tag:blogger.com,1999:blog-32902056.post6074702554876836972..comments2024-03-19T00:10:09.383-07:00Comments on Paul Goldberg: UK election gamePaul Goldberghttp://www.blogger.com/profile/10952445127830395305noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-32902056.post-2384440177546463542010-05-15T10:35:18.960-07:002010-05-15T10:35:18.960-07:00David, thanks, I just took a look at the wikipedia...David, thanks, I just took a look at <a href="http://en.wikipedia.org/wiki/Blotto_games" rel="nofollow">the wikipedia page</a>, they look similar; I will now study them further!Paul Goldberghttps://www.blogger.com/profile/10952445127830395305noreply@blogger.comtag:blogger.com,1999:blog-32902056.post-66292585811669279642010-05-15T00:55:12.658-07:002010-05-15T00:55:12.658-07:00Sounds quite related to Colonel Blotto Games. (I h...Sounds quite related to Colonel Blotto Games. (I hope I got the spelling right there.) Though if I remember right, there, you don't care about how many constituencies you win, just winning more than your opponents.David Kempenoreply@blogger.comtag:blogger.com,1999:blog-32902056.post-26194811826213924612010-05-14T06:03:59.196-07:002010-05-14T06:03:59.196-07:00To give the game away, so to speak, a party should...To give the game away, so to speak, a party should aim for the following: For each constituency, the amount spent on it should be uniformly distributed in the range [0,2<i>M</i>/<i>N</i>]. If this can be achieved, then for the other party, the probability of winning a given constituency is directly proportional to the amount it spends there, up to a limit of 2<i>M</i>/<i>N</i>, beyond which there is no incentive to spend more. One constituency's gain becomes another's loss, and you may as well use any payments that are all below 2<i>M</i>/<i>N</i>.<br /><br />To achieve this for the 3-constituency case, let's put <i>M</i>=1 without loss of generality; choose a random number <i>x</i> uniformly in [0,1/3] and spend it on constituency 1. Choose one of the other 2 constituencies at random and spend 2/3-<i>x</i>/2 on it, and spend the rest on the other one.<br /><br />Larger values of <i>N</i> are left as an exercise.Paul Goldberghttps://www.blogger.com/profile/10952445127830395305noreply@blogger.comtag:blogger.com,1999:blog-32902056.post-73223025950436565972010-05-13T14:39:45.498-07:002010-05-13T14:39:45.498-07:00In reply to comment 3: I agree, that breaks the su...In reply to comment 3: I agree, that breaks the suggestion of comment 1. But the idea of comment 1 is roughly correct, at least it becomes a good approximate Nash equilibrium for large <i>N</i> (error proportional to 1/<i>N</i>). Tomorrow I will give a 3-constituency solution...<br /><br />To Ashley, thanks for the link, I spent some time reading the survey this morning and will continue! I would have guessed that the sensitivity to noise was about that same.Paul Goldberghttps://www.blogger.com/profile/10952445127830395305noreply@blogger.comtag:blogger.com,1999:blog-32902056.post-56896482395351200742010-05-13T11:53:42.487-07:002010-05-13T11:53:42.487-07:00@Anonymous
For M=N=3 I can beat your random permu...@Anonymous<br /><br />For M=N=3 I can beat your random permutation of {0,1,2} by taking a random permutation of {1.49, 1.49, 0.02} (I always beat your 0, always lose to your 2, and beat your 1 two-thirds of the time)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-32902056.post-59764020184992648522010-05-13T00:25:38.883-07:002010-05-13T00:25:38.883-07:00Another fun and topical game to play is to work ou...Another fun and topical game to play is to work out, using the techniques of Section 3.4 of <a href="http://www.cs.cmu.edu/~odonnell/papers/analysis-survey.pdf" rel="nofollow">this survey</a> or similar, how sensitive the UK election is to noise (i.e. mistakes in tallying ballots). If I haven't made a mistake, for small levels of noise, our current constituency system is about 20x more likely to give the "wrong" answer than straight majority voting...Unknownhttps://www.blogger.com/profile/00984900489341123357noreply@blogger.comtag:blogger.com,1999:blog-32902056.post-58333656131908555952010-05-12T16:41:25.588-07:002010-05-12T16:41:25.588-07:00My rough guess
choose a random permutation of the...My rough guess<br /><br />choose a random permutation of the constituencies and allocate 2(i-1)M/((N(N-1)) to the i-th constituency.<br /><br />If both parties follow this, each should win (N-1)/2 constituencies in expectation. Additionally, in expectation, one constituency will be a tie.Anonymousnoreply@blogger.com