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The 18181 problem revised: 2 of 4 LEGISLATIVE races: 1 out of 1,601,666!

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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Tue Sep-02-03 09:02 PM
Original message
The 18181 problem revised: 2 of 4 LEGISLATIVE races: 1 out of 1,601,666!
Interested DUers: please comment. Does it make sense to revise the original problem: 3 out of 30 elections have the same result (18181) to the following:

Determine the probability that 2 out of 4 races for Senator or Representative will have the same number of Republican winning votes?

I have calculated this to be: 1 out of 1,601,666!

Step 1: get the data for 4 Texas legislative elections in Comal county.

Step 2: determine the range of Republican votes - 3100.

Step 3: Calculate the probability that at least TWO of the 4 county-wide elections will have the IDENTICAL NUMBER OF REPUBLICAN VOTES WITHIN THE 3100 RANGE, ASSUMING ALL POSSIBLE DUPLICATES ARE EQUALLY LIKELY (A UNIFORM DISTRIBUTION). In this case the number was 18,181, but it could just as well be 18,182, etc. Of course, 18,181 reversed is still 18181 (a palindrome)- but that is NOT going to be an issue here.

Eliminate all the races except the legislative ones for U.S. and Texas Reps (2) and Senators (2).

Race Republican Democrat Other Total
1. U.S. Senator 18156 5696 350 24202
2. U.S. Rep. District 21 19066 4627 371 24064
19. State Senator District 25 18181 4988 723 23892
20. State Rep. District 73 18181 5303 * 23484

The other race having 18181 votes was for a non-legislative office
25. County Judge 18181 5547 * 23728
Disregard this one.

Step1: Note that there are 6 combination pairs in which 2 out of 4 elections can be identical, as follows:
1,2
1,3
1,4
2,3
2,4
3,4

Step 2: Calculate the probability as before, but with the number of elections now equal to 4. Note that the probability of a single combination of two races having the same vote count over a possible range of 3100 outcomes is
1/(3100*3100) = 1/9,610,000

Step 3: This result can occur in 6 different ways, for each combination, so the probability = 6/(9,610,000)= 6.2435E-07

or 1 out of 1,601,666 !.

Compare this result to the original, where 3 out of 30 elections were the focus. The chances for that occurrence is just 1 out of 2,500. Quite a difference.

Maybe our suspicions were legitimate, after all.






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ferg Donating Member (873 posts) Send PM | Profile | Ignore Tue Sep-02-03 09:33 PM
Response to Original message
1. nope
The probability of a single combination of two races with the same vote count is 1/3100, because there are 3100 possible results for the first race.

Remember, there are 3100 possible values for the first election. So the odds of the second matching the first is 1/3100.

The odds of any two (or more) being the same is:

1 - (3099/3100)*(3098/3100)*(3097/3100)

Which is 1 in 517.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Tue Sep-02-03 09:46 PM
Response to Reply #1
2. You are right. My bad. Sorry, folks, but can we rephrase the prob?
Edited on Tue Sep-02-03 10:42 PM by TruthIsAll
Perhaps the problem should be rephrased to:

What is the probability (that 3 out of 30 races would have identical Repub winning votes) AND also that 2 of the three would be for legislative office?

Could it be 1 out of (1/2500)*1/567?
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Tue Sep-02-03 09:58 PM
Response to Reply #2
3. Perhaps the problem should be redefined as:
Edited on Tue Sep-02-03 10:57 PM by TruthIsAll

What is the JOINT probability (that 3 out of 30 races would have identical Repub winning votes) AND (that at least 2 of these three would be for legislative office)?

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