Lies, Damned Lies, and Statistics
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Ok, I give it a try myself
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We have the starting position, you either have the illness or your don't, with a 0.1% chance that you have it.
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The test always has a result, but there's a 1 % chance the result is the exact opposite.
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it is asked only for the cases that the test says "You have it"
- you have it 0.001 and the test shows it 0.99 => 0.00099
- you don't have it 0.999 but the test says you have it 0.01 => 0.00999
=> 0.01098 ~ 1.1 % chance you're diagnosed with the illness when only 0.1% off all people have it ?
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@J.Hilk
The question posed is: "Given that your result is reported as positive, what is the probability that you actually do have the disease?"Are you claiming that the answer to that is your "1.1%"? I say it's ~ 1 in 11, more like "9.09%".
@JonB said in Lies, Damned Lies, and Statistics:
@J.Hilk
The question posed is: "Given that your result is reported as positive, what is the probability that you actually do have the disease?"Are you claiming that the answer to that is your "1.1%"? I say it's ~ 1 in 11, more like "9.09%".
well it is 0.099 % you have it and it is diagnosed
to 0.999 % you don't have it and it is diagnosed=> ~10% chance you actually have it, when it is diagnosed ?
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@JonB said in Lies, Damned Lies, and Statistics:
@J.Hilk
The question posed is: "Given that your result is reported as positive, what is the probability that you actually do have the disease?"Are you claiming that the answer to that is your "1.1%"? I say it's ~ 1 in 11, more like "9.09%".
well it is 0.099 % you have it and it is diagnosed
to 0.999 % you don't have it and it is diagnosed=> ~10% chance you actually have it, when it is diagnosed ?
@J.Hilk
Well, your "~ 10%" is not far off my "~ 9.1%", so we're close, though I'll stick (as per my black-balls-in-bag) to my 9.1% being closer than 10%.BTW, your:
well it is 0.099 % you have it and it is diagnosed
is slightly off. We know (before the test) there is a 0.1% chance you have the ailment, and that the test "correctly diagnoses all [actual] cases". So, depending on your phrasing, this should remain at 0.1%. The bit where you originally wrote:
you have it 0.001 and the test shows it 0.99 => 0.00099
should have read:
you have it 0.001 and the test shows it 1.0 => 0.001
Then you have:
The test always has a result, but there's a 1 % chance the result is the exact opposite.
Not quite. It does not do "the exact opposite". There is a 1% chance it reports positive when it should be negative. But the opposite is not the case: it does not report negative when it should be positive ever.
- There is a 0.01% I have the disease, in which case I will deffo be told I do.
- There is a 0.1% I don't have the disease, but will be told I do.
- [Note that the above 2 cases are mutually exclusive, with no dependencies.]
- The test will report 0.11% total positives. 11 people out of 1,000. 10 will be incorrect, 1 will be correct. You have a 1 in 11 chance of being the positive one, and a 10 in 11 chance of being one of the false ones. Period.
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JonB nailed it. His analysis was spot-on, with one minor nit: The problem stated:
Someone devises a test for this disorder which, in correctly diagnoses all cases, but also reports a false positive exactly 1% of the time. As stated, the false positive rate is given without regard to any true positives -- it occurs at a rate of 1%. In a population, it will "lie" about 10 individuals. So, the correct answer is exactly (not nearly) 1 in 11.
Regarding the "everything is 50-50" assertion: this has its roots in philosophy as much as it does in probability, but it's still valid IMO. I'd love to hear how Mike Caro
would respond to your interesting point. -
@mzimmers
Right, let's start my logical analysis :)First, let me see if I've got the figures from what you have said:
- Out of every 1,000 people, 1 has the affliction.
- The test will always identify that one person as being afflicted.
- Additionally, the test will report 10* other people as being afflicted who in fact are healthy.
[* Actually, the remaining population is 999, so really 9.99 rather than 10.0. This would affect my final figure, but I imagine you're not looking for that degree of accuracy, so my answer will be right to nearest couple of decimal places!]
Obviously I have misunderstood them I reserve the right to be corrected by you and then re-analyse! Otherwise, please continue....
