JSSM- 2006, Vol.5, Issue 4, 567

If Z = X₁ + X₂ +……… + X_N where X_i are i.i.d. then κ_Z(t) = κ_N(κ_X(t)). Taking the derivatives of the result and setting t = 0 gives the following useful results in terms of cumulants:

k⁽¹⁾_Z = k⁽¹⁾_N k⁽¹⁾_X
k⁽²⁾_Z = k⁽²⁾_N [k⁽¹⁾_X ]² + k⁽¹⁾_N k⁽²⁾_Xk⁽³⁾_Z = 3k⁽¹⁾_X k⁽²⁾_N k⁽²⁾_X + [k⁽¹⁾_X ]³k⁽³⁾_N + k⁽¹⁾_N k⁽³⁾_X
k⁽⁴⁾_Z = 3k⁽²⁾_N [k⁽²⁾_X ]² + 6[k⁽¹⁾_X ]²k⁽²⁾_X k⁽³⁾_N + 4k⁽¹⁾_X k⁽²⁾_N
k⁽³⁾_X + [k⁽¹⁾_X ]⁴k⁽⁴⁾_N + k⁽¹⁾_N k⁽⁴⁾_X

For example the mean number of points in a tiebreaker match, M^pmT, with the associated variance, V^pmT, can be calculated from the cumulant generating function as:

M^pmT = M^psTM^sm
V ^pmT = V ^sm(M^psT )² +M^smV ^psT

where: M^psT represents the mean number of points in a tiebreaker set. M^sm represents the mean number of sets in a tiebreaker match. V^psT represents the variance of the number of points in a tiebreaker set. V^sm represents the variance of the number of sets in a tiebreaker match

Let M^pm , U^pm , C^pm , S^pm and K^pm represent the mean, standard deviation, and coefficients of variation, skewness and kurtosis for the number of points in an advantage match. Let M^pmT, U^pmT, C^pmT, S^pmT and K^pmT represent the mean, standard deviation, and coefficients of variation, skewness and kurtosis for the number of points in a tiebreaker match. Tables 3 and 4 represent the exact parameters of the distributions for an advantage and tiebreaker match for different values of p_A and p_B. The results agree with Pollard (1983) for a best-of-three sets tiebreaker match. It shows that the mean, standard deviation, coefficients of variation, skewness and kurtosis of the number of points played are greater for an advantage match, compared to a tiebreaker match. Also included in the tables are the probabilities of the match lasting for at least n points, represented by P(n) for an advantage match and Q(n) for a tiebreaker match. These probabilities were calculated using the NP-expansion technique (Pesonen, 1975). Notice that when p_A and p_B become “large”, the probability of playing at least 400 points in an advantage match is considerably greater than for a tiebreaker match. This is some justification as to why an advantage match can seemingly never end with two strong servers.

Table 5 represents the exact parameters of distributions for a tiebreaker and an advantage match using 50-40 games, along with the probability of a match going beyond 300 points. For an extreme case, when p_A = p_B = 0.75, the probability of an advantage match going beyond 300 points is 0.06. In comparison to Tables 3 and 4, the probability of an advantage or tiebreaker match going beyond 300 points is 0.38. This shows that replacing standard ‘deuce’ games with 50-40 games, substantially decreases the likelihood of long matches occurring.

It is often the case that by shortening the length of matches, decreases the probability of winning for the better player. However this is not necessarily the case as shown by replacing standard ‘deuce’ games with 50-40 games. Table 6 represents the probabilities of winning under four different scoring systems, for different values of p_A and p_B. Notice when p_A=0.75 and p_B = 0.70, the probability of player A (the stronger player) winning using 50-40 games is greater than using standard ‘deuce’ games.