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The Venezuelan Presidential Vote -- What is the Probability That It Could Have Been Stolen?

Written by CEPR

Friday, 26 April 2013 11:45

Opposition candidate Henrique Capriles is currently “boycotting” a second audit of the voting results for the April 14 presidential election, which the National Electoral Council has agreed to undertake.  Capriles claims that the election was stolen through fraud. In a CEPR press release we note that it is practically impossible to have obtained the results of the audit that took place after the polls closed on April 14, if the election were actually stolen through fraud.  When the polls closed, a random sample of 53 percent[i] of all the machines (20,825 out of 39,303) was chosen, and a manual tally was made of the paper receipts.  This “hot audit” was done on site, in the presence of the observers from both campaigns, as well as witnesses from the community.  There were no reports from witnesses or election officials on site of discrepancies between the machine totals and the hand count. The following is a calculation of the probability of auditing 20,825 machines and finding zero errors when there are actually 50 among all 39,303 (this means that there are 50 machines with errors among the ones that were not audited).  The assumption here is that there would have to be at least 50 bad machines -- i.e. where the machine count did not match the paper ballot – in order to reverse a margin of 272,000 votes. This assumption is of course understating the number of bad machines that would be necessary to reverse the result. The average machine has only about 360 votes, and the maximum was about 564. And here we are assuming the election is stolen by moving about 2700 votes per machine from Capriles to Maduro,  on 50 machines.  If more machines were bad, then the probability below gets even (vastly) smaller.  So the calculation below is actually a very high estimate of the probability of obtaining the April 14 audit results, if the election were stolen. The probability of auditing 20,825 machines and finding zero errors when there are 50 machines with errors  among all 39,303 machines  is p = (50 choose 0) x (39253 choose 20825) / (39303 choose 20825)  (50 choose 0) is 1, so p = (39253! / (20825! 18428!)) / (39303! / (20825! 18478!) = (39253! / 39303!) x (18478! / 18428!)  = (18478 x 18477 x 18476 ... x 18429) / (39303 x 39302 x 39301 ... x 39254)  = (18478 / 39303) x (18477 / 39302) x (18476 / 39301) ... x (18429 / 39254)  because there are 50 such products, each less than the first (18478 / 39303) p < (18478 / 39303)^50 = 4 x 10^-17 In other words, the probability of getting an audit result that was obtained on April 14, if there were enough machine errors to reverse the result, is far less than 1 in 25,000 trillion.  It is therefore practically impossible that a further audit of the remaining machines could reverse the result of the election. [i] Another 1 percent was audited the next day.