Appendix: Proofs of the theorems
We start by introducing some additional notation and remarks. Given a competition (s, N) and an assignment \({\textbf{A}}\in {\mathcal {A}}^{\left( s,N\right) }\), \(M({\textbf{A}})\) stands for the set of matches induced by \( {\textbf{A}}\). No ties are allowed and the player who wins the match \(m\in M( {\textbf{A}})\) is denoted by \(w_{m}({\textbf{A}})\). Player \(i\in N\) is a winner of the competition (s, N) at \({\textbf{A}}\in {\mathcal {A}}^{\left( s,N\right) } \) if \(\left| \left\{ m\in M({\textbf{A}}):w_{m}({\textbf{A}})=i\right\} \right| \ge \left| \left\{ m\in M({\textbf{A}}):w_{m}({\textbf{A}} )=j\right\} \right| \) holds for all \(j\in N\). Thus, given \({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\), \(w({\textbf{A}})=\left( w_{m}({\textbf{A}})\right) _{m\in M({\textbf{A}})}\) stands for the vector of match winners, whose dimension is s. Given \({\textbf{A}}\in {\mathcal {A}}^{\left( s,N\right) }\) and a probability matrix \({\textbf{P}}\), the probability \(pr(w({\textbf{A}}))\) of \(w( {\textbf{A}})\) occurring is given by \(pr(w({\textbf{A}}))=\Pi _{m\in M({\textbf{A}} )}\varphi _{w_{m}({\textbf{A}})}({\textbf{A}},{\textbf{P}})\). Thus, for each \(i\in N\), \(\varphi _{i}({\textbf{A}},{\textbf{P}})\) can be expressed by \(\varphi _{i}( {\textbf{A}},{\textbf{P}})=\sum \limits _{w({\textbf{A}})\in W_{i}({\textbf{A}})}pr(w( {\textbf{A}}))=\sum \limits _{w({\textbf{A}})\in W_{i}({\textbf{A}} )}\prod \limits _{m\in M({\textbf{A}})}\varphi _{w_{m}({\textbf{A}})}({\textbf{A}}, {\textbf{P}})\), where \(W_{i}({\textbf{A}})\) stands for the set of all vectors of winners in the matches induced by \({\textbf{A}}\) for which i is a final winner of the competition. Additionally, when the probability matrix \( {\textbf{P}}\) is such that \(p_{ij}=0.5\) holds for all \(i,j\in N\), we have \( pr(w({\textbf{A}}))=\left( 0.5\right) ^{s}\) for each vector of match winners and thus, \(\varphi _{i}({\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{i}({\textbf{A}})\right| \) holding for each \(i\in N\).
Proof of Theorem 1
Let (s, N) be a league competition with either \(2s=\left| N\right| \) or \(2\,s>\left| N\right| =2\). We show first that (s, N) satisfies ET. Let \({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\) be an assignment and \({\textbf{P}}\) the probability matrix with \(p_{ij}=0.5\) for all \(i,j\in N\). Notice then that either each player in N plays exactly one match (when \(2s=\left| N\right| \)) or there are only two players who play all matches (when \(2s>\left| N\right| =2\)). Consider the following cases separately.
Case 1 (\(2s>\left| N\right| =2\) and s is odd). Let \(N=\left\{ 1,2\right\} \) and notice that in this case we have \(\varphi _{1}({\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{1}( {\textbf{A}})\right| \) and \(\varphi _{2}({\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{2}({\textbf{A}})\right| \). Because s is odd, \(w({\textbf{A}})\in W_{1}({\textbf{A}})\) implies \(w({\textbf{A}})\notin W_{2}({\textbf{A}})\), and \(w({\textbf{A}})\in W_{2}({\textbf{A}})\) implies \(w( {\textbf{A}})\notin W_{1}({\textbf{A}})\). Thus, \(W_{1}({\textbf{A}})\cap W_{2}( {\textbf{A}})=\emptyset \). Finally, given that \(N=\left\{ 1,2\right\} \) and \( W_{1}({\textbf{A}})\cap W_{2}({\textbf{A}})=\emptyset \), we can define a bijection \(f:W_{1}({\textbf{A}})\rightarrow W_{2}({\textbf{A}})\) by just replacing 1 by 2 and 2 by 1 as match winners at each \(w({\textbf{A}})\in W_{1}( {\textbf{A}})\) as to get \(f(w({\textbf{A}}))\in W_{2}({\textbf{A}})\). Thus \( \left| W_{1}({\textbf{A}})\right| =\left| W_{2}({\textbf{A}} )\right| \) and \(\varphi _{1}({\textbf{A}},{\textbf{P}})=\varphi _{2}({\textbf{A}},{\textbf{P}})\) holds. We conclude then that (s, N) satisfies ET.
