and so on. But how would it work without power-matching? What if teams were paired at random? The easy part is using the laws of probability to figure out which matches happen by chance. That's listed in column F.
The hard part is figuring out which team wins. If both teams have the same record, then whichever team wins, the outcome is the same. For example, in round two, the 25 teams in 1-0 vs. 1-0 rounds (ignore the fact that this is odd--it makes no difference in the end) and the 25 teams in 0-1 vs. 0-1 rounds guarantees that 12.5 teams will have a 2-0 record; 25 will be 1-1; and 12.5 will be 0-2. These guaranteed outcomes are listed in column I.
But what happens if the two teams have different records? One possibility is that there are no upsets at all. For example, in round two, of the 50 teams in 1-0 vs. 0-1 rounds, exactly half are 1-0s. These 25 teams might all win--no upsets--and become 2-0s. The 25 teams that are 0-1s all become 0-2s. These no-upset results are listed in column J.
The other possibility is that all rounds with mixed records have upsets. In round two, of the 50 teams with 1-0 vs. 0-1 rounds, the 25 teams that are 0-1s could all win, becoming 1-1s, while the 25 teams with 1-0s all lose, become 1-1s. Thus all 50 teams end up 1-1. These all-upset results are listed in column K.
Of course, neither no-upsets or all-upsets is realistic. From other research I've done, it turns out the upset rate is more like 20%, so I blended the two results 80:20 no-upsets:all-upsets in column L. As you can see, the ultimate outcome is that each record is nearly balanced with the others, though slightly more in the mediocre results. For example, after three rounds, a 20% upset rate results in about 17 teams that are 4-0s; 22 teams that are 3-1s; 23 teams that are 2-2s; etc.
Yet the 20% upset rate is probably conservative. It is unlikely that an 0-3 team has a 20% chance against a 3-0 team. As the teams are farther apart in record in later rounds, the overall upset rate must drop. If this is so, the final outcomes flatten. It turns out that if the upset rate is 1/6 for round two, drops to 1/8 for round three, and further drops to 1/10 for round four, then the final outcome is that exactly 20 teams are 4-0s; 20 are 3-1s; etc.
What happens if teams are paired at random? It depends on the upset rate. If it's exactly 50% (which is far too high), then the final outcomes look exactly like it would with power-matching:
If the upset rate is a more realistic, empirically justified 20%, then the outcomes are much flattened and nearly equally distributed:
Here's the sheet for anyone who'd like to play around with it.
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