that you have fallen down stairs may be larger than both - but the probability of you falling down stairs AND breaking your leg is still smaller than either you falling down stairs or breaking your leg.
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby wisconsin_cur » Tue 06 Jan 2009, 05:20:50

Not my strong suite but if we were to say (for the sake of illustration) that you would fall down the stairs once out of every six descents. On any given descent you would have a 1/6 chance of falling. Now if we say that every time you fall there is another one in six (1/6) chance of you breaking a leg. To find out the chance of falling down and breaking your leg we would multiply 1/6 x 1/6 and find out that the answer is 1/36.
http://www.thenewfederalistpapers.com
User avatar
wisconsin_cur
Light Sweet Crude
Light Sweet Crude
 
Posts: 4576
Joined: Thu 10 May 2007, 03:00:00
Location: 45 degrees North. 883 feet above sealevel.
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 05:26:17

$this->bbcode_second_pass_quote('wisconsin_cur', 'N')ot my strong suite but if we were to say (for the sake of illustration) that you would fall down the stairs once out of every six descents. On any given descent you would have a 1/6 chance of falling. Now if we say that every time you fall there is another one in six (1/6) chance of you breaking a leg. To find out the chance of falling down and breaking your leg we would multiply 1/6 x 1/6 and find out that the answer is 1/36.


Exactly - in other words:
P(A) = Probability of falling down stairs = 1/6
P(B|A) = Probability of breaking leg given falling down stairs = 1/6.

Therefore P(A and B) = probability of falling downstairs and breaking leg = P(A)*P(B|A) = 1/36.
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby Carlhole » Tue 06 Jan 2009, 11:31:44

[web]http://en.wikipedia.org/wiki/Probability[/web]

Probability

It's all right there on Wikipedia.

Falling down stairs is an event. Breaking a leg in a fall down the stairs is a conditional event. You can't use the chances of breaking a leg for ANY reason in this calculation because breaking a leg is conditional on falling down stairs in this example.

Conditional Probability

You could make the subject simple by asking "What are the chances of rolling a dice and getting six twice in a row?" The conditional event is the first throw of six. Obviously, the chances of throwing two sixes in a row are lower than throwing just one six.

In the case of the WTC collapse, it seems pretty clear to me that NIST's Final Report declared that ordinary office fires were the sole cause of the collapse. So, if you wanted to analyze the chances of a high-rise office building collapsing due to fire (and you accepted NIST's conclusions), you would have to accept the usual probability of serious fires occurring in any given high-rise in the first place -- because the conditional event is serious fires (NOT terrorist strikes occurring nearby; NOT other buildings collapsing next door).

[The true origin of the fires in WTC7 remains a mystery. Were they deliberately set or were they started by the collapse of the Towers. We don't know.]

I find it pretty humorous that NIST characterizes the collapse of Building 7 as "extremely rare" -- given that, allegedly, it is the ONLY building in history ever to have suffered total collapse due to fire. Similarly, there had never been a Total Progressive Collapse of a high-rise structure for ANY reason prior to 911. All the building collapses on 911 were "extremely rare".

I'm sure you could call a building insurer and discover what the statistical chances are of a high-rise suffering a serious fire. But how would you determine the statistical probability of a unique event like a total collapse caused by fire? How would you determine the statistical probability of a Total Progressive Collapse of a steel-framed high-rise for ANY reason? I don't know. Those would be good questions to ask a statistician.

A whole separate statistical question from the above is: "What are the chances of a high-rise building totally collapsing due to fire AND a terrorist strike on two buildings next door leading to total progressive collapse of both buildings due to fire and damage?"

Obviously, that is a vanishingly small probability. NIST has admitted in its report that the collapse of WTC7 is virtually a separate event from the collapse of the Towers. And so analyzing the probabilities of these "extremely rare", independent events occurring simultaneously and in close proximity makes mathematical sense.
Carlhole
 
Top

Re: Statistics Question

Unread postby jbeckton » Tue 06 Jan 2009, 20:05:10

$this->bbcode_second_pass_quote('Koyaanisqatsi', '')$this->bbcode_second_pass_quote('jbeckton', '
')AGAIN, I will continue to debate your statistical ignorance if you start your own thread.
link

Here's the question:

$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of BOTH occurring must be even lower.


