# Best of 3 Probability

#### KFDE

##### New member
Hi, its been a long time since I was formally involved in math study, so I hope this is the right category.

I am exploring the use of a best-of-2 format for a game that is currently best-of-3. I have been given the results for a tournament in which the best-of-3 rules were used, and I'd like to see how the results would have changed for best-of-2. Unfortunately, the app that was used to run the tournament does not include all of the information I would need to be certain. I feel like there's a relatively simple formula I could use to figure out the likelihood of a different result, given a bit of information, but I can't figure out what it would be.

Info dump incoming:

1) Each match was played in a best of 3 games format, where each game's winner was determined by the total points scored in that game. This means it is possible that the loser of a match would have more total points than the winner, but I found this to be uncommon. (6 of 187)
2) Any match which was won 2-0 would not have the outcome altered by a best of 2 system.
3) Unfortunately, the app that ran the tournament does not keep track of the scores in individual games, only in the total score for a match and the number of games won by each player.
4) Therefore, I am only looking at matches that were won 2-1.
5) In order for a match to go 2-1, each player has to win one of the first 2 games. In a best-of-2 format, this would mean that the match would be determined by the total points scored across both games (as opposed to playing a 3rd game).
6) Since the app doesn't keep track of individual game scores, I can only estimate the likelihood that the losing player outscored the winning player in their first two games. This is what I'm trying to do. Given the total score at the end of the best of 3 match, is there an equation that I could use to estimate the likelihood of this happening?

I'll include some example scores below:

Round 1, Table 3: 38-22
Round 1, Table 35: 21-17
Round 3, Table 39: 43-36

Thanks!