Case 2 (\(2s>\left| N\right| =2\) and s is even). Notice that in this case \(W_{1}({\textbf{A}})\cap W_{2}({\textbf{A}})\ne \emptyset \). We have then that \(w({\textbf{A}})\in W_{1}({\textbf{A}})\setminus W_{2}({\textbf{A}})\) implies \(w({\textbf{A}})\notin W_{2}({\textbf{A}}){\setminus } W_{1}({\textbf{A}})\), and \(w({\textbf{A}})\in W_{2}({\textbf{A}}){\setminus } W_{1}( {\textbf{A}})\) implies \(w({\textbf{A}})\notin W_{1}({\textbf{A}}){\setminus } W_{2}( {\textbf{A}})\). Finally, we can define a bijection \(f:W_{1}({\textbf{A}} ){\setminus } W_{2}({\textbf{A}})\rightarrow W_{2}({\textbf{A}}){\setminus } W_{1}( {\textbf{A}})\) in the same way as in Case 1 and conclude that \(\left| W_{1}({\textbf{A}}){\setminus } W_{2}({\textbf{A}})\right| =\left| W_{2}( {\textbf{A}}){\setminus } W_{1}({\textbf{A}})\right| \) should follow and thus, \( \left| W_{1}({\textbf{A}})\right| =\left| W_{2}({\textbf{A}} )\right| \) also holds. We conclude that \(\varphi _{1}({\textbf{A}},{\textbf{P}})=\varphi _{2}({\textbf{A}},{\textbf{P}})\) and thus (s, N) satisfies ET also in this case.
Case 3 (\(2s=\left| N\right| \)). Because each player plays exactly one match in this case, for any vector \(w({\textbf{A}})\) of match winners, a player is the final winner of the competition only if she is the winner of the match which she is assigned by \({\textbf{A}}\). Thus, \( \left| W_{i}({\textbf{A}})\right| =\left| W_{j}({\textbf{A}} )\right| \) holds for all \(i,j\in N\). By \(\varphi _{i}({\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{i}({\textbf{A}})\right| \) holding for each \(i\in N\), (s, N) satisfies ET.
Assume now that (s, N) is a league competition satisfying ET. To show that either \(2s=\left| N\right| \) or \(2\,s>\left| N\right| =2\) holds, let \({\textbf{P}}\) be the probability matrix with \(p_{ij}=0.5\) for all \( i,j\in N\). It suffices to show that, for some assignment in \({\mathcal {A}} ^{(s,N)}\), \(2\,s>\left| N\right| >2\) leads to a contradiction (as \( \left| N\right| \le 2\,s\) follows from (s, N) satisfying ET and the definition of an assignment). Consider then the following possible cases.
Case 1 (\(\left| N\right| \) is even and \(2s=\left| N\right| +2\)). Let the assignment \({\textbf{A}}\) be such that \(a_{12}=2\) and \(\Sigma _{j\in N{\setminus } \left\{ 1,2\right\} }a_{ij}=1\) for all \(i\in N{\setminus } \left\{ 1,2\right\} \), and notice that \({\textbf{A}}\in {\mathcal {A}} ^{(s,N)}\).
