If you play games to $4$ points, where you have to win by $2$, you can assume the players play 6 points. If no player wins by $2$, then the score is tied $3-3$, and then you play pairs of points until one player wins both. This means the the chance to win a game to $4$ points, when your chance to win each point is $p$, is
$$p^6 + 6p^5(1-p) + 15p^4(1-p)^2 + 20 p^3(1-p)^3 \frac{p^2}{p^2 + (1-p)^2}$$.
In top level men's play, $p$ might be about $0.65$ for the server. (It would be $0.66$ if men didn't ease off on the second serve.) According to this formula, the chance to hold serve is about $82.96\%$.
Suppose you are playing a tiebreaker to $7$ points. You can assume that the points are played in pairs where each player serves one of each pair. Who serves first doesn't matter. You can assume the players play $12$ points. If they are tied at that point, then they play pair until one player wins both of a pair, which means the conditional chance to win is $p_sp_r/(p_sp_r + (1-p_s)(1-p_r))$. If I calculate correctly, the chance to win a tiebreaker to $7$ points is
$$ 6 p_r^6 ps + 90 p_r^5 p_s^2 - 105 p_r^6 p_s^2 + 300 p_r^4 p_s^3 -
840 p_r^5 p_s^3 + 560 p_r^6 p_s^3 + 300 p_r^3 p_s^4 - 1575 p_r^4 p_s^4 +
2520 p_r^5 p_s^4 - 1260 p_r^6 p_s^4 + 90 p_r^2 p_s^5 - 840 p_r^3 p_s^5 +
2520 p_r^4 p_s^5 - 3024 p_r^5 p_s^5 + 1260 p_r^6 p_s^5 + 6 p_r p_s^6 -
105 p_r^2 p_s^6 + 560 p_r^3 p_s^6 - 1260 p_r^4 p_s^6 + 1260 p_r^5 p_s^6 -
462 p_r^6 p_s^6 + \frac{p_r p_s}{p_r p_s + (1-p_r)(1-p_s)}(p_r^6 + 36 p_r^5 p_s - 42 p_r^6 p_s + 225 p_r^4 p_s^2 - 630 p_r^5 p_s^2 +
420 p_r^6 p_s^2 + 400 p_r^3 p_s^3 - 2100 p_r^4 p_s^3 + 3360 p_r^5 p_s^3 -
1680 p_r^6 p_s^3 + 225 p_r^2 p_s^4 - 2100 p_r^3 p_s^4 + 6300 p_r^4 p_s^4 -
7560 p_r^5 p_s^4 + 3150 p_r^6 p_s^4 + 36 p_r p_s^5 - 630 p_r^2 p_s^5 +
3360 p_r^3 p_s^5 - 7560 p_r^4 p_s^5 + 7560 p_r^5 p_s^5 -
2772 p_r^6 p_s^5 + p_s^6 - 42 p_r p_s^6 + 420 p_r^2 p_s^6 -
1680 p_r^3 p_s^6 + 3150 p_r^4 p_s^6 - 2772 p_r^5 p_s^6 + 924 p_r^6 p_s^6)$$
If $p_s=0.65, p_r=0.36$ then the chance to win the tie-breaker is about $51.67\%$.
Next, consider a set. It doesn't matter who serves first, which is convenient because otherwise we would have to consider winning the set while having the serve next versys winning the set without keeping the serve. To win a set to $6$ games, you can imagine that $10$ games are played first. If the score is tied $5-5$ then play $2$ more games. If those do not determine the winner, then play a tie-breaker, or in the fifth set just repeat playing pairs of games. Let $p_h$ be the probability of holding serve, and let $p_b$ be the probability of breaking your opponent's serve, which may be calculated above from the probability to win a game. The chance to win a set without a tiebreak follows the same basic formula as the chance to win a tie-breaker, except that we are playing to $6$ games instead of to $7$ points, and we replace $p_s$ by $p_h$ and $p_r$ by $p_b$.
The conditional chance to win a fifth set (a set with no tie-breaker) with $p_s=0.65$ and $p_r=0.36$ is $53.59\%$.
The chance to win a set with a tie-breaker with $p_s=0.65$ and $p_r=0.36$ is $53.30\%$.
The chance to win a best of $5$ sets match, with no tie-breaker in the fifth set, with $p_s=0.65$ and $p_r=0.36$ is $56.28\%$.
So, for these win rates, how many games would there have to be in one set for it to have the same discriminatory power? With $p_s=0.65, p_r=0.36$, you win a set to $24$ games with the usual tiebreaker $56.22\%$, and you win a set to $25$ game with a tie-breaker possible $56.34\%$ of the time. With no tie-breaker, the chance to win a normal match is between sets of length $23$ and $24$. If you simply play one big tie-breaker, the chance to win a tie-breaker of length $113$ is $56.27\%$ and of length $114$ is $56.29\%$.
This suggests that playing one giant set is not more efficient than a best of 5 matches, but playing one giant tie-breaker would be more efficient, at least for closely matched competitors who have an advantage serving.
