Voting geekery, -devel dropped. On Wed, May 21, 2003 at 12:57:29PM +0200, Matthias Urlichs wrote: > Nick Phillips wrote: > > If a winning option would be discarded due to quorum requirements, then > > I think the vote should probably be considered void. > That seems to be the best choice. If the option's discarded, it's not the winning option, by definition. Nullifying the entire vote, rather than choosing one of the other options is equivalent to continuing discussion, rather than agreeing on a compromise. Sometimes this is an appropriate thing to do, sometimes it's not. Continuing discussion is the same as choosing the default option (that's the entire point of the default option) - as described in the proposal, we allow it to win very rarely, in fact only when it's preferred to all other options on the table, which means we'll almost always get a resolution once we put things to a vote, which is, after all, the entire point. The only way for the scenario you suggest to happen  at all is for less than 2R votes to be submitted (or for there to be up to around 2R votes that vote for some other option, but don't express a preference between the potential winner and the default option). R is around 44 at the moment, so you have either less than 88 people participating, or really bizarre voting patterns. For comparison, we've never had that sort of bizarre voting pattern, and we've only once had less than 2R people vote . Cheers, aj  Proof: By hypothesis: W defeats D by w : w' (w = q-k; w' < w) X defeats D by x : x' (x = q+l; x' < x) W defeats X by m : n (m > n) So there are x votes of the form: X XWD XDW WXD x' votes of the form: D DWX DXW WDX w votes of the form: W WDX WXD XWD w' votes of the form: D DWX DXW XDW m votes of the form: W WDX WXD DWX n votes of the form: X XWD XDW DXW Since, x > w, X XWD XDW WXD > W WDX WXD XWD X XDW = W WDX + k X XDW XWD DXW = W WDX XWD DXW + k n = W WDX XWD DXW + k m = W WDX XWD DXW + k + l W WDX WXD DWX = W WDX XWD DXW + k + l WXD DWX = XWD DXW + k + l Since w < q, W WDX WXD XWD < q WXD XWD < q - (W WDX) Since w' < q, D DWX DXW XDW < q XDW < q - (D DWX DXW) Thus: X XWD XDW WXD = X (XDW) (WXD XWD) X XWD XDW WXD < X (q - D DWX DXW) (q - W WDX) D W WDX DWX DXW XWD XDW WXD < 2q But this is all possible votes, except for "X". How many X can we have? n < m X XWD XDW DXW < W WDX WXD DWX X < W WDX WXD DWX - (XWD XDW DXW) X < q-k + DWX - (XWD XWD XDW DXW) DWX is at most w' (which is less than w, which is less than q, so it's at most q-2). An example is thus: DWX q-2 WXD q-1 X 2q-4 W beats D q-1:q-2 X beats D 3q-5:q-2 W beats X 2q-3:2q-4 With 4q-7 votes expressing a preference amongst W, X and D. This is the maximum number of votes the can result in the scenario described above.  The Logo Swap vote: http://lists.debian.org/debian-vote/1999/debian-vote-199907/msg00074.html R = 68, according to vote.debian.org; 108 people voted, "Against" didn't make quorum, and didn't defeat the default option. -- Anthony Towns <firstname.lastname@example.org> <http://azure.humbug.org.au/~aj/> I don't speak for anyone save myself. GPG signed mail preferred. ``Dear Anthony Towns: [...] Congratulations -- you are now certified as a Red Hat Certified Engineer!''
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