So, I take the test, and it reports me positive. (I knew it! Just my luck :( This is about my smoking, isn't it?)
Well, in this case, the test has reported 11 people as positive. 1 is genuinely positive, while 10 are false positive.
My conclusion:
- Before the test result I had 1 in 1,000 chance of the terminal illness you are imposing.
- After the test I have a 1 in 11 chance of being the positive one, and a 10 in 11 chance of being one of the falsies.
If it helps any, you can also think of this as balls in a bag:
- There is 1 black ball, which has "You're toast" on a piece of paper inside it.
- There are 10 black balls, which have "Only kidding" on a piece of paper inside them.
- There are 989 white balls.
You put your hand in the bag and pull out a ball. It's black :( Given that, until you open the ball and look at the piece of paper, there's a 1 in 11 chance it contains the fateful news.
Right?
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Meanwhile....
You also wrote:As Mike Caro (a brilliant professional gambler) has observed, "in the beginning, everything was even money." In other words, lacking any other information, one's best guess as to the probability of ANYTHING is 50-50.
I don't know if there was a context in which he wrote this which you have omitted, but that's a very strange statement. Lacking any information at all, one's "best guess" of a probability should not be anything like "50-50". I can only think a gambler might think that way!
BTW, a quick analysis:
- I tell you I have a bag of balls, which you cannot see.
- I ask you to guess how many balls are in the bag.
- This is an example of "you have absolutely NO [other] information upon which to base an estimate".
- You say: There are 23 balls in the bag.
- According to you/him, the odds of this being correct are 0.5.
- You decide to guess again. This time you predict 587.
- Again, you/he claim the odds of this being right are 0.5.
- Finally, you decide to change your mind to 77.
- One more time, it's 0.5 likely you're right.
3 guesses, each of which has a 0.5 chance of being right? I don't think so!
Now, we could re-analyse precisely what you mean by "one's best guess as to the probability of ANYTHING is 50-50", because perhaps you didn't have just the case above in mind.
But the point is: "lacking any other information, one's best guess as to the probability of ANYTHING is 50-50." is not a "good guess". The correct answer is: "Lacking any information, a 'probability' is simply meaningless." Probability requires some information in order to have anything to say.
I like algebra, so here's some notation from probability theory:
- P(A): Probability that A occurs
- P(A ∩ B): Probability that A and B both occur
- P(A | B): Probability that A occurs, given that B occurs
In this example,
- A: Have the disease
- B: Get a positive test result
@JonB said in Lies, Damned Lies, and Statistics:
- Out of every 1,000 people, 1 has the affliction.
P(A) = 0.001- The test will always identify that one person as being afflicted.
P(B | A) = 1By the axiom of probability, P(B∩A) = P(B|A) * P(A) so
P(B ∩ A) = 0.001- Additionally, the test will report 10* other people as being afflicted who in fact are healthy.
[* Actually, the remaining population is 999, so really 9.99 rather than 10.0. This would affect my final figure, but I imagine you're not looking for that degree of accuracy, so my answer will be right to nearest couple of decimal places!]
P(B | ¬A) = 0.01Similarly to before,
P(B ∩ ¬A) = 0.00999Well, in this case, the test has reported 11 people as positive. 1 is genuinely positive, while 10 are false positive.
P(B) = P(B ∩ A) + P(B ∩ ¬A) so
P(B) = 0.01099My conclusion:
- Before the test result I had 1 in 1,000 chance of the terminal illness you are imposing.
Yep, before the test, nothing was given, so you can only use P(A). We already know
P(A) = 0.001.- After the test I have a 1 in 11 chance of being the positive one, and a 10 in 11 chance of being one of the falsies.
Given that you got a positive test result, what is the probability that you have the disease?
P(A | B) = P(A ∩ B) / P(B) = 0.001 / 0.01099 so
P(A | B) = 0.0909918...which is a teensy bit more than 1 in 11.If it helps any, you can also think of this as balls in a bag:
- There is 1 black ball, which has "You're toast" on a piece of paper inside it.
- There are 10 black balls, which have "Only kidding" on a piece of paper inside them.
