<div dir="auto">Hi Daniel,<div dir="auto"><br></div><div dir="auto">Though I'm sceptical of this, if we stick with the "vote on a machine but get a printout to cast" paradigm, I reckon we would have the machine do the count for us and could afford all sorts of complex rules. </div><div dir="auto"><br></div><div dir="auto">We only need to be able to audit the input ballots against the physical printouts, the actual count the system provides given the inputs can be very easily verified by anyone with a spreadsheet given the inputs are open sourced.</div><div dir="auto"><br></div><div dir="auto">Regards,</div><div dir="auto"><br></div><div dir="auto">Etjon</div></div><br><div class="gmail_quote gmail_quote_container"><div dir="ltr" class="gmail_attr">On Fri, 23 May 2025, 4:59 am Daniel Kirslis, <<a href="mailto:dankirslis@gmail.com">dankirslis@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">Hi Etjon,<div><br></div><div>This is an interesting question. I agree that the hand-countability of ballots, at least in the case of an audit, is an important practical feature of an election.</div><div><br></div><div>I wonder if the ballot design itself could be modified to suit Condorcet methods. So, you rank your candidates on the touch screen voting machine. Then, the voting machine prints out your ballot, as is the case now. However, rather than simply printing a piece of paper with your ranking, it prints out each pairwise preference separately. So, if your ranking was A > B > C > D, it would print out 6 ballots:</div><div><br></div><div>A>B</div><div>A>C</div><div>A>D</div><div>B>C</div><div>B>D</div><div>C>D</div><div><br></div><div>Ballots can then be sorted by type. That way, it is easy to tally the ballots into the Condorcet matrix, and any entry into the matrix is easy to double check. And, we can audit the count easily, as ballots should sum up to the total number of voters, i.e., (A>C + C>A + A=C) should equal the total number of voters, which should also equal (A>D + D>A + A=D), and so on. And, as is the case now, you would also have a computer count to check against.</div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Thu, May 22, 2025 at 6:41 AM Etjon Basha via Election-Methods <<a href="mailto:election-methods@lists.electorama.com" target="_blank" rel="noreferrer">election-methods@lists.electorama.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="auto">Good evening gentlemen,<div dir="auto"><br></div><div dir="auto">I've been pondering the above issue, and already consulted Gemini who disagrees with me on the practicality of pairwise matrices, so couldn't help a lot.</div><div dir="auto"><br></div><div dir="auto">I suspect that compiling pairwise matrices in the context of a hand counted election would be very time consuming, and quite prone to errors and challenges from all parties. </div><div dir="auto"><br></div><div dir="auto">Assuming we agree on this (which you might not) is there any practical Condorcet method can can be hand counted? </div><div dir="auto"><br></div><div dir="auto">I suspect Nanson is a reasonable candidate. Yes, it still requires log(candidates,2) counting rounds, and each of those rounds require sending a matrix of how many times each candidate was ranked in which position to a central location, so quite the bother indeed. </div><div dir="auto"><br></div><div dir="auto">Yet, I suspect this task can at least be completed within acceptable timeframes with an acceptable error rate by most volunteers.</div><div dir="auto"><br></div><div dir="auto">(Interestingly, Gemini considers Copeland easier to hand count than Nanson, which I disagree with)</div><div dir="auto"><br></div><div dir="auto">Are there any simpler methods I'm unaware off, despite any other shortcomings such a method might have?</div><div dir="auto"><br></div><div dir="auto">Best regards,</div><div dir="auto"><br></div><div dir="auto">Etjon</div><div dir="auto"><br></div></div>
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