
09082014, 05:21 PM

#1

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A Probability Question
I need to check my reasoning for another weird calculation in my book.
Assume that one person sneezes, on average, once every three days (not a real figure). He/she does not sneeze when sleeping.
Over the course of 70 years, what's the probability that two people will sneeze at the same time (where "same time" means within .1 second)?
Here's my calculation, please let me know if I'm thinking straight.
In three 16hour days there are 173,000 seconds, or 1,730,000 .1 second periods.
The chance that the two people will sneeze "simultaneously" in a threeday period is therefore one chance in 1,730,000.
In 70 years, there are 8,517 threeday periods. So the probability comes to 1 / 1,730,000 * 8517 = .0049 or
About one chance in 200.
Is my reasoning correct?
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09082014, 05:29 PM

#2

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This is off the top of my head, but no. I think it is one chance in (1.7M ^2).
Think about two coin tossers. Call a sneeze 'heads', and four tosses per day. 2 in 4 chance that they (edit for clarity  each one separately) will 'sneeze' in one day.
But two people tossing ends up:
A B
00
01
10
11
only 1 in 4 chance they both 'sneeze' on the same toss.
go from there...
EDIT  NOPE, changing my mind, see later post
ERD50
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09082014, 05:34 PM

#3

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Implicit in your calculations is an assumption that each person's sneeze is an independent event (the occurrence of one isn't correlated with the occurrence of the other event).
In reality, sneezes can be caused by dust, particulate matter, pepper, cold air, or even light.
As a result it's likely that sneezing events will tend to cluster around exposure events to dust, cold air, pepper, light, etc. So the odds of simultaneous sneezing will be significantly higher than you calculated.
Anecdotally, my family often has nearsimultaneous sneeze sessions. I guess DW and I have the photic sneeze reflex and passed that trait on to our kids. We can walk out of the house on a sunny day and 23 of us will sneeze within seconds of each other. I seem to recall simultaneous sneezes on multiple occasions.
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09082014, 05:56 PM

#4

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Quote:
Originally Posted by ERD50
This is off the top of my head, but no. I think it is one chance in (1.7M ^2).

Without considering Fuego's assertion, 1.73M x 17.3M sounds about right.
Anecdotal data ... DW and I lived together for 27 years + 1 year of dating. I don't recall we sneezed at the same time, ever. But my memory has been steadily degenerating. I will have to ask DW what is her recollection. That woman has an elephant's memory.
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09082014, 06:01 PM

#5

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If B catches A's cold, all bets are off!



09082014, 06:07 PM

#6

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Consider throwing two dice. What are the odds that they end up with the same number? One in six, right? So what if you throw them seven times? Are the odds 7*1/6 = more than 100%? No. To determine these odds it is easier to calculate the odds of it not happening, and subtract that from 1. So the odds for the dice are:
1(5/6)^7 = 0.72
...and for the sneezes are:
1(1727999/1728000)^8517 = 0.0036, or 1 in 300.



09082014, 09:00 PM

#7

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Changing from my first impulse...
It isn't one probability times the other  there are two factors.
1) It isn't just like throwing a die, or flipping a coin. The problem states the person will sneeze once in a three day period. While a coin or die may not come up in a given period. Heck, that's not even right  he just said 'on average'.
2) And since the second sneeze only has to match the first sneeze, and both will happen, it is just the odds of the second sneeze occurring at a given time. Or maybe not...
But I forget how the math works, not convinced it is just the 1/x either, I think there is still another factor? Need to think some more.
ERD50



09082014, 09:08 PM

#8

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As one who is allergic to math and probabilities this thread makes my nose itch
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09082014, 09:17 PM

#9

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Quote:
Originally Posted by ERD50
And since the second sneeze only has to match the first sneeze, and both will happen, it is just the odds of the second sneeze occurring at a given time.

Right. Sneeze #1 during any three day period is a given. The odds of sneeze #2 during that three day period matching (within 0.1 sec) sneeze #1 is random...or 1 in the number of 0.1 sec periods in 3 days (1728000 of them). So the odds of the sneezes coinciding during any three day period is 1/1728000, and the odds of them not coinciding is 1727999/1728000. The odds of them not coinciding in 70 years of three day periods (8517 of them) is (1727999/1728000) to the 8517 power. So 1 minus that number is the chance of the sneezes actually coinciding at least once during those 70 years.



09082014, 09:34 PM

#10

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Google "Birthday Paradox" and read the Wiki. It's a similar problem involving two events occurring.
Your current methodology is not correct.
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09082014, 09:56 PM

#11

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OK, to make the numbers more manageable, think of it as one sneeze on average per day, during waking hours, and say they just need to be in the same 8 hour period.
So each have a 1 in 3 chance of sneezing in any 8 hour period. The chance of both sneezing in that 8 hour period is 1/3 * 1/3 = 1/9. But there are three chances per day, so that takes you back to the 1 in 3 chance in any one day. So that's why it's not the square function I first thought it was, because there are also that many opportunities for a match, so that washes.
So I think TAl was right in the OP.
hehheh, good thing I had PhD statisticians to call on at work!
ERD50



09082014, 10:11 PM

#12

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Quote:
Originally Posted by youbet
Google "Birthday Paradox" and read the Wiki. It's a similar problem involving two events occurring.
Your current methodology is not correct.

