[Closed] Probability

Are they STW forum members? If so.....No bloody chance whatsoever 😯

Are the votes cast completely randomly because obv in real life, that's not usually the case...

Are the votes cast completely randomly because obv in real life, that's not usually the case

Speak for yourself.

It's a real life situation and yes, I'd expect some options to be more attractive but on a completely random basis is what I'd like to know.

I'm kind of embarrassed that I don't even know where to start.

Edit: not helpful - sorry.

It's actually quite complex here. It's not straightforward permutations or combinations as choices by one do not effect choices by another. Furthermore, majority test changes depending on the ratio of choices to voters and whether voters or choices are odd or even in numbers.

actually, thinking about what I just wrote, it's probably easier to work out how often one can get a tied result...

Depends what you mean by majority - do you mean over 50% of votes cast? In which case it's 8 votes.

And if so, you'd need to hit a 1 in 7 chance 8 times to achieve it so the odds would be 1/7 to the power of 8 which is going to be a very small chance (very roughly .00002% chance)

Maybe 😉

My stock answer would be 50/50. You either will or you won't. Much like my odds for winning the Euromillions tonight.

8 votes the same for a majority?

I thought I had an answer using combinations, but it fails the simple check, as I get a probability of more than 1 - I think the problem is actually that the combinations aren't independent.

[quote=nemesis ]And if so, you'd need to hit a 1 in 7 chance 8 times to achieve it so the odds would be 1/7 to the power of 8 which is going to be a very small chance (very roughly .00002% chance)
Maybe

Not exactly - even the probability of 8 voters all voting the same is only (1/7)^7 as you don't care which candidate they vote for.

I'm thinking the best thing to do is work with a simpler problem by reducing the numbers, one which can be done numerically and see if I can get a formula from that...

7 candidates? FIFA?

probability of a majority vote depends on financial incentives, not a-level maths 😉

Rather more than 14 voters in FIFA aren't there?

Actually, it's not what I said - thats the odds for one particular choice getting a majority. For any choice getting a majority, it's 7 times more than my result (which is still a very small number)

But that's the probability of the first 8 voters all voting the same way, nemesis. If they don't then the probability of 7 of them voting the same way and the 9th voting the same way as those 7 is 6*8/7 times that. I think I have a way forwards...

So what you are really asking is what the chances of 8 or more out of 14 voters picking any one of the 7 options... That sounds complex, and small. Much easier with just one or two choices. And the chances of random assortment of views in any real world situation is small...

... Which is a way of saying I don't know how 😉

for your particular Qu:

a non-majority result (i.e. a tie) can occur (nCr, n!/(n – r)!r!)

21 times, if 2 choices get 7 votes,
21 times, if 2 choices get 6 votes,
21 times, if 2 choices get 5 votes,
35 times, if 3 choices get 4 votes
35 times, if 4 choices get 3 votes (opposite of above)
and 1 time if each choice gets 2 vote.

a total of 134 times

The problem now is I dont know how to calculate the total number of possibilities available. I think it's a v big number.

Come on Stoner, you can do it!

Aracer... Exactly... And the answer also has to include the chances of 9, 10, 11,12, 13 & 14 votes for each of the 7..,
A lot of permutations...

[quote=Stoner ]a non-majority result (i.e. a tie) can occur

Hang on, there's all sorts of other non-majority results - if one candidate gets 7 votes that's still a non-majority. I'm thinking you may be on the right lines though, it might be easier to work out the answer for a non-majority.

there's all sorts of other non-majority results

only if you're being awkward 😉

it's clear we're talking about a "clear winner" in a FPTP kind of voting.

Now, put your mind to solving the divisor of my sum.

[quote=Stoatsbrother ]Aracer... Exactly... And the answer also has to include the chances of 9, 10, 11,12, 13 & 14 votes for each of the 7..,
A lot of permutations...

Yeah, and I can't see any way to do it without going through all of them - I'm fairly sure the answer for 9 voters the probability of one candidate getting 8 votes is 55/(7^8) and can't see any logical way to extend that for more voters.

Yeah, and I can't see any way to do it without going through all of them

hence solving the reciprocal instead!

[quote=Stoner ]Now, put your mind to solving the divisor of my sum.

I thought about doing it, but realised it's almost as tough a problem - waiting for the OP to advise on what "majority" means before wasting time on it!

hence solving the reciprocal instead!

I think I already agreed that was a good approach, and working on it...

Easy way would be a monte-carlo simulation

Waiting to hear from my colleague who thought I was smart enough to solve this.

