Dead heat: Obama and Clinton split the Syracuse vote 50-50

Seems to be there's either a math professor who should be fired for not knowing probability or a reporter who should be fired for misquoting. (See the comments for some more reasonable estimates.)

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very interesting problem, whose answer depends on how it is modeled. Say, there were 100,000 possible voters for Clinton or Obama, 50,000 of each, of which 12,002 made it to the polls. Assuming the 12,002 were selected randomly, what are the chances that 6001 come from each camp?

There are (50,000 choose 6001) ways to select 6001 voters for one camp. Square that to get the total number of ways to choose 6001 of each camp to vote.

Then, divide it by the (100,000 choose 12,002) ways the voters could have been selected.

According to Mathematica, this is a 9059-digit number over a 9061-digit number. Evaluated numerically gives about 0.00776368, or one in 128.8

It's not all that unlikely.

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Todd (Bowie, MD, US, North America, Earth, Sol System, Vega region, Local Bubble, Orion arm, Milky Way Galaxy, Local Group, Virgo A Cluster, Virgo supercluster, the universe in which spock is clean shaven)

Quidquid latine dictum sit, altum sonatur.

personal page: http://blog.astrosketches.info

Indeed.

Okay. The number of voters who actually go to the polls for Clinton is x. What are the odds the number of voters who actually go the polls for Obama will also be x?

One in 50,000, isn't it?

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2009 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

Bad maths in media? Shock! Horror!

Truth!

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"Any technology, no matter how primitive, is magic to those who don't understand it." - Florence Ambrose

With 12002 votes, there are ways the election can go. Of these, there are
ways to pick 6001 votes, so the chances against it happening is

I leave the actual calculation to the student as a trivial exercise.

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‘To those who regard “crime fiction” as some sacred icon which must follow a rigid formula, I will always be the man who writes 18-syllable haiku.’Trying to make sense of computers, The Error Log.

I see a question mark in the formula

Besides, there are only 2¹²⁰⁰¹ ways that the election can go, because I'm one of the voters and there's no way in h*ck that I'm going to vote for **** (name removed because of the ban on politcal discussion).

The calculations you guys have made seem fine to me, but I wonder whether there isn't another way to look at this. You're finding the probability of such an implausible event in one election, if I well understood. It's quite low, though not astronomically so.

But perhaps it would make more sense to think about the probability of this rare event ever being observed in at least one of the many elections that the average person has lived through. The expected number of such ties increases linearly with the number of elections, which makes the event considerably less unlikely.

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0.992717

or about 1 chance in 137.3 of it happening.


Not too far from mine.

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Todd (Bowie, MD, US, North America, Earth, Sol System, Vega region, Local Bubble, Orion arm, Milky Way Galaxy, Local Group, Virgo A Cluster, Virgo supercluster, the universe in which spock is clean shaven)

Quidquid latine dictum sit, altum sonatur.

personal page: http://blog.astrosketches.info

'One in a million' is contemporary shorthand for 'unusual'.

Although I'd have thought with inflation the default would be at least one in a billion.

What are the odds of getting ANY arbitrary result?

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The probability is 1 (the odds are infinite, or infinity-to-one).

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"A witty saying proves nothing" Voltaire.
"All your bias are belong to us" Ara Pacis.

I saw another "1 in a million" recently--frozen flushed remains from an airplane toilet crashes into house, causing significant damage. FAA person said it was a one in a million event.

Ok--one in a million houses are damaged that way? probably not or about 100 or so houses in the US will have been damaged that way. One in a million flights damage a house that way? hmm..I recall there are about 40,000 flights per day on average, so every 25 days a house would be damaged. I don't think that's likely. Or--one in a million flushes damages a house? that's even less likely.

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Todd (Bowie, MD, US, North America, Earth, Sol System, Vega region, Local Bubble, Orion arm, Milky Way Galaxy, Local Group, Virgo A Cluster, Virgo supercluster, the universe in which spock is clean shaven)

Quidquid latine dictum sit, altum sonatur.

personal page: http://blog.astrosketches.info

One in a million is also a figure of speech. Given the context I don't think that particular use was literal. It's like hearing someone say "wait a minute" and then counting sixty seconds.

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"One-in-a-million chances come up 9 times out of 10"

Isn't the question 'what is the total possible number of combinations possible when six candidates split up 12,342 votes?'

If only two candidates were running and 12,002 votes were cast, the probability of any particular set of totals would be 1/12,002, wouldn't it?

12,002 to 0
12,001 to 1
.....
1 to 12,001
0 to 12,002

Like playing cards. The probability of drawing any arbitrary hand is about 2,600,000:1

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If they can get you asking the wrong questions, they don't have to worry about the answers.

What's the probability of any particular set of totals on two dice?

If you say, 1/11 (2,3,4,5,6,7,8,9,10,11,12), then I say, hey c'mon over I got a game on!

(I preface this with the warning that I can never keep the formulas for permutations, subluxations and combinations straight. Although I approve of using factorials in formulae as a way of making math more exciting.)

I think using a pair of dice as illustration is a different problem, since there are six ways to get a total of 7. But that is not the question here.

My example: a and b are running and six people vote. The final tally has to be one of the following:

a b
6 0
5 1
4 2
3 3
2 4
1 5
0 6

There are 7 possible vote totals, of which 1 is a tie. Probability of a tie is 1/7.

In a race between a and b as shown, I note that since the total number of votes is given a and b are not independent variables. Given the known total, once the value for a or b is known the other is also determined. By my thinking, if the race is only between a and b and 12,000 votes are cast, of the 12,000 possible outcomes (Wait. 12,001) one is a tie. so, by chance alone the probability of a tie score is 1/12,001.