Here is an excerpt from my March 2013 GammonVillage column, "Game, Set, and Match." I considered coin flips with a fixed advantage ($51\%$) and asked whether it is more efficient to play one big match or a series of shorter matches:
... If a best of three is less efficient than a single long match, we
might expect a best of five to be worse. You win a best of five $13$
point matches with probability $57.51\%$, very close to the chance to win
a single match to $45$. The average number of matches in a best of five
is $4.115$, so the average number of games is $4.115 \times 21.96 = 90.37$. Of
course this is more than the maximum number of games possible in a
match to $45$, and the average is $82.35$. It looks like a longer series
of matches is even less efficient.
How about another level, a best of three series of best of three
matches to $13$? Since each series would be like a match to $29$, this
series of series would be like a best of three matches to $29$, only
less efficient, and one long match would be better than that. So, one
long match would be more efficient than a series of series.
What makes a series of matches less efficient than one long match?
Consider these as statistical tests for collecting evidence to decide
which player is stronger. In a best of three matches, you can lose a
series with scores of $13-7 ~~ 12-13 ~~ 11-13$. This means you would win $36$
games to your opponent's $33$, but your opponent would win the series.
If you toss a coin and get $36$ heads and $33$ tails, you have evidence
that heads is more likely than tails, not that tails is more likely
than heads. So, a best of three matches is inefficient because it
wastes information. A series of matches requires more data on average
because it sometimes awards victory to the player who has won fewer
games.
Best Answer
How odds are set is a really interesting subject that I have done some research into, and in a similar way sports analytics.
The first paper I would refer to covers the NFL specifically "Why are Gambling Markets organised so differently from Financial Markets", Steven.D.Levitt (The Economic Journal 2004). This illustrates that the odds on the NFL are rarely set to generate 50/50 action because the bookmaker can exploit "square" action by skewing odds against their traditional bias (i.e. the point made above about the Ohio State Buckeyes - if the bookmaker is aware that they are going to take a larger % of the bets, they can either adjust the odds or the spread so the better has to pay a premium to bet the Buckeyes - e.g. the -7.5 or more than one touchdown instead of -6.5 - especially if the true rating for the game was around -5 or -6). It also makes the point that bookmakers/sportsbooks rarely make the odds themselves they are usually paying influential odds makers who set the line for a lot of events. The bookmaker will then rarely adjust these odds greatly as they will effectively handicap the market against other bookmakers and sportsbooks (generating a profitable opportunity for "sharp" action).
In the case of the game quoted by the OP, the prices quoted by Bet 365 is consistent with the over-round % that they have run on most football games this season of between 105-107% (I have an interest in this - their over-round% on the English Premier League is typically 5-6%). That 5-7% margin will look after them in the long run as it increasingly means unsophisticated gamblers have to be more right than average in the long run to make a a sustained profit. How the actual odds are generated is another matter in the case of Bet365 a lot of their competitors use the Bet Genius group for Odds Data (e.g. Sportingbet, Paddy Power, Sky Bet). They will probably then make small adjustments to this based on their typical clients betting preferences (e.g. what type of action they take and biases).
For a lot of sports the Cantor Fitzgerald group have created the Midas Algorithm to set up odds in the same way they would deal on Wall Street and they are getting an increasing presence in Las Vegas running several sports books - http://m.wired.com/magazine/2010/11/ff_midas/all/1. This has allowed them to set spreads for the entire NFL season (http://www.grantland.com/blog/the-triangle/post/_/id/27740/nfl-win-totals-hot-off-the-sportsbook-press) before the pre-season has taken place (which is not a typical case as most bookmakers seems to react on week to week action and player injuries and performance).
How are the actual odds generated? This is the more difficult question. Going on Mathletics (Wayne L.Winston 2009), some sports e.g. the NFL can be governed by a simple least squares algorithm based on margins of victory and points scored which can then be finessed (e.g. to give more weight to recent games). This can then be used to generate win percentages based on the ratings derived. In the case of the NFL, Hal Stern "On the Probability of Winning an American Football Game" (American Statistician 45, 1991) showed that the probability of the final margin of victory for home NFL team can be well approximated by a normal random variable with mean = home edge+home team rating-away team rating and a standard deviation of 13.86. Plug the ratings generated by your least squares work in and you have a set of percentages against a given spread. I believe that this can also be applied to a lot of other sports (e.g Australian Rules Football). In the case of football though I believe that oddsmakers also have done some regression analysis into player statistics to allow them to make a more rational rating based on the players that will actually be on the pitch rather than past team performance in terms of margins of victory (e.g. the Dtech group who analyse European football for the Times Newspaper base their ratings on the Team shots and goals data http://www.dectech.org/football/help_info.php - rather than a least squares model based purely on margin of victory). Given that sports could and should be viewed as an academic subject, I believe this is why we have seen increases in the number of the groups such as the Accuscore group who have a largely academic background (from interviews on the ESPN Behind the Bets Podcast) and have used their knowledge to generate opportunities from odds skewed to exploit gamblers that bet with pre-conditioned biases (e.g. the home team favourite wins more than 50% of games). If you can remove bias from the team that you pick, I believe this will generate opportunity.