Thursday, April 25, 2019

How many EVMs should the EC check?

How many EVMs should be manually checked (i.e., VVPAT count compared with that of the machine) to have sufficiently high (>99%) confidence in EVMs?
Let's take an assembly election, say in Karnataka, which has around 5crore voters and 224 seats. This means that each constituency has ~2 lakh votes and given the EC recommendation of ~1400 people per booth (urban), around 130 booths per constituency.
Clearly, if we check all the EVMs in a constituency, we'll have 100% confidence in the outcome. But what is the minimum number of EVMs you need to check to be >99% confident?
Is it 50%, as CBN told Rajdeep Sardesai? Is it 90%? Or is it 99%? Well, the answer may surprise you.
But first, let's see how many EVMs need to be tampered in a close election. Let's assume two parties, and a vote split of 51-49. [Typical margins are way higher than this.] So, in our model constituency, this is a difference of 4000 votes. If the tampering party needs to overcome this margin without raising suspicion, the margin must be split across many polling booths. Given that each booth has 1500 votes and that voting percentages are typically around 60%, there are around 900 votes per booth. Let's say that to avoid raising suspicions, the number of tampered votes per booth won't exceed 10-20% of the overall vote (i.e., 100-150 votes). The tampering party therefore needs to tamper at least 4000 / 150 = ~ 27 booths. Which means, out of the 130 EVMs in the constituency, they have to tamper at least 27. So, we have ~80% good EVMs and ~20% bad ones per constituency.
OK, so let's say that we pick 5 machines (as per the SC order) for verification. Even if we find one bad EVM amongst the five, we can countermand the election in the constituency. Therefore, if all the five machines we select turn out to be good, we would have failed to detect the bad EVMs and the tamperers go scot-free. [Remember, there are 20% bad EVMs in the constituency]. What is the probability of this happening? Well, we have five selections to perform. In the first selection, the probability of selecting a good EVM = 0.8. On the second try, because we have removed this EVM, the probability = 0.79, and so on until the probability is 0.76 on the
5th selection. But what is the probability that all our selections are good? It is : 0.8 X 0.79 X 0.78 X 0.77 X 0.76 = ~0.288. So we have a 29% chance that our test will fail and a 71% chance that we'll detect a tampered EVM in a single constituency.
But remember, these tests are being done across the state! Even a single tampered EVM across the state will put the entire election into question. For simplicity let's assume that EVMs were tampered in 10 seats. So, what is the probability that our test will fail? It is 0.29^10, which is 0000042, which means our test will work 99.9995% of the time. And like the opposition if you assume that EVMs have been tampered in each constituency, the probability of our test failing is 0.29^224; a number that is closer to zero than even the mass of some elementary particles! [Of course, I'm making some simplifying assumptions, like the percentage of tampered machines being the same across constituencies.]
So one can see that manually verifying only 5 machines per constituency can give a near 100% confidence in our voting system. This even if we ignore the logistical difficulties and systemic safeguards that would make such large scale tampering near-impossible in the first place! Too bad simple mathematics does not find place in our national media narrative. 

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