Well, about one-third of those 24 players will survive and proceed to Round 2 with their weight unchanged (since they all will have equal weights in the Round 1, I will have to roll the dice for each pair, and one-third of the outcome will mean both players surviving). So, under this scenario, in Round 2 we will see 8 players with weight 50 g (survivors of the past episode), 8 players with weight 100 g (half of the remaining minnows from Round 1, who eliminated their competition), and one previously unmatched minnow with weight 50 g. So the unmatched fellow’s chances are actually pretty decent under this model. Does this make sense to you?
Welcome to the game! ;)
Alright that makes sense :D Thanks for reply