> At the start, everyone has shaken zero hands, so you all start at an even number. Let’s say Anna and Byron take the initiative and shake each other’s hands. Now they are at one handshake each.
> Caitlin needs to join in, so she shakes Byron’s hand, giving her one handshake, an odd number. But now Byron has two handshakes, and so he’s back to an even number.
The first blunder of the game. Byron was already in a winning position and was under no obligation to shake a hand, but he has allowed Caitlin to capitalise on his stupidity.
The solution, when dealing with an odd number of folks, is to do a 'three-way' handshake. Some times you have to think outside the box.
Isn't the simplest solution that everybody shakes hand with Byron?
Everybody but Byron has one handshake and Byron has n-1, with n the number of people
While the title is entertaining this doesn’t seem to be a very good candidate for a game at a party.
Title is backwards. graph theory tells us about the party game.
Seems like a introduction to the 'Handshaking lemma'?
https://en.wikipedia.org/wiki/Handshaking_lemma