From: Ron Crane <voting_at_lastland_dot_net>

Date: Wed Mar 08 2006 - 19:14:12 CST

Joseph Lorenzo Hall wrote:

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.

------

Date: Wed Mar 08 2006 - 19:14:12 CST

Joseph Lorenzo Hall wrote:

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.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

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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