Consider a competition system \((s-1,N)\) and the assignment \({\textbf{A}} ^{\prime }\in {\mathcal {A}}^{(s-1,N)}\) such that \(a_{12}^{\prime }=1\) and \( a_{ij}^{\prime }=a_{ij}\) for all \(i,j\in N{\setminus } \left\{ 1,2\right\} \). Observe that, by \(2(s-1)=\left| N\right| \) and as argued in Case 3 above, \(\left| W_{1}({\textbf{A}}^{\prime })\right| =\left| W_{3}( {\textbf{A}}^{\prime })\right| \). Take \(w({\textbf{A}}^{\prime })\in W_{1}( {\textbf{A}}^{\prime })\) and note that this implies that player 1 wins his single match in \(M({\textbf{A}}^{\prime })\) against player 2. Thus, for each \( w^{*}({\textbf{A}}^{\prime })\in W_{1}({\textbf{A}}^{\prime })\) there are exactly two vectors \(w^{\prime }({\textbf{A}}),w^{\prime \prime }({\textbf{A}} )\in W_{1}({\textbf{A}})\) of match winners for (s, N) where, all else being equal, either player 1 wins his other match against player 2 or player 2 wins it. On the other hand, for \(w^{**}({\textbf{A}}^{\prime })\) to belong to \(W_{3}({\textbf{A}}^{\prime })\) player 3 must necessarily win her single match. Thus, for each \(w^{**}({\textbf{A}}^{\prime })\in W_{3}( {\textbf{A}}^{\prime })\) there is a unique vector \(w^{\prime \prime \prime }( {\textbf{A}})\in W_{3}({\textbf{A}})\) of match winners for (s, N) where, all else being equal, the winners of the two matches between players 1 and 2 are different. We conclude then that \(\left| W_{1}({\textbf{A}})\right| >\left| W_{3}({\textbf{A}})\right| \) holds and thus, by \(\varphi _{1}( {\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{1}({\textbf{A}})\right| >\left( 0.5\right) ^{s}\cdot \left| W_{3}({\textbf{A}} )\right| =\varphi _{3}({\textbf{A}},{\textbf{P}})\), we reach a contradiction to (s, N) satisfying ET.
Case 2 (\(\left| N\right| \) is even and \( 2\,s>\left| N\right| +2\)). Let \({\textbf{A}}\) be such that \(a_{12}=\frac{ 2\,s-\left| N\right| }{2}+1\) and \(\Sigma _{j\in N{\setminus } \left\{ 1,2\right\} }a_{ij}=1\) for all \(i\in N{\setminus } \left\{ 1,2\right\} \), and notice that \({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\). Fix a vector \(w({\textbf{A}})\) of match winners and note that, given the definition of \({\textbf{A}}\), player 3 is a final winner in \(w({\textbf{A}})\) only if, for each \(i\in N\), player i wins at most one match. However, by \(\frac{2\,s-\left| N\right| }{2} +1>2\), either player 1, or player 2 or both must win more than one match. We conclude then that \(W_{3}({\textbf{A}})=\emptyset \) should hold. Meanwhile, \( W_{1}({\textbf{A}})\ne \emptyset \) because player 1 is a final winner of the competition if she wins all her matches. We have then \(\varphi _{1}({\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{1}({\textbf{A}} )\right| >0=\left( 0.5\right) ^{s}\cdot \left| \mathbf {\emptyset } \right| =\left( 0.5\right) ^{s}\cdot \left| W_{3}({\textbf{A}} )\right| =\varphi _{3}({\textbf{A}},{\textbf{P}})\) in contradiction to (s, N) satisfying ET.
Case 3 (\(\left| N\right| \) is odd). Let \( N=\left\{ 1,\ldots ,n\right\} \) and take \({\textbf{A}}\) defined as follows: \( a_{12}=\frac{2s-\left| N\right| +1}{2}\), \(a_{1n}=1\), and \(\Sigma _{j\in N{\setminus } \left\{ 1,2,n\right\} }a_{ij}=1\) for all \(i\in N{\setminus } \left\{ 1,2,n\right\} \). Notice that \({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\). We proceed by considering the following three possible sub-cases.
