# Is this mathematically/statistically possible?
**URL:** https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376
**Category:** Factual Questions
**Created:** [February 1, 2012, 1:47am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376 "2012-02-01T01:47:21Z")
**Posts on this page:** 13
**Page:** 2
### Author:  [@TATG](https://boards.straightdope.com/u/TATG)
#### Post date: [February 1, 2012, 4:08am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/21 "2012-02-01T04:08:25Z")
> [@Indistinguishable](#):
>
> In a group with 0 people, it is simultaneously true that the number of 1s is 0% of everyone, 100% of everyone, 50% of everyone, 51% of everyone, and every which other thing.
So if all 4 groups have zero members, their agregate has any percentage of 1s that you like, including less than 50%.
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### Author:  [@Manlob](https://boards.straightdope.com/u/Manlob)
#### Post date: [February 1, 2012, 5:26am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/22 "2012-02-01T05:26:43Z")
Say a room has 103 people, 51 of them men, and 52 women. As a whole 49.5% are men. Number those people so that number 1 through 51 are the men, and number 52 to 103 are the women.
Group A has all people wearing yellow shoes and contains #1 and #100: 50% men.
Group B has everyone who speaks English and incudes everyone but #101, #102, and #103: 51% men
Group C has everyone with a pierced nose and consists of #2 and #99: 50% men
Group D has everyone who fluent in Russian and contains #3 and #98: 50% men
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### Author:  [@Giles](https://boards.straightdope.com/u/Giles)
#### Post date: [February 1, 2012, 5:31am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/23 "2012-02-01T05:31:39Z")
**Manlob** , I think there’s an assumption in the OP that the four groups are disjoint, i.e., no two of them have any members in common. Without that assumption, as you demonstrate, the average of the aggregate need not be between the top and bottom average.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [February 1, 2012, 5:53am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/24 "2012-02-01T05:53:18Z")
> [@Giles](#):
>
> **Manlob** , I think there’s an assumption in the OP that the four groups are disjoint, i.e., no two of them have any members in common. Without that assumption, as you demonstrate, the average of the aggregate need not be between the top and bottom average.
Exactly. This is what’s known as Simpson’s paradox in statistics.
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### Author:  [@TriPolar](https://boards.straightdope.com/u/TriPolar)
#### Post date: [February 1, 2012, 6:00am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/25 "2012-02-01T06:00:22Z")
> [@TATG](#):
>
> So if all 4 groups have zero members, their agregate has any percentage of 1s that you like, including less than 50%.
I’m waiting for someone to address this.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [February 1, 2012, 6:21am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/26 "2012-02-01T06:21:13Z")
> [@TriPolar](#):
>
> I’m waiting for someone to address this.
What’s to address? It’s just correct.
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### Author:  [@TriPolar](https://boards.straightdope.com/u/TriPolar)
#### Post date: [February 1, 2012, 6:33am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/27 "2012-02-01T06:33:40Z")
> [@Indistinguishable](#):
>
> What’s to address? It’s just correct.
So the groups A, B, C, D aggregated can have less than 50% 1s then. This was contraindicated earlier in the thread.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [February 1, 2012, 7:30am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/28 "2012-02-01T07:30:48Z")
> [@TriPolar](#):
>
> So the groups A, B, C, D aggregated can have less than 50% 1s then. This was contraindicated earlier in the thread.
Alright, good call. It always has to be the case that the aggregate percentage is at least 50%; it just may also happen (if, and only if, there are 0 people) that the aggregate percentage is also, separately, a value less than 0%. This is the problem with division by zero.
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### Author:  [@Norse](https://boards.straightdope.com/u/Norse)
#### Post date: [February 1, 2012, 9:43am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/29 "2012-02-01T09:43:32Z")
> [@Cagey\_Drifter](#):
>
> Is it possible for Groups A, B, C, D aggregated to be less than 50%? Is it still true if Group B is 50% instead of 51%?
Depends on whether your percentage numbers are accurate or rounded.
If your real, perfectly accurate, percentages are 49.5, 50.5, 49.5 and 49.5 (rounded to 50, 51, 50 and 50), the aggregate percentage will be 49.75%. However, even if this is less than 50%, you should report the result with the same precision as your other numbers, i.e. as 50%.
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### Author:  [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)
#### Post date: [February 1, 2012, 2:12pm UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/30 "2012-02-01T14:12:49Z")
> [@ultrafilter](#):
>
> Exactly. This is what’s known as Simpson’s paradox in statistics.
That’s not what Simpson’s paradox describes. It describes relative percentages, yes, but not aggregate percentages. Note that in the [Wiki examples](http://en.wikipedia.org/wiki/Simpson%27s_paradox#Examples), all the aggregate percentages are still between the two original values. In other words, when a and b are averaged to x, it’s still true that a\
### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [February 1, 2012, 7:17pm UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/31 "2012-02-01T19:17:33Z")
At any rate, though, it is true that with overlapping groups, the aggregate percentage need not be a weighted average of the in-group percentages. For example, consider Barack Obama, Hilary Clinton, and Mitt Romney. Out of the two men, the percentage of Obamas is 1/2, and out of the two Democrats, the percentage of Obamas is 1/2. Overall, however, the percentage of Obamas is 1/3.
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### Author:  [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)
#### Post date: [February 1, 2012, 8:20pm UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/32 "2012-02-01T20:20:13Z")
> [@Cagey\_Drifter](#):
>
> Let’s say there are 4 groups of people, with an unknown number of people in each group. Within each group, every person is either a 1 or a 0. The percentages below describe the percentage of people who are 1s.
>
> Given:
> Group A: 50%
> Group B: 51%
> Group C: 50%
> Group D: 50%
>
> Is it possible for Groups A, B, C, D aggregated to be less than 50%? Is it still true if Group B is 50% instead of 51%?
Are the groups disjoint or not?
Group A: Alex and Bill are 1, Cathy is 0 (67%)
Group B: Alex and Bill are 1, Damian is 0 (67%)
Group C: Alex and Bill are 1, Elizabeth is 0 (67%)
Group D: Alex and Bill are 1, Frank is 0 (67%)
But the percentage of 1’s in the aggregate is 33%.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [February 2, 2012, 1:57am UTC](https://boards.straightdope.com/t/is-this-mathematically-statistically-possible/611376/33 "2012-02-02T01:57:57Z")
Even more simply, and in keeping with the OP:
Group A: Obama is 1, Alex is 0 (50%)
Group B: Obama is 1, Bill is 0 (50%)
Group C: Obama is 1, Cathy is 0 (50%)
Group D: Obama is 1, Damien is 0 (50%)
Yet the percentage of 1s in the aggregate is 1/5 = 20%.
But as said before, if the groups are disjoint and exhaustive (so everyone is counted just once as we go through each group), and there’s at least one person around (so we don’t worry about division by zero), then the groups cannot all have percentage 50% or greater while the aggregate percentage is less than 50%. [And same for any other number in place of 50%]
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