Re: another use for idea futures
[EMAIL PROTECTED] (Wei Dai) writes: >On Tue, Oct 19, 2004 at 11:32:16AM -0700, Peter C. McCluskey wrote: >> What I really want is something which would quantify the probability my >> vote will affect what interest groups future candidates pander to. I suspect >> this is higher than the chance of affecting the identity of the winner. > >The probability of that is zero, because election results are not >broken down by interest groups. (Am I missing something?) I'd think >that politicans instead decide which groups to pander to based on their >own polling. They can't afford many polls, and election results provide information about voter desires. For example, if I vote for a typical Democrat, that's weak evidence that I want trial lawyers to get more business, whereas if I vote Libertarian that's stronger evidence against that. -- -- Peter McCluskey | To say that President George W. Bush has been www.bayesianinvestor.com | spending money like a drunken sailor is an insult to | drunken sailors. - Nick Gillespie, Reason Magazine
Re: another use for idea futures (fwd)
> On Wed, Oct 20, 2004 at 02:13:23AM -0400, Robert A. Book wrote: > > I think what you want is the Banzhaf Power Index, developed by Banzhaf > > (surprise!) in the 1960s. I forwarded your post to a friend of mine > > who's done some work on this, and discovered he's giving a talk on > > this very topic on Friday at the GWU math department. > > > > His summary, with a link to a more detailed web page, is below. > > I read his web page quickly, but did not find it particularly relevant. > The Banzhaf Power Index is apparently about figuring out how much power > each voter has in a block voting system, where everyone is not equal in > the sense that your vote has a different probability of influencing the > election depending on where you live. But he starts off by assuming that > every voter has an independent .5 probability of voting for each > candidate. That makes the analysis useless for the purpose of computing > the expected utility of voting, because it ignores all of the relevant > information that we actually have about the likely outcome of the > election, such as the IEM market data. > Funny you should mention that -- I thought of the same thing after I posted Mark's message to the list, and I think I have a solution. The Banzhaf model could be tweaked to take this into account by weighting each coalition by (p^i)*((1-p)^(n-i) where p is the probability of a given voter being for candidate #1, n being the number of voters, and i being the number voting for candidate #1 in the particular coalition. This is the binonmal distribution with parameters (n,p), which for large n (and p not too close to 0 or 1) (or more precisely, when np(1-p)>5) can be approximated very well by a normal distribution with mean np and variance np(1-p). The result should be that the power of an individual vote should drop as p gets farther from 0.5, in either direction. That is, an individual vote in Nevada is more likely to be decisive than one in Utah even though both have 5 electoral votes -- because in Utah Bush is ahead by 37 points (in some poll anyway) and in Nevada Kerry is ahead by 1 point (again, in some poll). So a vote in Utah will not be decisive at all, but a vote in Nevada might be. I would suspect that a vote in Florida (25 electoral votes, 1 point difference) is more powerful than a vote in California (55 electoral votes, Kerry leading by 8 points). I think that in real elections, a model that takes current polls into account would be more useful -- it would tell candidates where to campaign. They should campaign where the poll-adjusted Banzhaf index is higher. Who knows, maybe they have figured this out already --Robert
Re: another use for idea futures (fwd)
On Wed, Oct 20, 2004 at 02:13:23AM -0400, Robert A. Book wrote: > I think what you want is the Banzhaf Power Index, developed by Banzhaf > (surprise!) in the 1960s. I forwarded your post to a friend of mine > who's done some work on this, and discovered he's giving a talk on > this very topic on Friday at the GWU math department. > > His summary, with a link to a more detailed web page, is below. I read his web page quickly, but did not find it particularly relevant. The Banzhaf Power Index is apparently about figuring out how much power each voter has in a block voting system, where everyone is not equal in the sense that your vote has a different probability of influencing the election depending on where you live. But he starts off by assuming that every voter has an independent .5 probability of voting for each candidate. That makes the analysis useless for the purpose of computing the expected utility of voting, because it ignores all of the relevant information that we actually have about the likely outcome of the election, such as the IEM market data.
