In his letter regarding the Alternative Vote system, (Oxford Mail, May 2) John Saunders has the following: First phase: Candidate A = 4,500 votes; Candidate B = 4,000 votes; and Candidate C = 1,500 votes; so A wins with 45 per cent. This is okay so far.
He then states that in phase 2, 300 votes are not used and A receives 900 votes and B 300, giving Candidate A = 4,800 votes and Candidate B = 4,900 votes.
Now, I’m no mathematician but even I can see that A would have 5,400 votes to B’s 4,300 and would therefore have over the required 50 per cent to win.
Simon Jaggs, Bourne Close, Bicester
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