True or False?


Lets put this back into perspective along with the root of the analogy.

P(A)=probability of me falling down the stairs (WTC7 on fire)
P(B)=probability of me breaking my leg (WTC7 collapsing from fire)

P(A and B) is not P(A)*P(B) because the events are dependent.

Now what do I mean with the anology?

The probability of me breaking my leg is INCREASED once 'A' has occurred [P(B)<P(B|A)] . Therefore the probability of me breaking my leg is not "even lower" after I have fallen down the stairs [P(B|A)], it's actually higher! (i.e. the probability of the building collapsing from fires is not lower than it was before the fire occurred as was implied IMO.

Therefore:

$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of BOTH occurring must be even lower.


False

Perhaps I should have specified that by “both” I meant P(B|A), but after I realized that you either misunderstood me, or were perhaps trying to take my statement out of context I quickly verified my point:

$this->bbcode_second_pass_quote('jbeckton', 'D')o you not agree that P(B)<P(B|A)?

I didn’t think it was important because no one disputes that WTC 7 was on fire, only if it collapsed due to the fire is disputed, i.e. no one is concerned with P(B), only P(B|A).
Those that cannot do..... teach. Those that cannot teach......teach gym.-Jack black
User avatar
jbeckton
Expert
Expert
 
Posts: 2082
Joined: Fri 05 Jan 2007, 04:00:00
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 20:37:21

$this->bbcode_second_pass_quote('Carlhole', '
')In the case of the WTC collapse, it seems pretty clear to me that NIST's Final Report declared that ordinary office fires were the sole cause of the collapse...


Thanks, Carlhole, but my intent here is to answer the question in the OP first, which is independent of what happened on 9/11 (even though of course the 9/11 question is the important issue that motivated this discussion to begin with).
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 21:08:09

Let's leave WTC7 out of this for now, as that's not what the OP is asking.

$this->bbcode_second_pass_quote('jbeckton', '
')Perhaps I should have specified that by “both” I meant P(B|A),


But you didn't so the statement is wrong. "Both" means "A and B" not "A given B". As I continually said in that thread, you were confusing P(B|A) with P(A and B), which only now you have conceded. Thanks for finally admitting it.

Furthermore, if that was actually what you meant, then statement would have read:

$this->bbcode_second_pass_quote('', 'T')his is like saying that there is a low probability of me falling down the stairs and there is also a low probability of me breaking my leg so the probability of breaking my leg given falling down stairs is even lower.


which is definitely NOT what I said (which was about the probability of major terrorist attack AND fire-induced progressive collapse, not about the probability of fire-induced progressive collapse given office fire) [okay, I just broke my own rule about not talking about 9/11].

$this->bbcode_second_pass_quote('jbeckton', '
')I didn’t think it was important because no one disputes that WTC 7 was on fire, only if it collapsed due to the fire is disputed, i.e. no one is concerned with P(B), only P(B|A).


Uh-huh, just like I didn't think it was important to distinguish between the formulas

P(A and B) = P(A)*P(B) [independent events]
P(A and B) = P(A)*P(B|A) [dependent events]

BECAUSE THEY BOTH YIELD THE SAME ANSWER, since both P(B) and P(B|A) are very small numbers. If anything you have been taking my comments out of context by saying my argument is invalid rather than me taking your comments out of context.
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby jbeckton » Tue 06 Jan 2009, 21:35:45

$this->bbcode_second_pass_quote('Koyaanisqatsi', 'L')et's leave WTC7 out of this for now, as that's not what the OP is asking.

$this->bbcode_second_pass_quote('jbeckton', '
')Perhaps I should have specified that by “both” I meant P(B|A),


But you didn't so the statement is wrong.


No, only your interpretation was wrong. I didn't know that I had to break it down for "Barney Style" for you.

