Re: minority disenfranchisement via machine?

From: Ron Crane <voting_at_lastland_dot_net>
Date: Wed Mar 08 2006 - 19:14:12 CST
Joseph Lorenzo Hall wrote:
That's really interesting... wouldn't you think the easiest way to do
this would be if the local election official (who also knows the lay
of the land, so to speak) knew beforehand which machines were more
likely to fail (e.g., the older ones or ones that have been doctored)
and directed certain machines to certain polling places?  That is, not
necessarily programmed them differently (although with the
AccuVote-[TS|OS] would seem the easiest to tamper with given deep
access and poor procedures). -Joe
  
Both crooked vendors and crooked officials can perpetrate this denial-of-service attack. A crooked vendor or official could ship "bad" machines to precincts favoring one party, and "good" machines to the other precincts. They might select "good" and "bad" machines by testing and choosing, by mishandling (e.g., overvolting) some machines, or they might instrument the software to make the machines fail at the desired rates.

I just sent the California SoS a brief analysis of this fraud along with my comments on the proposed certification of some new machines. Here's the analysis, which expands upon the one I posted to the OVC list earlier today.

------

Imagine, for example, that a crooked vendor instruments its machines to fail more frequently as the proportion of party A votes to party B votes increases. By "fail", I mean lock up so as to require servicing by a vendor's repairperson, and thereby become unusable for the remainder of the affected election. Imagine further that the probability of failure at a 50%/50% party A/party B ratio is 0.01% per ballot cast, and that it increases linearly to 0.2% per ballot cast at a 60%/40% party A/party B ratio. Now imagine two precincts, each of which has 4 machines and 500 voters who will come to the polls (= 125 voters/machine). In precinct 1, voters vote at a 50%/50% ratio, and in precinct 2, they vote at a 60%/40% ratio.


The probability that any one particular machine will fail in each precinct is:


P(precinct 1 failure) = 1 - (1 - 0.0001)^125 = 0.0124

P(precinct 2 failure) = 1 - (1 - 0.002)^125 = 0.221

The probability that exactly one machine will fail in each precinct is:

 

P(exactly 1 failure) =

P(a single machine fails) *

P(the 3 others don’t fail) *

the number of ways this can happen (number of combinations)

 

P(single precinct 1 failure) = 0.0124*(1-0.0124)^3 * 4 = 0.0478

P(single precinct 2 failure) = 0.221*(1-.0.221)^3 * 4 = 0.418

 

Assuming that each voter takes 5 minutes on average to vote, that the polls are open 14 hours, and that only a single machine fails, we can compute what that difference in failure rates does (on average) to the wait times at the two precincts. At 500 voters per precinct, voters will arrive at the polls at an average rate of 500/(14*60) =  0.595/minute. Since voting takes 5 minutes, each machine can handle 1/5 = 0.20 voters/minute. Assuming that both the voter arrival rates and the voting rates follow a Poisson distribution (the usual for problems of this nature), the average wait times are:

 

With 4 operating machines at all times                                    2.4 minutes

With 3 operating machines at the start of polling                         191 minutes

With 3 operating machines where one fails after 1/4 have voted            174 minutes

With 3 operating machines where one fails after half have voted           142 minutes

With 3 operating machines where one fails after 3/4ths have voted         86 minutes

 

So the difference between all the machines operating and only one failing is dramatic, and is 0.418/0.0478 = 8.7 times as likely to happen at the precinct under attack. Weighting the wait times for the probabilities of failure, and assuming failures occur after half the voters have voted, we get:

 

Precinct 1 average wait time (1-0.0478)*2.4 minutes + 0.0478*142 minutes       = 9 minutes

Precinct 2 average wait time (1-0.418)*2.4 minutes + 0.418*142 minutes          = 61 minutes

 

This difference in average wait times cannot help but dramatically lower the turnout in the precinct under attack, but the general public has no good way to determine whether a crooked vendor (or a crooked official) is mounting this attack. And the general public has no good way to determine how much the attack, if mounted, lowered the turnout in the affected precincts. Further, the voters who saw the line and turned away from the polls have no good legal grounds to contest the election.


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Received on Fri Mar 31 23:17:02 2006

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