Counting in the Rajya Sabha elections in Gujarat did not go into the second round. Three of the four candidates were elected on the basis of first preference votes after two votes were invalidated. Had these votes not been invalidated, would Ahmed Patel of the Congress still have won? An examination.
Ahmed Patel of the Congress may have won the Rajya Sabha election in Gujarat even if the two invalidated votes cast in favour of the Bharatiya Janata Party had been allowed, given the peculiarities of the system of single transferable vote.
In fact, what gave Mr. Patel the edge was his unassailable lead over his BJP rival. If he had got one vote less, it was possible that the third BJP nominee would have sailed through, depending on how many second preference votes he gathered.
There were four candidates in the fray for three seats, resulting in the necessity for polling. As many as 176 votes were cast by Gujarat legislators, and two of them being declared invalid by the Election Commission. As a result, the election was over in the first round of counting itself, with BJP leaders Amit Shah and Smriti Irani polling 46 votes each, and Ahmed Patel 44. The losing BJP candidate Balwantsinh Rajput received 38 votes.
How was the result arrived at?
As 176 votes were cast, the qualifying number for a candidate to win on the basis of first preference votes would have been 45. But this came down to 44 because the total number of valid votes was reduced to 174. Therefore, Mr. Patel sailed through with the requisite votes in the first round itself.
Now, what would have been the scenario had the two votes been allowed to be counted?
The election would have gone into the second round, and the counting process in the system of proportional representation based on single transferable vote would have kicked in. This is how the system works.
The votes are given a value of 100 in the first round. The total votes cast, therefore, would be worth 17,600. To qualify, a candidate needs to get one point more than the quotient obtained by dividing the total value by the number of seats at stake plus one.
That is, 17600/4 = 4400 + 1= 4401. A candidate, thus, needs a value of 4,401 to win.
Going by the results, Mr. Shah and Ms. Irani received 4,600 points each and won comfortably. If Mr. Rajpat had got 40 votes instead of the final figure of 38, his votes would have been worth 4,000. Mr. Patel, with 4,400, would have been ahead of him, but a single point short of qualifying.
This would have taken the counting into the second round. In this round, the ‘surplus’ votes of the two winning candidates will have to be distributed to the remaining candidates. Mr. Shah and Ms. Irani each had a surplus of 199 votes (4,600 as against the required 4,401). How are these votes distributed?
The second or third preference votes, if any, found on the ballot papers in which they were marked as first choice, would be taken into account. Assuming that all the 92 ballots that marked Mr. Shah and Ms. Irani as their first preference contained a second preference, it is possible that these were in favour of Mr. Rajput. The surplus of one candidate (chosen by draw of lots, as the original votes of the first two candidates are the same) will be transferred first, and, if no one has reached the qualifying value yet, the second candidate’s surplus will also be transferred.
The value of a transferred vote, it must be pointed out, is not the same 100 given to an original first preference vote. It is calculated on a different formula. It is arrived at by dividing the surplus value by the number of ‘unexhausted ballot papers’. That is, if all 46 of Mr. Shah’s principal voters had a second preference for another candidate, each of that will have a value of 199/46 = 4.32. Under election rules, the remainder is to be ignored; so each vote has a value of 4. The value will be the same for the surplus transferred from Ms. Irani to Mr. Rajput (again assuming all 46 marked a second preference for Mr. Rajput).
If even one of them had marked a second preference for Mr. Patel, he would go through as his tally would go up from 4,400 to 4,404, as against the required 4,401.
However, this is not a likely scenario as all of Mr. Shah’s principal voters can be expected to have favoured Mr. Rajput as their second choice to boost his prospects and to defeat Mr. Patel.
So, when the first lot of 46 second preference votes are transferred to Mr. Rajput, he would receive 46 x 4 = 184 points. A similar lot drawn from the votes cast principally for Ms. Irani with Mr. Rajput as second choice, would also have the same value.
Therefore, Mr. Rajput gains 184 + 184 = 368 points. This will be added to his original vote value.
So his tally would be 4,000 + 368 = 4,368. But, his vote will still be short of Mr. Patel’s vote value of 4,400.
The scenario is explained in the table below:
In a single transferable vote, it is possible to have any number of rounds of counting, but in this case, it has to be ended here, as the ballots of the two qualified candidates have likely been exhausted. Mr. Patel would be ahead of Mr. Rajput, and with all ballots getting exhausted, he would have been declared elected.
Therefore, Mr. Rajput has to be eliminated from the contest, leaving Mr. Patel the winner of the third seat, with a tally of 4,400 against the losing candidate’s 4,368.
What is the highest possible surplus value the BJP could have created?
The answer is 396. This is why the four-vote or 400-point lead that Mr. Patel had in the original votes proved to be unassailable.
How is the figure of 396 arrived at?
Let us take a scenario in which only one second preference vote has gone to Mr. Rajput in the Amit Shah parcel of votes. Mr. Shah, remember, has a surplus value of 199. There is a crucial counting rule that says the value of a transferred vote remains the same if the total value of such transferred votes is equal to or less than the surplus.
This means that one vote could be worth 100, but two votes cannot be 200, which is higher than the surplus value of 199.
Therefore, if there is only one second preference vote in Mr. Shah’s parcel (the rest not marking any second preference at all), its value will be 100, and only this value will be transferred to the recipient of the second choice vote.
If there are two second preference votes in Mr. Shah’s parcel, its value will be calculated by dividing the surplus votes by the number of unexhausted votes. 199/2 = 99.5. Ignoring the remainder, the two transferred second preference votes will have a value of 99x2 = 198.
If it is assumed that only two of the 92 principal voters of both Mr. Shah and Mr. Irani have second preference votes in favour of Mr. Rajput, he will get 198 x 2 = 396 points.
If it is three, each vote will be valued at 199/3 = 66, and if it is four, it will be 199/4 = 49, and so on. In all these scenarios, the total value of second preference votes transferred from both candidates will be 396 or less.
This is why the 400-point lead of Mr. Ahmed Patel became unassailable. However, if he had obtained even one vote less than the 44 he got, that is, if he had a total value of 4,300, the transferred votes would have enabled Mr. Rajput to finish with more than 4,300 and thereby win the seat. Ultimately, after the invalidation of two votes, Mr. Patel had a clear six-vote or 600-point lead.
