The expectation might be the same, but the variance certainly isn't. Definitely a new number each round please.
Disclaimer: I'm not taking part in the tournament.
You can certainly construct it (quite easily) the way that it is equivalent (up to the fact that you know beforehand). The way it was proposed it's probably not, as if you go first somehow depends on who's advancing.
But you could e.g. just do all the coin tosses beforehand, and than construct the bracket so that it is equivalent. 4 player example:
Take player A vs B and C vs D, and 3 cointosses. First take the coin for the final, if winner of A vs. B goes first, set first bit of A's and B's number to 0, otherwise 1. C and D get the opposit. Now take A and B, toss coin, if A goes first, second bit of A's number is 0, otherwise 1. B the opposite. C and D analogue. Now if you just reveal the last remaining bit each round, from the perspective of the players it looks like you would toss the coin for each round. But nevertheless you have done all coin tosses beforehand.