- There are 989 white balls.
You put your hand in the bag and pull out a ball. It's black :( Given that, until you open the ball and look at the piece of paper, there's a 1 in 11 chance it contains the fateful news.
Right?
Haha, awesome analogy!
As Mike Caro (a brilliant professional gambler) has observed, "in the beginning, everything was even money." In other words, lacking any other information, one's best guess as to the probability of ANYTHING is 50-50.
I don't know if there was a context in which he wrote this which you have omitted, but that's a very strange statement. Lacking any information at all, one's "best guess" of a probability should not be anything like "50-50". I can only think a gambler might think that way!
I don't think that "50-50" means "Each answer has a 50% chance to be correct". Rather, it means "Each answer has the same chance of being correct as every other possible answer".
So, if you're guessing heads or tails, there are only 2 possible answers so you have a 50% chance of getting it right. However, with the bag of balls, if the bag is big enough to hold 99 balls then there are 100 possible answers, so you have a 1% chance of getting it right.
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JKSH got it right as well (with the same very minor glitch as JonB).
@J.Hilk said in Lies, Damned Lies, and Statistics:
well it is 0.099 % you have it and it is diagnosed
Actually 0.1% (as discussed above).
to 0.999 % you don't have it and it is diagnosed
1%.
=> ~10% chance you actually have it, when it is diagnosed ?
10 false positives, one real positive: your chances are 1 in 11, or about 9%. You were pretty close.
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Since you guys did so well on that one, here's another: I hand you a bag, inside which are three coins. The coins appear identical, but while two are "fair," one will always land heads-up.
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
(Those who get this right might be ready for the extremely unintuitive Monte Hall problem...)
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I like algebra, so here's some notation from probability theory:
- P(A): Probability that A occurs
- P(A ∩ B): Probability that A and B both occur
- P(A | B): Probability that A occurs, given that B occurs
In this example,
- A: Have the disease
- B: Get a positive test result
@JonB said in Lies, Damned Lies, and Statistics:
- Out of every 1,000 people, 1 has the affliction.
P(A) = 0.001- The test will always identify that one person as being afflicted.
P(B | A) = 1By the axiom of probability, P(B∩A) = P(B|A) * P(A) so
P(B ∩ A) = 0.001- Additionally, the test will report 10* other people as being afflicted who in fact are healthy.
[* Actually, the remaining population is 999, so really 9.99 rather than 10.0. This would affect my final figure, but I imagine you're not looking for that degree of accuracy, so my answer will be right to nearest couple of decimal places!]
P(B | ¬A) = 0.01Similarly to before,
P(B ∩ ¬A) = 0.00999Well, in this case, the test has reported 11 people as positive. 1 is genuinely positive, while 10 are false positive.
P(B) = P(B ∩ A) + P(B ∩ ¬A) so
P(B) = 0.01099My conclusion:
- Before the test result I had 1 in 1,000 chance of the terminal illness you are imposing.
Yep, before the test, nothing was given, so you can only use P(A). We already know
P(A) = 0.001.- After the test I have a 1 in 11 chance of being the positive one, and a 10 in 11 chance of being one of the falsies.
Given that you got a positive test result, what is the probability that you have the disease?
P(A | B) = P(A ∩ B) / P(B) = 0.001 / 0.01099 so
P(A | B) = 0.0909918...which is a teensy bit more than 1 in 11.If it helps any, you can also think of this as balls in a bag:
- There is 1 black ball, which has "You're toast" on a piece of paper inside it.
- There are 10 black balls, which have "Only kidding" on a piece of paper inside them.
- There are 989 white balls.
You put your hand in the bag and pull out a ball. It's black :( Given that, until you open the ball and look at the piece of paper, there's a 1 in 11 chance it contains the fateful news.
Right?
Haha, awesome analogy!
As Mike Caro (a brilliant professional gambler) has observed, "in the beginning, everything was even money." In other words, lacking any other information, one's best guess as to the probability of ANYTHING is 50-50.
I don't know if there was a context in which he wrote this which you have omitted, but that's a very strange statement. Lacking any information at all, one's "best guess" of a probability should not be anything like "50-50". I can only think a gambler might think that way!