Does the 'Birthday Paradox' really fit? I've seen that applied to figure the probability of a match between any two people of a larger group. Each person in the group would have a birthday that could be represented by a number between 1 and 365 (excluding those pesky leap year babies).
But here, we are looking for match between two, and only two people.
Or do you view the 0.1 second increments as the 'people' and the sneezes as the values? Hmmm, my brain hurts, I'm going to bed!
ERD50



09092014, 09:29 AM

#13

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Quote:
Originally Posted by scrinch
Right. Sneeze #1 during any three day period is a given. The odds of sneeze #2 during that three day period matching (within 0.1 sec) sneeze #1 is random...or 1 in the number of 0.1 sec periods in 3 days (1728000 of them). So the odds of the sneezes coinciding during any three day period is 1/1728000, and the odds of them not coinciding is 1727999/1728000. The odds of them not coinciding in 70 years of three day periods (8517 of them) is (1727999/1728000) to the 8517 power. So 1 minus that number is the chance of the sneezes actually coinciding at least once during those 70 years.

Thanks, Scrinch.
When I run those numbers I get:
1727999/1728000 = 0.99999942129
0.99999942129 ^ 8517 = 0.99508325247
1  0.99508325247 = 0.00491674753
1,000 / 0.00491674753 = 203.386485455762256

Now, that's approximately the same result I got with my initial calculation, which is, what, coincidence? Too much thinking here. This is how I'll put it in the book, leaving some assumptions out, so no one can catch me if I'm wrong.

Wunderkind Alex Carter was working on an electronics project with his equally precocious twin, Martin, when the sneeze hit them both. They laughed.
“What are the chances of that?” For normal people, it would be a rhetorical question, but Alex knew he’d get an answer.
“Let’s see.” Martin continued soldering a capacitor to the circuit board while he talked. “Let’s say you and I sneeze, on average, once every … three days.”
“And don’t sneeze in our sleep … and the sneezes are independent.”
“Right. So, three sixteenhour days comes out to about … 173,000 seconds. Hand me the Tronex cutters.”
“Get them yourself. They’re on your side. And we define simultaneous as …”
“Say, within .1 seconds of one another.” Martin picked up the cutters, and snipped the capacitor leads.
Alex released a diode from its packaging. “Okay, so the chances that we’ll sneeze at the same time, for a threeday period is—”
“One chance in 1,730,000.”
The twins continued their soldering and reasoning, and estimated that there was a one in 200 chance of them both sneezing simultaneously at some point during their lifetimes
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09092014, 09:48 AM

#14

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Quote:
Originally Posted by TromboneAl
Thanks, Scrinch.
When I run those numbers I get:
1727999/1728000 = 0.99999942129
0.99999942129 ^ 8517 = 0.99508325247
1  0.99508325247 = 0.00491674753
1,000 / 0.00491674753 = 203.386485455762256

Now, that's approximately the same result I got with my initial calculation, which is, what, coincidence?

Not coincidence, math again.
When you work with low odds simple multiplication of those odds is a good approximation for the right way (as done above).
With higher odds (or likewise higher number of experiments) the multiplication overestimates more and more, until it breaks down completely.
Devious isn't it



09092014, 10:23 AM

#15

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Quote:
Originally Posted by scrinch
Right. Sneeze #1 during any three day period is a given. The odds of sneeze #2 during that three day period matching (within 0.1 sec) sneeze #1 is random...or 1 in the number of 0.1 sec periods in 3 days (1728000 of them).

This sounds wrong to me. Sneeze #2 can occur 0.1 second before sneeze #1, or it can exactly coincide with sneeze #1, or it can occur 0.1 second after sneeze #1. In all three cases the two sneezes are to be regarded as occurring at the same time, based on the problem description. So three out of the 1728000 intervals will result in simultaneous sneezes.



09092014, 10:33 AM

#16

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Quote:
Originally Posted by TromboneAl
...
The twins continued their soldering and reasoning, and estimated that there was a one in 200 chance of them both sneezing simultaneously at some point during their lifetimes

In that context, I do think FUEGO's observation of sneezing being affected by the environment (solder fumes?) comes into play and your readers will likely pick up on that. And also Meadbh's observation  they are brothers in the same environment, they would be exposed to the same germs and likely have similar response systems.
I hadn't mentioned it, since you didn't provide context, I thought maybe the people were unrelated, and in separate rooms, countries, planets, solar systems, dimensions (it is science fiction, right?)...
And though I didn't completely analyze it, I do think it is just two different approaches to the same math, and not a coincidence that they are the same (close) answers. The rounding could work in different directions, but similar results.
ERD50



09092014, 10:41 AM

#17

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Quote:
Originally Posted by karluk
This sounds wrong to me. Sneeze #2 can occur 0.1 second before sneeze #1, or it can exactly coincide with sneeze #1, or it can occur 0.1 second after sneeze #1. In all three cases the two sneezes are to be regarded as occurring at the same time, based on the problem description. So three out of the 1728000 intervals will result in simultaneous sneezes.

Well, 'within 0.1 sec' does imply plus or minus 0.1 sec  looks like we all missed that. But that would be two 0.1 sec intervals, not three? The 'exact' would be included in one or the other boundaries.
ERD50



09092014, 11:21 AM

#18

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Quote:
Originally Posted by ERD50
Well, 'within 0.1 sec' does imply plus or minus 0.1 sec  looks like we all missed that. But that would be two 0.1 sec intervals, not three? The 'exact' would be included in one or the other boundaries.
ERD50

Yes, that sounds right. There is only a range of 0.1 second to +0.1 second for sneeze #2 to be considered simultaneous with sneeze #1. That's 2 out of the 1728000 0.1 second intervals in the three day period. Plugging this number into scrinch's calculation, I get
1  (1727998/1728000)^8517 = 0.98%
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