Case 3.1 (\(\frac{2\,s-\left| N\right| +1}{2}=1\)). In this case player 1 plays exactly one match against player 2 and exactly one match against player n. Denote these matches by \(m_{12}\) and \(m_{1n}\), correspondingly. There are four possible combinations of winners in \(m_{12}\) and \(m_{1n}\): (i) 1 wins in both matches; (ii) 1 wins against 2 and loses against n; (iii) 1 loses against 2 and wins against n; and (iv) 1 loses in both matches. Notice that each player from \(N\setminus \{1,2,n\}\) can win at most one match. Therefore, for each of the vectors of match winners in \(M( {\textbf{A}})\setminus \{m_{12},m_{1n}\}\), three out of the four possible combinations of winners in \(\{m_{12},m_{1n}\}\) give as a result a vector of match winners in \(W_{1}({\textbf{A}})\) and only two give as a result a vector of match winners in \(W_{n}({\textbf{A}})\). We conclude then that \(\left| W_{1}({\textbf{A}})\right| >\left| W_{n}({\textbf{A}})\right| \) holds and thus, by \(\varphi _{1}({\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{1}({\textbf{A}})\right| >\left( 0.5\right) ^{s}\cdot \left| W_{n}({\textbf{A}})\right| =\varphi _{n}({\textbf{A}}, {\textbf{P}})\), we reach a contradiction to (s, N) satisfying ET.
Case 3.2 (\(\frac{2\,s-\left| N\right| +1}{2}=2\)). Denote by \(m_{12}^{\prime }\) and \(m_{12}^{\prime \prime }\) the two matches between players 1 and 2, and let \(m_{1n}\) be the one match played by player n (against player 1). Analogously to Case 3.1, it is easy to compute that, for each of the vectors of match winners in \(M({\textbf{A}}){\setminus } \{m_{12}^{\prime },m_{12}^{\prime \prime },m_{1n}\}\), three out of the eight possible combinations of winners in \(\{m_{12}^{\prime },m_{12}^{\prime \prime },m_{1n}\}\) give as a result a vector of match winners in \(W_{1}( {\textbf{A}})\) and only two give as a result a vector of match winners in \( W_{n}({\textbf{A}})\). Therefore \(\left| W_{1}({\textbf{A}})\right| >\left| W_{n}({\textbf{A}})\right| \) and thus, by \(\varphi _{1}( {\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{1}({\textbf{A}})\right| >\left( 0.5\right) ^{s}\cdot \left| W_{n}({\textbf{A}} )\right| =\varphi _{n}({\textbf{A}},{\textbf{P}})\), we again reach a contradiction to (s, N) satisfying ET.
Case 3.3 (\(\frac{2\,s-\left| N\right| +1}{2}>2\)). Fix \(w({\textbf{A}})\) and note that, given the definition of \({\textbf{A}}\), \(w( {\textbf{A}})\in W_{n}({\textbf{A}})\) only if, for any \(i\in N\), \(w_{m}({\textbf{A}})=i\) holds for at most one match \(m\in M({\textbf{A}})\). However, by \(\frac{ 2\,s-\left| N\right| +1}{2}>2\), the latter condition is violated for either player 1, or player 2, or for both players. We conclude then that \( W_{n}({\textbf{A}})=\emptyset \) should hold. Meanwhile, \(W_{1}({\textbf{A}})\ne \emptyset \) because player 1 is the winner of the competition if she wins all the matches that she plays. We have then \(\varphi _{1}({\textbf{A}}, {\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{1}({\textbf{A}} )\right| >0=\left( 0.5\right) ^{s}\cdot \left| \mathbf {\emptyset } \right| =\left( 0.5\right) ^{s}\cdot \left| W_{n}({\textbf{A}} )\right| =\varphi _{n}({\textbf{A}},{\textbf{P}})\) in contradiction to (s, N) satisfying ET. \(\square \)
Proof of Theorem 2
Let (s, N) be a league competition with either \(2s=\left| N\right| \) or \(2\,s=k\left( \left| N\right| -1\right) \left| N\right| \) for some integer \(k\ge 1\). Because ET is stronger than WET, it follows from Theorem 1 that (s, N) satisfies WET when \(2s=\left| N\right| \).
