EPTSo, let's start with two Players, A and B. Lets also say we can quantify A's chances of beating B as equal to X, and his chances of losing equal to (1 - X), with all external factors being equal. How can we figure out his chances of beating B in a Bo1 or Bo3?
We'll, if it's a Bo1, his chances of winning are simply equal to X. So, if we think a has a 60% chance (ie, that he'd win 6 out of 10 games against B) we'd expect him to win a Bo1 60% of the time, again with all things being equal.
For a Bo3, it's a bit more complicated. Bo3's can end with 4 different scores, 2-0, 2-1, 1-2, or 0-2. We need to figure out the odds for all of these outcomes occurring to know A's overall chances.
For a 2-0, there is only one possible way this can happen, with A winning both games (WW). So. the odds of that are equal to X * X. Again, using our 60% favored number, A has a 36% chance of this happening.
Likewise, 0-2 works the same way. Since there is only one way this can happen (LL), we can solve the odds easily as (1-X) * (1-X). For 60%, this equals only a 19.2% chance.
For three game series, it's only a bit more complicated.
There are two possible ways a 2-1 can happen, WLW or LWW. WWL is obviously not a possibility, since with two wins the series is over. So, the odds are equal to 2 * (X * X * (1-X) ). For 60%, this gives A a 28.8% chance of winning in three games.
Same is true for a 1-2, the two possibilities are LWL and WLL. So, the odds are equal to 2 * ( (1 - X) * (1 - X) * X). A 60% favored player has a 19.2% chance of losing in three games.
Overall, the 60% player has 64.8% chance of winning a Bo3, hardly better than his Bo1 odds.
It's a few minutes work to plug this all into a spreadsheet, and get the numbers for any given odds for player A. As a quick correctness checker, the total odds should always add up to 1. I'll save you the effort though, and post some for you.
50% in Bo1 = 50% in Bo3
55% in Bo1 = 57.74% in Bo3
60% in Bo1 = 64.8% in Bo3
65% in Bo1 = 71.8% in Bo3
70% in Bo1 = 78.4% in Bo3
75% in Bo1 = 84.4% in Bo3
80% in Bo1 = 89.6 in Bo3
As you can see, in fairly close matchups the odds of a player winning a Bo3 aren't much better than their odds in a Bo1. Obviously though, as the matchup becomes more and more favored for Player A, the effect of a Bo3 becomes more significant.
However, the major limit to these numbers is that they assume the odds of A beating B are the same on Maps 1, 2 and 3 in a Bo3. Obviously this isn't the case. Not only do certain maps favor certain races and players, but this doesn't account for the mind games that can occur in a series. The latter number isn't really possible to predict in advance, so we can't account for it. The maps problem, however, is a much bigger deal. Heavily biased maps swing the odds of single games fairly dramatically, and are something a tournament should try and be aware of. However, If the maps are close to even in a given matchup, then their effect is negligible. If the maps suck, blame the people who picked them, not the format.
Finally, it's worth pointing out that this has nothing to do with the viewing experience. Personally, as a BW fan used to this format i like the tension and (slightly greater) potential for upsets created by the OSL group stage format. Likewise, as an adult with other interests and commitments i like that it takes less time to watch all the games. People who are coming from a Sc2 background, on the other hand, might miss the mind games more than i do, and might simply want to see more games played. Regardless of where you stand on this issue, however, the numbers don't care.
TLDR: For closely matched players on mostly balanced maps, the odds of a player winning a Bo3 are barely different from them winning a Bo1. However, Bo3's benefit heavily favored players quite significantly, and highly biased maps are far more significant in a Bo1 than a Bo3. Read it, and learn some basic math.