Re: another use for idea futures (fwd)
Wei Dai writes: > I think I finally solved a problem that's been bugging me since high > school: how do you actually compute the probability that your vote > will be decisive, in other words, the probability that without your > vote the election will be a tie or your side will lose by one vote? I think what you want is the Banzhaf Power Index, developed by Banzhaf (surprise!) in the 1960s. I forwarded your post to a friend of mine who's done some work on this, and discovered he's giving a talk on this very topic on Friday at the GWU math department. His summary, with a link to a more detailed web page, is below. --Robert Book [EMAIL PROTECTED] - Forwarded message from Mark A. Livingston - Date: Tue, 19 Oct 2004 18:14:31 -0400 (EDT) From: Mark A. Livingston To: Robert A. Book Subject: Re: another use for idea futures (fwd) Robert, [ Feel free to post most of this if you like. The web page is already public, so you are welcome to post its location. ] This is exactly the stuff I did in 1997 or so with the Banzhaf Power Index. And the same stuff on which I am giving a talk at the GWU math department on Friday. No, really. I am told that Prof. Banzhaf will be unable to attend, but he *is* currently on the GW Law School faculty. My web page is still at UNC <http://www.cs.unc.edu/~livingst/Banzhaf/>. I haven't updated it much since 2000. I did not read the Slate articles, but they wouldn't be the first to get it wrong, either. Any Google search will quickly turn up badly mangled attempts at similar calculations, and I note an erroneous ca. 1995 Scientific American article on my web page. As for a few factual responses, it looks like he sets up more or less the right problem. Basically, if you want the probability that your vote will be decisive, then you want the probability that all other votes (again, for now ignoring the block voting) will be split evenly. So if there are N + 1 voters (for simplicity, we can assume N is even without loss of generality), then you want N-choose-(0.5*N) divided by (2*2^N). The combination is the number of ways in which the other voters can split 50-50, and the exponential is the total number is ways they can vote (times two for your own choice). The combination is approximated by Stirling's formula, which swamps the need for (i.e. approximation requiring) N to be even. All this is what Banzhaf and others did in the 1960s. (Sorry, the posted solution is neither novel, nor did it begin in 1975.) Banzhaf published in various law journals; Shapley and Shubik in various RAND Corp. memos. (There have been numerous articles in various game theory journals since.) The Shapley-Shubik method is slightly different, but uses similar principles. Not to offend the original poster, but to a mathematician, the solution is relatively obvious. But then I aced my probability course (and lived to tell). And all the other engineers taking it with me were barely passing. So he should be proud for figuring it out on his own anyway. All the above is on my web page, including citations for the various references. But wait, there's more! :-) Banzhaf (and the others) went further and incoporated the block structure. Banzhaf's method not only computes probabilities that any given individual within a block (i.e. a state or DC) would cast the one deciding vote within that state, but also the probability that a state would be "critical" to the winner of the Electoral College. That is, that without that state, the candidate would not have won the election. (e.g. Florida was critical to Bush in 2004; had he lost in Florida, Gore wins the Electoral College.) Shapley-Shubik uses a different definition of "critical," but that's not material at this level. Again, it is on my web page. Banzhaf actually took a shortcut. I have notes on this, but they are not yet incorporated on my web page. Maybe I will get to that soon. L'hit, Mark - End of forwarded message from Mark A. Livingston -
Re: another use for idea futures
On Tue, Oct 19, 2004 at 11:32:16AM -0700, Peter C. McCluskey wrote: > I think it's harder than this to adequately model p(x), because it acts > differently in close races because the incentive for the losing side to > cheat is highest when it's most likely to change the result. Yes, to be more realistic, we need to compute the probability that my vote discourages the non-preferred candidate from cheating, or encourages the preferred candidate to cheat. I think it can be argued that after taking this into account, the probability of my vote deciding the outcome is still close to p(0), where p is the probability function for the actual vote without cheating (presumably with a normal distribution), not the final certified result. > What I really want is something which would quantify the probability my > vote will affect what interest groups future candidates pander to. I suspect > this is higher than the chance of affecting the identity of the winner. The probability of that is zero, because election results are not broken down by interest groups. (Am I missing something?) I'd think that politicans instead decide which groups to pander to based on their own polling.
Re: another use for idea futures
[EMAIL PROTECTED] (Wei Dai) writes: >To handle the case when it's not good enough to assume that p(x) is flat >in the range -.02n to .02n, for example when one candidate has a >clear lead, we can instead assume that p(x) has a normal distribution, and >find the normal distribution that best fits the 4 data points offered by >IEM. I think it's harder than this to adequately model p(x), because it acts differently in close races because the incentive for the losing side to cheat is highest when it's most likely to change the result. Another relevant contract is the Tradesports contract on "Presidential Election is Certified on or before Dec 13 2004" which is trading around 87 to 89%. What I really want is something which would quantify the probability my vote will affect what interest groups future candidates pander to. I suspect this is higher than the chance of affecting the identity of the winner. -- -- Peter McCluskey | To say that President George W. Bush has been www.bayesianinvestor.com | spending money like a drunken sailor is an insult to | drunken sailors. - Nick Gillespie, Reason Magazine
another use for idea futures
I think I finally solved a problem that's been bugging me since high school: how do you actually compute the probability that your vote will be decisive, in other words, the probability that without your vote the election will be a tie or your side will lose by one vote? (Well, solved if we ignore the electoral college and assume that only the popular vote matters.) This number is of course needed to compute an expected utility of voting. We all know it's small, but how small? Slate magazine published two recent attempts at this calculation: http://www.slate.com/id/2107240/ by Stephen E. Landsburg and http://www.slate.com/id/2108029/ by Jordan Ellenberg. Both attempts are very much flawed. The second article says what's wrong with the first one, and it should be pretty obvious what the second article's own problem is. (And although the two authors don't mention it, both of their approaches can be found in the academic literature, starting in 1975. See http://www.independent.org/pdf/tir/tir_06_4_bohanon.pdf for the references. I think these professors need to be more careful about giving proper attribution when writing popular articles.) Here's my approach, which hopefully is novel. Let n be the number of people who will vote for either Bush or Kerry, and assume that it's even. Let p(x) be the probability that Bush receives exactly n/2 + x votes. We want to compute p(0). It's well known that the the Iowa Electronic Markets runs a futures market (with real money) on US presidential elections, but perhaps it's less well known that it has separate bets on whether each candidate will win over 52% of the two-party popular vote. This gives us s = sum of p(x) for x > .02n and x < -.02n. Now if we assume that p(x) is more or less flat in the range -.02n <= x <= .02n, then p(0) must be approximately (1-s) / (.04n). Using s = .488 (IEM averages on 10/16), and n = 105586274 (turnout for the 2000 election), p(0) is about 1.2e-7. To handle the case when it's not good enough to assume that p(x) is flat in the range -.02n to .02n, for example when one candidate has a clear lead, we can instead assume that p(x) has a normal distribution, and find the normal distribution that best fits the 4 data points offered by IEM.
