<TEI xmlns="http://www.tei-c.org/ns/1.0" xmlns:py="http://codespeak.net/lxml/objectify/pytype" py:pytype="TREE"><text><body><div type="translation" n="urn:cts:greekLit:tlg0062.tlg063.perseus-eng3" xml:lang="eng"><div type="textpart" subtype="section" xml:base="urn:cts:greekLit:tlg0062.tlg063.perseus-eng3" n="41"><sp><speaker>LYCINUS</speaker><p>Stop there. This is just what I wanted. Now, suppose they are nine in number and they have all drawn and are holding their lots. You go round (I want to make you a National Judge instead of a spectator) and inspect the letters. I fancy you will not learn in advance who has been given a bye, unless you go to every one of them and pair them.</p></sp><sp><speaker>HERMOTIMUS</speaker><p>What do you mean by this, Lycinus?</p></sp><pb n="v.6.p.341"/><sp><speaker>LYCINUS</speaker><p>It is impossible immediately to find that letter which gives you the bye, or perhaps you could find the letter, but you will certainly not know if it is that one, for there is no prior declaration of kappa or mu or iota as the letter which chooses the bye. When you find alpha, you look for the competitor who has the other alpha, and, when you find him, you have already paired them. Then again, when you come on beta, you look for the other beta, the counterpart of the one you have found, and so with all of them, until you are left with the competitor who has the only letter that has no counterpart.</p></sp></div></div></body></text></TEI>