I don't think that "50-50" means "Each answer has a 50% chance to be correct". Rather, it means "Each answer has the same chance of being correct as every other possible answer".
So, if you're guessing heads or tails, there are only 2 possible answers so you have a 50% chance of getting it right. However, with the bag of balls, if the bag is big enough to hold 99 balls then there are 100 possible answers, so you have a 1% chance of getting it right.
@JKSH
Two quick observations:P(A | B) = P(A ∩ B) / P(B) = 0.001 / 0.01099 so P(A | B) = 0.0909918... which is a teensy bit more than 1 in 11.
There is still something wrong here with where you go about calculating these figures, but I'm too tired to spot it. @mzimmers said of my solution above:
His analysis was spot-on, with one minor nit:
[...]
So, the correct answer is exactly (not nearly) 1 in 11.In my first attempt, at the end I stated:
After the test I have a 1 in 11 chance of being the positive one, and a 10 in 11 chance of being one of the falsies.
And in my second clarification earlier, I had come to the same conclusion when I wrote:
The test will report 0.11% total positives. 11 people out of 1,000. 10 will be incorrect, 1 will be correct. You have a 1 in 11 chance of being the positive one, and a 10 in 11 chance of being one of the false ones. Period.
That was my attempt to say ("Period") that I had realized my previous talk about "999" & "roundings" was unnecessary & inaccurate. Like @mzimmers I conclude the chance is exactly 1 in 11.
I don't think that "50-50" means "Each answer has a 50% chance to be correct". Rather, it means "Each answer has the same chance of being correct as every other possible answer".
The second sentence might be a better way of phrasing it. Which, certainly to my mind/understanding, should never be referred to as "50-50".
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Since you guys did so well on that one, here's another: I hand you a bag, inside which are three coins. The coins appear identical, but while two are "fair," one will always land heads-up.
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
(Those who get this right might be ready for the extremely unintuitive Monte Hall problem...)
@mzimmers said in Lies, Damned Lies, and Statistics:
(Those who get this right might be ready for the extremely unintuitive Monte Hall problem...)
Darn, I was going to quote that one! :) (If you do, I won't say a word, till it's solved by someone who doesn't know.)
P.S.
Are you old enough to have watched the show live in the USA? ;-) -
Since you guys did so well on that one, here's another: I hand you a bag, inside which are three coins. The coins appear identical, but while two are "fair," one will always land heads-up.
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
(Those who get this right might be ready for the extremely unintuitive Monte Hall problem...)
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@mzimmers said in Lies, Damned Lies, and Statistics:
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
8 in 10@JonB said in Lies, Damned Lies, and Statistics:
@mzimmers said in Lies, Damned Lies, and Statistics:
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
8 in 10Correct (though I would have said 4 in 5). Care to share with the other students how you arrived at this answer?
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@JonB said in Lies, Damned Lies, and Statistics:
@mzimmers said in Lies, Damned Lies, and Statistics:
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
8 in 10Correct (though I would have said 4 in 5). Care to share with the other students how you arrived at this answer?
@mzimmers
I chose to write "8 in 10" rather than "4 in 5" deliberately, because of the way I reached the figure mentally.I thought I would not explain, at least for now, so that others might have their opportunity to think it through and see what they came up with. Like you did for the other one, perhaps I should wait for 24 hours before explaining! BTW, I found this one easier to think through than the first one, for some reason --- perhaps because the other one gave me medical frights? ;-)
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Hah...fair enough, though I'm now curious as to how you ended up at 8 in 10...but if everyone else can can wait for the answer, I suppose I can wait for the explanation.
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Since you guys did so well on that one, here's another: I hand you a bag, inside which are three coins. The coins appear identical, but while two are "fair," one will always land heads-up.
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
(Those who get this right might be ready for the extremely unintuitive Monte Hall problem...)
@mzimmers said in Lies, Damned Lies, and Statistics:
JonB nailed it. His analysis was spot-on, with one minor nit: The problem stated:
Someone devises a test for this disorder which, in correctly diagnoses all cases, but also reports a false positive exactly 1% of the time. As stated, the false positive rate is given without regard to any true positives -- it occurs at a rate of 1%. In a population, it will "lie" about 10 individuals.
