Suppose you have two players competing against one another in two possible formats.

The two formats are:

(1) The first of the two players to win 55 games wins the competition.

(2) The first of the two players to win 11 sets, with each set being determined by the first player to win 5 games, wins the competition.

If player A is a 55% favorite to win each individual game, what are the chances of player A winning the overall competition in format (1) versus the chances of player A winning the overall competition in format (2)?

It seems that the second scenario is slightly better for player A. This problem can be solved through the negative binomial distribution, which measures the probability of having n successes before k failures. In this case, we need to calculate how much is likely that A's opponent wins 0 through 54 games before A wins his 55th game. This can be achieved with a single line in R:

For the second scenario, we evaluate the probability of A to win a set and then in the same fashion the probability of winning 11 sets:

A good 2% improvement for player A.

Very interesting. Thanks.

Even without math, it makes sense that scenario 2 would be less bad for the underdog. He can lose more games overall and still be the first to win 11 sets. The favorite will have more 4-1 and 5-0 sets than the underdog, but gets no extra points for doing that.

Actually, the second scenario is worse for the underdog. Consider also that winning 55 games will basically never suffice; you'll end up playing much more games in the second scenario and so enhancing A's edge.

Oh good point. For some reason I was thinking series of 5. "First to 5" means it could take 9 games, "best of 9". So yeah, the match will last more than 55 games on average and it will be a "longer run", magnifying the favorite's odds.

Series of 5 would be no different (odds wise) than "best of" 5, in which case the average length would be less than 55 and it would be a shorter run. That's why it would help the underdog (maybe also for the reason I said before but I'm not sure).

I agree with NickTheGreek's calculation. You can do this with the cumulative binomial instead of the negative binomial. In Excel it's Binomdist(54,109,0.55,true) for the probability that player A loses. You pretend they play 109 games instead of stopping when one player gets 55. You then ask for the probability that player A will win 54 or fewer of those games.

For the second scenario, you do it in two steps, as NickTheGreek did. First Binomdist(4,9,0.55,true) gives the probability of player A losing a set. Call that A1, then Binomdist(10,21,1-A1,true) is the probability of player A winning 10 or fewer of the first 21 sets.

In general, if p is the probability of A winning, S is the number of games in a set and N is the number of sets, Binomdist(S*N-1,2*S*N-1,p,true) is greater than or equal to (meaning the better player has less of an advantage) than Binomdist(N-1,2*N-1,1-Binomdist(S-1,2*S-1,p,1),1).