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.
If you assume that the weaker players won't gang up on the stronger player (a very strong assumption!), then a reasonable model would be the following. (I'm following the notation of the "theory" section of the Wikipedia article on ELO.)
- let $R_A, R_B, R_C$ be the ratings of the three players.
- let $Q_A = 10^{R_A/400}$; define $Q_B, Q_C$ similarly.
- the probability of $A$ winning is $Q_A/(Q_A + Q_B + Q_C)$ and similarly for the other two players.
With the numbers you gave, $C$ has a probability of about $0.613$ of winning, and $A, B$ each have probability $0.194$ of winning.
This seems like the "obvious" generalization of the Elo math. The most obvious problem, to me, is that I wouldn't know how to update these ratings after a multi-player game is played.
Best Answer
Your question seems to concern the inductive bias that you have to make in order to be able to predict anything meaningful at all. Without assumptions there is nothing you can learn (no free lunch theorem). So what do you assume?
Probably something like this: both players have their own skill level that is constant over the whole year, and whenever they play, the outcome is independent of all previous games. Then you can think of this as a Bernoulli experiment (flipping a coin), biased towards the better player. So, simply check who of the both players won more often, and then always predict him. However, probably you also assume that their skills vary over time? Then you should ignore the first 11 months, and always predict the better of both from the last 30 days - or 3 months? You can also take the approach described by Huanaphu, where the time window length of the dependency on the past is implicitly modelled by the choice of the resolution. However, probably you further suppose that there are even other effects with even less independence between the games? (e.g., whenever Danny looses, he will be extremely motivated the single next time and then win with much higher probability). Now you need a more advanced model that conditions your prediction on the outcome of the last game.
So, first, think carefully about your model assumptions, second, use standard techniques that fit your model in order to predict the "most likely" outcome.