So you meant "1% of the whole population receives a false positive" (
P(B ∩ ¬A) = 0.01).I thought you meant "1% of the healthy people receive a false positive" (
P(B | ¬A) = 0.01).@JonB said in Lies, Damned Lies, and Statistics:
P(A | B) = P(A ∩ B) / P(B) = 0.001 / 0.01099 so P(A | B) = 0.0909918... which is a teensy bit more than 1 in 11.
There is still something wrong here with where you go about calculating these figures, but I'm too tired to spot it.
It boiled down to the interpretation of the false-positive rate (see above). With the correct interpretation, we have:
P(A) = 0.001(0.1% of the population have the disorder)P(B | A) = 1(The test detects the disorder 100% of the time)P(B ∩ ¬A) = 0.01(The test has a 1% false positive rate within the whole population)
Finding intermediate parameters,
- P(B∩A) = P(B|A) * P(A) ⇒
P(B ∩ A) = 0.001(0.1% of the whole population have the disorder AND get a positive result) - P(B) = P(B∩A) + P(B ∩ ¬A) ⇒
P(B) = 0.011(1.1% of the whole population get a positive test result)
Finally,
- P(A | B) = P(A∩B) / P(B) ⇒
P(A | B) = 1/11(Given that I got a positive result, I have 1 in 11 chance of having the disorder)
All good! :-D
@mzimmers said in Lies, Damned Lies, and Statistics:
Since you guys did so well on that one, here's another: I hand you a bag, inside which are three coins. The coins appear identical, but while two are "fair," one will always land heads-up.
You pull a coin from the bag, and toss it three times. You get a head every time. What are the chances you pulled the unfair coin?
I used the same method as my first attempt. Same equations, just different starting numbers.
P(X|Y) = 0.8where- X: Got the unfair coin
- Y: Flipped 3 times and got 3 heads
P.S. Thanks for the fun puzzles, @mzimmers! I used to do them in school/university but haven't done any in a while.
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Hah...fair enough, though I'm now curious as to how you ended up at 8 in 10...but if everyone else can can wait for the answer, I suppose I can wait for the explanation.
I don't use @JKSH 's equations --- too much brainache!
The method is just:
- There are 8 permutations from flipping a coin 3 times.
- The unfair coin produces 3 heads in all of its permutations.
- The fair coins each produce 1 set of 3 heads in each of theirs.
- Thus of the possible 24 outcomes, there are 10 with all heads, and of those 2 are produced by the fair coins while 8 are produced by the weighted one.
Hence my initial writing of
8 in 10, rather than simplifying :)You should probably now throw Monte Hall at @JKSH :)
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Well done, and well presented. When I was faced with this problem, I did it slightly differently (1/3 * 100%) vs. (2/3 * 12.5%). The underlying logic is the same.
JKSH's notations are just a formal representation of what we're doing. Given that I took my only statistics class nearly 40 years ago, I've forgotten all the notation, though I remember most of the principles. As long as we all get to the right answers, the various approaches are equally valid.
I'll bring up Monte Hall if KJSH chimes in. And yes, I can remember watching that show live...good entertainment (if you're 12 years old).
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Well done, and well presented. When I was faced with this problem, I did it slightly differently (1/3 * 100%) vs. (2/3 * 12.5%). The underlying logic is the same.
JKSH's notations are just a formal representation of what we're doing. Given that I took my only statistics class nearly 40 years ago, I've forgotten all the notation, though I remember most of the principles. As long as we all get to the right answers, the various approaches are equally valid.
I'll bring up Monte Hall if KJSH chimes in. And yes, I can remember watching that show live...good entertainment (if you're 12 years old).
Given that I took my only statistics class nearly 40 years ago, I've forgotten all the notation, though I remember most of the principles
In that case, please remind me what the "Chi squared" test thingy is? I remember the teacher banging on about that one. And no, you are not allowed to look it up. :)
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Chi squared...ew.
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