tag:blogger.com,1999:blog-264226589944705290.post6335600020794916929..comments2024-09-25T08:51:01.854-07:00Comments on God Plays Dice: 51,199,463,116,367: A fuller solution to Tuesday's electoral vote problemMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-264226589944705290.post-71621990795050822882008-06-16T11:21:00.000-07:002008-06-16T11:21:00.000-07:00The real issue is not how well Obama or McCain mig...The real issue is not how well Obama or McCain might do in the closely divided battleground states, but that we shouldn't have battleground states and spectator states in the first place. Every vote in every state should be politically relevant in a presidential election. And, every vote should be equal. We should have a national popular vote for President in which the White House goes to the candidate who gets the most popular votes in all 50 states. <BR/><BR/>The National Popular Vote bill would guarantee the Presidency to the candidate who receives the most popular votes in all 50 states (and DC). The bill would take effect only when enacted, in identical form, by states possessing a majority of the electoral vote -- that is, enough electoral votes to elect a President (270 of 538). When the bill comes into effect, all the electoral votes from those states would be awarded to the presidential candidate who receives the most popular votes in all 50 states (and DC). <BR/><BR/>The major shortcoming of the current system of electing the President is that presidential candidates have no reason to poll, visit, advertise, organize, campaign, or worry about the voter concerns in states where they are safely ahead or hopelessly behind. The reason for this is the winner-take-all rule which awards all of a state's electoral votes to the candidate who gets the most votes in each separate state. Because of this rule, candidates concentrate their attention on a handful of closely divided "battleground" states. Two-thirds of the visits and money are focused in just six states; 88% on 9 states, and 99% of the money goes to just 16 states. Two-thirds of the states and people are merely spectators to the presidential election.<BR/><BR/>Another shortcoming of the current system is that a candidate can win the Presidency without winning the most popular votes nationwide.<BR/><BR/>The National Popular Vote bill has been approved by 18 legislative chambers (one house in Colorado, Arkansas, Maine, North Carolina, Rhode Island, and Washington, and two houses in Maryland, Illinois, Hawaii, California, and Vermont). It has been enacted into law in Hawaii, Illinois, New Jersey, and Maryland. These states have 50 (19%) of the 270 electoral votes needed to bring this legislation into effect. <BR/><BR/>See http://www.NationalPopularVote.com <BR/><BR/>susanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-88581519734646168242008-06-13T14:57:00.000-07:002008-06-13T14:57:00.000-07:00Jeff,you're right -- that would be a better way to...Jeff,<BR/><BR/>you're right -- that would be a better way to do the calculation. The solution I present here basically reproduces the way I got to the answer the first time, so it's not surprising that there are some inefficiencies.Michael Lugohttps://www.blogger.com/profile/15671307315028242949noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-15358726843049631712008-06-13T14:46:00.000-07:002008-06-13T14:46:00.000-07:00Never mind, the number is right. I misunderstood w...Never mind, the number is right. I misunderstood what you meant when you said you repeated the process with 4, 5, 6, etc as the minimum. Obviously for 4 you added in the 273 coefficient as well as the 270, 271 and 272 ones.<BR/><BR/>That being said, I think you made the process a bit harder than you needed to. Instead of doing the subtraction, just add the 270, 271, 272 coefficients for the full generating polynomial, then the 273 coefficient for the 4+ EV polynomial, the 274 coefficient for the 5+ EV polynomial, and so on.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-70143770339059590862008-06-13T13:47:00.000-07:002008-06-13T13:47:00.000-07:00Ok, now I do actually have a disagreement with the...Ok, now I do actually have a disagreement with the result. What you're computing is not the number of minimal winning sets. <BR/><BR/>For the 270, 271, or 272 EV totals, it doesn't matter what the smallest state is, because if you remove any of those you lose. Instead you want to add up:<BR/><BR/>1) All ways to get 270<BR/>2) All ways to get 271<BR/>3) All ways to get 272<BR/>4) All ways to get 273 without any 3 EV states<BR/>5) All ways to get 274 without any 3 or 4 EV states<BR/>etc...<BR/><BR/>The generating function approach works very nicely for this as well, but the total is going to be different.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-43082295794017417632008-06-13T12:56:00.000-07:002008-06-13T12:56:00.000-07:00Jeff,the number of EVs needed to win is 270, and n...Jeff,<BR/><BR/>the number of EVs needed to win is 270, and no state has less than 3 EV. So 270, 271, and 272 are all essentially the same, in that if a candidate gets any of those numbers they need every state they got.Michael Lugohttps://www.blogger.com/profile/15671307315028242949noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-56367420064524004962008-06-13T12:39:00.000-07:002008-06-13T12:39:00.000-07:00Most of this looks good, but I'm a little confused...Most of this looks good, but I'm a little confused by something. I get why you want to remove the solutions for 271 and 272 with no 3-EV states because you could then swap a state for a lower valued one. But why are you removing those solutions for the 270 total? If you're exactly at 270, pretty much by definition there are no excess votes no matter what the combination of states is.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-51361781365726104342008-06-13T06:04:00.000-07:002008-06-13T06:04:00.000-07:00Thanks for the post. I've written up the dynamic ...Thanks for the post. I've written up the dynamic programming approach <A HREF="http://research.swtch.com/2008/06/electoral-programming.html" REL="nofollow">here</A>.rschttps://www.blogger.com/profile/09576271159839887762noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-47374749147936543602008-06-13T02:45:00.000-07:002008-06-13T02:45:00.000-07:00Congratulations to all! I would have guessed a muc...Congratulations to all! I would have guessed a much smaller number, until the comment on the size of 2^52.CarlBrannenhttps://www.blogger.com/profile/17180079098492232258noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-90359099971420336522008-06-12T22:41:00.000-07:002008-06-12T22:41:00.000-07:00I'm the anonymous author of the Perl dynamic progr...I'm the anonymous author of the Perl dynamic programming solution posted as a comment to your earlier post. Just wanted to add two points: I think it's surprising how quick a good dynamic solution can run. The Perl runs nearly instantaneously on any modern computer. Second, it is critical to memoize (which is, of course, sound dynamic programming practice). I don't think everyone who attempted a solution remembered to do that, even if they were otherwise on the right track in terms of how to divide and conquer the problem.<BR/><BR/>Oh, and, my first thought about your solutoin, as a programmer, was exactly what you expected: that the polynomial calculation was basically just a framework for employing a dynamic programming tool provided by your computer algebra software of choice. But seriously, it was interesting to see how you approached the problem, and this has been one of my personal favorites of your posts.Anonymousnoreply@blogger.com