Since there were actually six names on the ballot it gets messier, of course (Perhaps a Messier Number could be used here).

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If they can get you asking the wrong questions, they don't have to worry about the answers.

I disagree. The context implies a literal interpretation.

In the case of the OP, yes.

In the other example tdvance gave (and to which noclevername was responding), the "figure of speech" interpretation is more likely.

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2009 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

What if there were six voters and two candidates? There's only one way to get the 6-0 tally, have all six vote for the first candidate. But there's more than one way to get 3-3, including just having them switch votes!

I guess I should've done it for two voters, two candidates
way to swing it to astronomy

I would say that each way people vote are not equally likely--hence my original model of choosing 12002 voters randomly from a pool that was already evenly split between two candidates--which seems to be close to reality. If it were something like 25% Obama 75% Hillary overall, the 6001-6001 split would be far less likely. And a 3001-9001 split would be far more likely.

Of course, as I said, it all depends on how you model it--a true model would consider neurons in people, their likelihood of having an ancestor moving to Syracuse in the past few centuries, weather patterns, etc. etc....one reason math whizzes get hung up on probability (and even some famous ones have screwed up in the past....) is probability questions aren't generally stated mathematically, but you have to find a mathematical model that matches the question somewhat, making assumptions not stated in the question, etc. It's an art.

Tangential: I recall an argument between my high school Trig teacher (we had a probability month during the Trig year, as well as a few other random subjects since Trig doesn't require the full year) and myself--I still think I'm right, and I bet he still thinks he's right Here is how it went.

You have a keychain. You put 5 different keys on it. How many ways can you do it? (really, a combinatorics question, but same kind of deal).

He said: doesn't matter how you put the first key on (by symmetry). 4 ways to choose the next key in line, 3 for the next, 2 for the next, 1 for the last, total: 4*3*2*1=24.

I said: doesn't matter how you put the key on (by symmetry). 4 for the next key, 3, then 2, then 1, just as before, but now if you had put the keys on in the reverse order, it would be the same arrangement! (just flip the keychain around). Every arrangement has a corresponding symmetric arrangement, so you divide by two, giving 12 possibilities. (so I got a grade of 99 on that test as a result.)

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Todd (Bowie, MD, US, North America, Earth, Sol System, Vega region, Local Bubble, Orion arm, Milky Way Galaxy, Local Group, Virgo A Cluster, Virgo supercluster, the universe in which spock is clean shaven)

Quidquid latine dictum sit, altum sonatur.

personal page: http://blog.astrosketches.info

With 6 votes cast for two different candidates, you have 2⁶ possible elections.
Of these, there are 6!/((6-n)!n!) ways to get n votes.

a b #of ways to get this
6 0 1 aaaaaa
5 1 6 aaaaab aaaaba aaabaa aabaaaa abaaaa baaaaa
4 2 15 aaaabb aaabab aaabba aabaab aababa aabbaa abaaab abaaba ababaa abbaaa baaaab baaaba baabaa babaaa bbaaaa
3 3 20 aaabbb aababb aabbab aabbba abaabb ababab ababba abbaab abbaba abbbaa repeat with a's and b's swapped to get the rest
2 4 15 bbbbaa bbbaba bbbaab bbabba bbabab bbaabb babbba babbab bababb baabbb abbbba abbbab abbabb ababbb aabbbb
1 5 6 bbbbba bbbbab bbbabb bbabbb babbbb abbbbb
0 6 1 bbbbbb

so given no voter bias, the chances are ~1.5% for one candidate to get all votes, ~9.5% for one to get all but one vote, 23.4% for one to get all but 2, and a whopping 31.25% or almost 1/3 change to get a tie.

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‘To those who regard “crime fiction” as some sacred icon which must follow a rigid formula, I will always be the man who writes 18-syllable haiku.’Trying to make sense of computers, The Error Log.

If you look at the slightly more complicated version of 6 voters, with equal chance to vote a, vote b and abstain, you get 3⁶ possible outcomes with lots of combinations possible.
Looking only at the ties
a b # of ways
0 0 1 (ways to pick 0 out of 6 * ways to pick 0 out of 6=1*1)
1 1 30 (ways to pick 1 out of 6 * ways to pick 1 out of 5=6*5)
2 2 90 (ways to pick 2 out of 6 * ways to pick 2 out of 4=15*6)
3 3 20 (ways to pick 3 out of 6 * ways to pick 3 out of 3=20*1)
or 141/729 ~= 19% for a tie.

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‘To those who regard “crime fiction” as some sacred icon which must follow a rigid formula, I will always be the man who writes 18-syllable haiku.’Trying to make sense of computers, The Error Log.

Henrik: I figured that after doing all the theoretical calculations some real numbers might be interesting, so for fun I went to the historical records of all US presidential elections from 1856 to 2004 and got the Dem/Rep vote percentages. I normalized those to 100% of the vote to minimize the effect of odd elections (like 1860), then took the ratio of Dem to Rep.

For 38 elections, the mean was 0.994 with S.D. 0.29. On a histogram with bin size 0.2 I got a distribution looking pretty much normal, considering the small population size:

Bin/Frequency
0/0
0.2/0
0.4/0
0.6/2
0.8/7
1/11
1.2/11
1.4/3
1.6/2
1.8/1
2/1

Throw the numbers in a quick Excel plot and it's pretty neat.

This suggests to me that the large majority of US presidential elections are determined by chance.

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While technically John Edwards had dropped out of the race, he was still on the ballot and he did receive votes. His influence needs to be factor in as well.