$this->bbcode_second_pass_quote('Koyaanisqatsi', 'P')(A and B) = P(A)*P(B) [independent events]
P(A and B) = P(A)*P(B|A) [dependent events]

BECAUSE THEY BOTH YIELD THE SAME ANSWER, since both P(B) and P(B|A) are very small numbers.


"very small"?

One could make the argument that every probability is a very small number since it is always < or =1!

This is why there are different rules for dependent and independent events. Choosing to ignore (or not knowing) those rules is a mistake, a mistake you made when you claimed that P(A and B)=P(A)*P(B).

I have clearly spelled out your mistake and my explanation of your mistake (now available Barney style) so please forgive me if I don't respond to any further whining. If it makes you feel any better you can declare yourself “victorious”!

I know its embarrassing but at least you learned something!

Cheers
Those that cannot do..... teach. Those that cannot teach......teach gym.-Jack black
User avatar
jbeckton
Expert
Expert
 
Posts: 2082
Joined: Fri 05 Jan 2007, 04:00:00
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 22:05:59

$this->bbcode_second_pass_quote('jbeckton', '')$this->bbcode_second_pass_quote('Koyaanisqatsi', 'L')et's leave WTC7 out of this for now, as that's not what the OP is asking.

$this->bbcode_second_pass_quote('jbeckton', '
')Perhaps I should have specified that by “both” I meant P(B|A),


But you didn't so the statement is wrong.


No, only your interpretation was wrong. I didn't know that I had to break it down for "Barney Style" for you.


No, you used it incorrectly:

What is the probability that it will BOTH rain AND flood today? [P(A AND B)]

It's raining today. What is the probability that it will flood? [P(B|A)].

And if my interpretation was wrong, then your characterization of my argument was totally inaccurate: I did not say that the probability of a fire-induced progressive collapse of an office building given a major terrorist attack in NYC is smaller than the probability of a fire-induced progressive collapse of an office building. I said that the probability of both a major terrorist attack on NYC and a fire-induced progressive collapse of an office building together is smaller than the probability of either alone. Admit it - if your claim that you meant P(B|A) is true, then your characterization of my position is false.

$this->bbcode_second_pass_quote('jbeckton', '
')One could make the argument that every probability is a very small number since it is always < or =1!


This is bogus.

P=99.9999999999999999999999999999999%: large
P=00.0000000000000000000000000000001%: small

Get real.

$this->bbcode_second_pass_quote('jbeckton', '
')This is why there are different rules for dependent and independent events. Choosing to ignore (or not knowing) those rules is a mistake, a mistake you made when you claimed that P(A and B)=P(A)*P(B).


Bogus again:
P(fire-induced progressive collapse) = unique event (very small probability)
P(fire-induced progressive collapse|fire) = unique event (very small number)
P(9/11) = unique event (very small number)
Thus:

P(9/11 AND fire-induced progressive collapse) = P(9/11)*P(fire-induced progressive collapse|fire) = an even smaller number

$this->bbcode_second_pass_quote('', '
')I have clearly spelled out your mistake and my explanation of your mistake (now available Barney style) so please forgive me if I don't respond to any further whining. If it makes you feel any better you can declare yourself “victorious”!

I know its embarrassing but at least you learned something!

Cheers

It's laughable, time and again on the 9/11 thread you said that you would be happy to continue this discussion if I started a different thread on it. Now that I've done so you are dodging and weaving yet again. Like I've said, you can never admit to being wrong about anything!
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Tue 06 Jan 2009, 22:08:40

$this->bbcode_second_pass_quote('jbeckton', '
')AGAIN, I will continue to debate your statistical ignorance if you start your own thread.
link
:razz:
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby jbeckton » Wed 07 Jan 2009, 21:35:30

$this->bbcode_second_pass_quote('Koyaanisqatsi', '
')It's laughable, time and again on the 9/11 thread you said that you would be happy to continue this discussion if I started a different thread on it. Now that I've done so you are dodging and weaving yet again.


1) I was interested in having a discussion so that I may explain to you exactly what I meant and that you had clearly misinterpreted my statement (as you saw early in this thread many posters didn’t see it the way you did either). If after this step by step "barney style" explanation you wish to continue ranting about who is "right" and who is "wrong" this is clearly no longer a discussion. It's whining like a child.

