Jedan meč do 21pts ili na dva dobivena do 9pts?

Izračun da li je veća vjerojatnost da bolji igrač pobjedi slabijega u jednom meču do 21pts ili u ogledu na dva dobivena meča do 9pts i za koju od ove dvije opcije je potrebno više vremena.

According to the Elo formula as in extremegammon the probability of the better player having Elo difference +100 from his opponent in a 21 point match is 62.89%. 

Let's see how much is the probability of the better player winning two out of three 9 point matches: This can be done in three ways (WW, WLW, LWW) with relevant probabilities [given that in a 9 point match the better player's winning probability is 58,55%] (34.28%, 14.21%, 14.21%). Adding those three numbers gives us 62.7%. 
  
Now about the time consumed: Assuming 2 minutes per point, the total time in the clocks of the players in the 21 point match is 84 minutes. Playing two out of three 9 point matches, a third match will be necessary in the two cases WL and LW in the two first matches with relevant probability two times 24.27%=48.54%. That means that 51.46% of the time a third match will be necessary and on average, 36+36+36X51.46=90.5 minutes will be set on the clocks. One can obtain similar results with other match lengths or/and different Elo differences. 

One question is: will weaker players accept the single match schedule, if they are convinced about the correctness of the above analysis? I think "yes, gleefully", for in bg as in many other mental sports, everyone thinks he's much better than his equals. 

I am going to prove that two out of three 9 point matches is less favorable a schedule for the better player than a single 21 point match and more time consuming on average as well.