Maybe I'm overthinking the total number of voting options and it's just 14!-7! which is 87,178,286,160

Easy way would be a monte-carlo simulation

That's what I was thinking. Leave it running for a few hours and you'll get a number, but not a 'real' proof.

ok, the requirement is to get a winner, doesn't need to have more than 50% of the votes

good. So I can tell you that there are 134 voting combinations that do NOT get a winner. You can do the rest 🙂

(unless I'm right, and it's 1-(134/87,178,286,160) = 0.9999999985%)

ah, that's a different proposition as someone could win with only 3 votes.

That gives even more combinations which can win...
Can we agree that there are 7 to the 14 overall possible permutations?
An online calculator suggested to me that there were 45325 possible >50% combos...

Is the number of combinations not as simple as 7^14 / 14! ?

That may not be quite right, but maybe something similar...

Aracer - I think this is the same problem (for total voting combinations)

[i]how many ways there are to put n indistinguishable balls into k distinguishable bins.[/i]

https://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

but I dont understand the notation. You probably do.

[quote=Stoner ]for your particular Qu:
a non-majority result (i.e. a tie) can occur (nCr, n!/(n – r)!r!)
21 times, if 2 choices get 7 votes,
21 times, if 2 choices get 6 votes,
21 times, if 2 choices get 5 votes,
35 times, if 3 choices get 4 votes
35 times, if 4 choices get 3 votes (opposite of above)
and 1 time if each choice gets 2 vote.
a total of 134 times

What about 2 choices get 4 votes, the other 6 votes spread out, 2 choices get 3 votes the other votes spread out or 3 choices get 3 votes...

[s]you dont have a majority winner, you have 2 candidates in a dead heat[/s]

EDIT: I see what you mean. There's a few more combinations. but not many.

We need to use the same formula as for the total No. of combinations. But I dont understand binomial coefficients or how to calculate from them.

There will be an algebraic solution (as per @Stoners posts) but I also agree a monte-Carlo simulation will give a pretty reasonable answer within a few minutes on a PC as its not a complex multi-stage problem. I'd also say you are asking a theoretical question as in most real world situations people's selections from the 7 choices will not be uniformly distributed, eg think UK general election.

[quote=Stoner ]but I dont understand the notation. You probably do.

I thought you were the one who did probability as part of your job (well I do very occasionally, though one of the first things I got paid to do was monte-carlo!)? Though that looks like set theory - you're probably right, but it's not something I've done for an awfully long time.

Though I'm currently sidetracked because I think I have a solution to what I thought the problem was - had already coded a brute force method (though that won't solve the original problem this year, just ones with less voters) before the clarification, which has clarified my thoughts.

edit: actually brute force will work, and also I think for the actual problem as it's working quicker than I expected (time was a fag packet calc and clearly out by an order of magnitude or two), though it might be an hour or two

Rarely do stoch stuff. I used to do a bit of monte carlo but it's [s]wasted on my clients[/s] over complicated for most purposes in my industry.

As you say, I think the notation is more set theory and way beyond my knowledge.

MC wouldn't be great as the probability of a majority is small (so you'd need a lot of samples to get close to the right answer). I could answer the question for you but it's too boring, unless someone's paying.

Well I think I understand the stars and bars thing. We have 14 stars for the 14 votes and 6 bars to separate the 7 candidates (bins). There are 15 possible positions for the bars, as they can go on either side of any star (ie on each end of the row of stars, or between any two stars). Hence you get the binomial:

[code](15)
( 6)[/code]

which is saying you have 6 things to put in 15 positions, and you don't care about the order of the things, so it's a standard combination thing.

That evaluates as: 15! / ((15-6)! * 6!) = 5005 different voting combinations, which is less than you might have thought.

Stoner, could you add in the number of times those other options occur? You seem to be on top of that, and my brain is a bit frazzled with working out what the stars and bars thing means (though it's actually a lot simpler than it seems). We'll then have an answer...

Definitely doable from 5005 possibilities by brute force, provided you can manage an efficient way of enumerating those possibilities.

Isn't it 7 times 1/(7*7*7*7*7*7*7*7) ?
Once upon a time I could do maths, now it's all gone. 🙁

...oh, but 5005 seemed far too small, and I realised why, and why the stars and bars thing is useless. That only counts one for each voting possibility, which doesn't work for doing the probabilities as there are different probabilities of each voting possibility (clearly there is only one way for a single candidate to get 14 votes, but lots of different ways for each candidate to get 2 votes). Unfortunately Stoner's options occurrence suffers from a similar problem, so back to the drawing board...

...so the actual number of possibilities is 7^14 and the question remains; out of those possibilities how many times do you get 2 candidates with 7 votes each, etc.

Though actually that one's fairly easy: 7 * 6 * (14! / (7! * 7!)) = 144144

just 7 other options to go

...by jove I think I've got this:

possibilities for 2 candidates with 6 votes each: 7 * 6 * (14! / (6! * 8!)) * (8! / (6! * 2!)) = 3531528