2) I have over 2000 posts over the last 2 years on this board. After being in the conversation for about 10 pages and 15 posts you make the claim:

$this->bbcode_second_pass_quote('', 'L')ike I've said, you can never admit to being wrong about anything!


Cheers
Perhaps you need a refresher on sample sizes (and confidence intervals) as well as statistical probability!
Those that cannot do..... teach. Those that cannot teach......teach gym.-Jack black
User avatar
jbeckton
Expert
Expert
 
Posts: 2082
Joined: Fri 05 Jan 2007, 04:00:00
Top

Re: Statistics Question

Unread postby jbeckton » Wed 07 Jan 2009, 21:36:12

DP
Those that cannot do..... teach. Those that cannot teach......teach gym.-Jack black
User avatar
jbeckton
Expert
Expert
 
Posts: 2082
Joined: Fri 05 Jan 2007, 04:00:00
Top

Re: Statistics Question

Unread postby emeraldg40 » Wed 07 Jan 2009, 23:20:50

$this->bbcode_second_pass_quote('jbeckton', 'D')P


So, what are the chances you would fall out of your ski lift and have to hang upside down 1/2 naked?
User avatar
emeraldg40
Lignite
Lignite
 
Posts: 221
Joined: Sat 24 May 2008, 03:00:00
Top

Re: Statistics Question

Unread postby Koyaanisqatsi » Thu 08 Jan 2009, 03:56:36

$this->bbcode_second_pass_quote('jbeckton', '')$this->bbcode_second_pass_quote('Koyaanisqatsi', '
')It's laughable, time and again on the 9/11 thread you said that you would be happy to continue this discussion if I started a different thread on it. Now that I've done so you are dodging and weaving yet again.


1) I was interested in having a discussion so that I may explain to you exactly what I meant and that you had clearly misinterpreted my statement


And you posted one post here and in the second post you bailed. You already made multiple posts on this topic on the 9/11 thread - did you really need to start an entirely new thread to make only one post? What kind of a "discussion" is that? Dodge and weave!

$this->bbcode_second_pass_quote('jbeckton', '
')(as you saw early in this thread many posters didn’t see it the way you did either).


Since when did we start voting on the truth? As I said, even if you say you meant both=P(B|A), then the statement doesn't map onto my argument about the 9/11 probabilities, so your claim that this invalidates the argument is itself invalid. I would think that someone who genuinely wants to discover the truth would follow an argument to its logical conclusion, but you clearly aren't interested in the truth.

$this->bbcode_second_pass_quote('jbeckton', '
')2) I have over 2000 posts over the last 2 years on this board. After being in the conversation for about 10 pages and 15 posts you make the claim:

$this->bbcode_second_pass_quote('', 'L')ike I've said, you can never admit to being wrong about anything!



And point me to any one of those posts where you admit to being wrong about something. (Don't let my join date fool you - I lurked on this board for a very long time before joining.)

$this->bbcode_second_pass_quote('jbeckton', '
')Perhaps you need a refresher on sample sizes (and confidence intervals) as well as statistical probability!

Right back at'cha, "Engineer". :lol:
User avatar
Koyaanisqatsi
Lignite
Lignite
 
Posts: 351
Joined: Mon 17 Sep 2007, 03:00:00
Location: Pacific NW
Top

Re: Statistics Question

Unread postby jbeckton » Thu 08 Jan 2009, 19:41:00

$this->bbcode_second_pass_quote('Koyaanisqatsi', '
')And you posted one post here and in the second post you bailed.


Yes, I was interested in having a discussion to clarify my (and your) point, not to listen to you whine about who was right and who was wrong.

There is a difference.
Those that cannot do..... teach. Those that cannot teach......teach gym.-Jack black
User avatar
jbeckton
Expert
Expert
 
Posts: 2082
Joined: Fri 05 Jan 2007, 04:00:00
Top
Next
Post a reply

Return to Open Topic Discussion

Who is online

Users browsing this forum: No registered users and 2 guests



cron