Jump to content

A Maths Problem!


Chinahand
 Share

Recommended Posts

surely with 7 voters and six possible choices you are going to get 1 definate winner - which gives them 1/7 chance x 6 gives 42 combinations!

 

Close - there's one winner, so 6 options.

Link to comment
Share on other sites

Problem is more complex than that?

 

if A gets 2 votes and B gets 2 votes and C, D, and E get 1 vote each - then that is surely NOT a 'Result' but an 'Outcome'. Another election would have to occur?

 

To get a 'Result' you'd have to take out all of the 'Outcomes' that do not form an electoral 'Result'?

 

So IMO the only possible number of results can be 1. i.e. A winner of the election. Though there will be lots of possible outcomes!

 

Is this a trick question?

Link to comment
Share on other sites

It's not a trick question - I suppose I am after how many outcomes there are.

 

ie

1. All 7 vote for A;

2. 6 vote for A 1 votes for B

3. 6 vote for A 1 votes for C

etc

n-? 2 vote for A, 2 vote for B, 2 vote for C, 1 votes for D

etc

etc until

n. All 7 vote for F

 

How big is n. when it includes every possible permutation.

 

As I've said this is an analogy for what I'm doing - it's even more complex - look up Condorcet's voting paradox for a flavour of the details!

 

Link to comment
Share on other sites

Is this a trick question?

 

Sort of - READ THE QUESTION. That's what you have to do, in the end one of six wins (unless there's a Mexican standoff). I bet it's out there via Google somewhere.

Link to comment
Share on other sites

To give some more details each of the 6 votes is in fact how you'd rank three candidates

 

ie ABC, ACB, BAC, BCA, CAB, CBA

 

The candidates then stand in run off elections - ie AvB, BvC, CvA

 

ie if a voter had the preference ABC they'd vote for A in the first run off, B in the second and A in the third.

 

Even though each person ranks the candidates rationally you can get results where you get a paradox - A will beat B in the first vote, B will beat C, but C will beat A. The results are non-transitive.

 

For example if the first voter had the preference ABC, the second ABC the third ABC, the fourth CBA, the fifth BCA, the sixth CAB and the seventh CBA then A would beat B 4 votes to 3, B would beat C 4 votes to 3, and C would beat A 4 votes to 3.

 

What I'd really like to know is what percentage of elections result in this paradox - I think it's around 3%, but to do that calc I've assumed voting order is relevent 10500 out of 279936 elections which isn't true!

Edited by Chinahand
Link to comment
Share on other sites

This is a tricky problem, and neither raising 6 to the power 7 nor calculating 6 factorial, or even finding permutations and combinations will help you.

 

What you need is to calculate the fixed set for each member of the permutation group, which unfortunately has order 720. From this you can use the counting theorem (it's a bit of group theory) to find the result I think you're after, if I understand the question corectly.

 

I suspect a suitably programmed computer would be required. Whereas you can use the counting theorem to calculate how many different ways there are to colour in a dodecahedron with a certain number of colours, for example, this is because its symmetry group is simpler. For the general permutation group S6 I can't see an easy way to do it.

 

VinnieK where are you?

Link to comment
Share on other sites

sorry try again ! there is 1/7 chances of all 6 voters which gives 42

there are 2/7 chances of getting 2 voters x 6 gives 24

there are 3/7 chances of getting 3 votes x 6 combinations gi es you 10.5

4/7 gives 8.5

5/7 gives 7

6/7 gives 6 total 60?

 

 

are we getting warmer?

Link to comment
Share on other sites

WTF the point is it is perfectly possible to have a vote where it isn't a tie, but no-one wins - the choice of winner is arbitary and paradoxical - A beats B, B beats C, C beats A. You can arrange those results to make anyone (or noone) the winner!

Link to comment
Share on other sites

If it's a vote there's a winner eventually, the fact that how a tie is decided has not been described doesn't matter.

 

One of six candidates wins this first past the post election.

Link to comment
Share on other sites

I've got a maths problem at work which has me stumped. I was really poor at permutations and combinations at school and since then haven't found a nice guide to improve my skills (any recommedations gratefully recieved!)...

 

HeliX is right; you're just trying to calculate a combination here. An equivalent problem would be: I want to construct 'words' of r letters which I choose from a set of size n, where neither repetition nor order matter. The number of all such combinations is given by

 

(n+r-1)! / (r! (n-1)!)

 

So for n=2 and r=3 you would have 4!/ (3!) =4.

 

That this is correct can be checked by the following list:

 

AAA (three votes for A)

------

ABA

AAB

BAA (one vote for B, two for A)

------

BBA

BAB

ABB (two votes for B, one for A)

------

BBB (three votes for B)

 

so there are indeed four possible 'outcomes' (in the sense of post #19). Actually, from this example it's easier to see the whole 'word' analogy: every word of length 3 consisting of at most 2 letters can be expressed as a vote and vice versa, hence they're (for our purposes) the same thing.

 

Now, in your example n is the number of candidates and r is the number of voters, so you get 12!/(7!5!) = 792 different combinations.

 

Is this what you were looking for? If, on the other hand, you're asking for how many ways can a given candidate win the election, you have to get a little (though not massively) more creative.

 

EDITED: Corrected a mix up with r's and n's.

 

EDITED AGAIN TO ADD: The 'just' up there looks a wee bit pompous, but I just mean that the problem isn't quite as difficult as the question seems at first. The actual physical problem is very tricksy at first glance, and I certainly got misled at first into thinking that it demanded a much more complicated answer.

Edited by VinnieK
  • Like 1
Link to comment
Share on other sites

24 squared - because a drake cannot lay eggs, survivors don't get buried and there is no steam from an electric train that left the station at noon - four hours after the train which was 2 hour late heading South.

  • Like 1
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...