Consider now the case of \(2\,s=k\left( \left| N\right| -1\right) \left| N\right| \) for some integer \(k\ge 1\) and let \({\textbf{A}}\) be the assignment where each player plays k matches against every other player from N. By \(2\,s=k\left( \left| N\right| -1\right) \left| N\right| \), \({\textbf{A}}\in {\mathcal {A}}^{\left( s,N\right) }\) follows. We show now that (s, N) satisfies WET with respect to \({\textbf{A}}\). To that end, let \({\textbf{P}}\) be the probability matrix with \(p_{ij}=0.5\) for all \( i,j\in N\), and recall that \(\varphi _{q}({\textbf{A}},{\textbf{P}})=\left( 0.5\right) ^{s}\cdot \left| W_{q}({\textbf{A}})\right| \) holds for each \(q\in N\). It suffices then to prove that the cardinality of \(W_{q}( {\textbf{A}})\) is the same for all \(q\in N\). We proceed as follows.
Take
\(i,j,\ell \in N\) and for
\(x\in \left\{ 1,\ldots ,k\right\} \), denote by
\(m_{i\ell }^{x}\) (
\(m_{j\ell }^{x}\)) the
x-th match of player
i (
j) against player
\(\ell \). Denoting by
\(W({\textbf{A}})\) the set of all vectors of winners in the matches induced by
\({\textbf{A}}\), define
\(f:W_{j}({\textbf{A}} )\rightarrow W({\textbf{A}})\) as follows:
(1)
For \(\ell =i\) and each \(x\in \left\{ 1,\ldots ,k\right\} \), set \( f(w_{m_{j\ell }^{x}}({\textbf{A}}))\ne w_{m_{j\ell }^{x}}({\textbf{A}})\);
(2)
For each \(\ell \in N\setminus \left\{ i,j\right\} \) and each \(x\in \left\{ 1,\ldots ,k\right\} \):
-
if \(w_{m_{j\ell }^{x}}({\textbf{A}})=j\) and \(w_{m_{i\ell }^{x}}({\textbf{A}} )=\ell \), set \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=\ell \) and \(f(w_{m_{i\ell }^{x}}({\textbf{A}}))=i\);
-
if \(w_{m_{j\ell }^{x}}({\textbf{A}})=\ell \) and \(w_{m_{i\ell }^{x}}({\textbf{A}})=i\), set \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=j\) and \(f(w_{m_{i\ell }^{x}}( {\textbf{A}}))=\ell \);
-
if \(w_{m_{j\ell }^{x}}({\textbf{A}})=\ell \) and \(w_{m_{i\ell }^{x}}({\textbf{A}})=\ell \), set \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=\ell \) and \(f(w_{m_{i\ell }^{x}}({\textbf{A}}))=\ell \);
-
if \(w_{m_{j\ell }^{x}}({\textbf{A}})=j\) and \(w_{m_{i\ell }^{x}}({\textbf{A}} )=i \), set \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=j\) and \(f(w_{m_{i\ell }^{x}}( {\textbf{A}}))=i\);
(3)
For each \(q\in N\setminus \left\{ i,j\right\} \), each \(r\in N{\setminus } \left\{ q\right\} \), and each \(x\in \left\{ 1,\ldots ,k\right\} \): set \( f(w_{m_{qr}^{x}}({\textbf{A}}))=w_{m_{qr}^{x}}({\textbf{A}})\).
Notice that, by construction, the number of matches won by
i at
\(f(w( {\textbf{A}}))\) is the same as the number of matches won by
j at
\(w({\textbf{A}})\) and vice versa. Moreover, also by construction, for each
\(q\ne i,j\), the number of matches won by
q is the same in both
\(w({\textbf{A}})\) and
\( f(w({\textbf{A}}))\). Hence,
\(w({\textbf{A}})\in W_{j}({\textbf{A}})\setminus W_{i}( {\textbf{A}})\) implies
\(f(w({\textbf{A}}))\in W_{i}({\textbf{A}}){\setminus } W_{j}( {\textbf{A}})\), while
\(w({\textbf{A}})\in W_{j}({\textbf{A}})\cap W_{i}({\textbf{A}} ) \) implies
\(f(w({\textbf{A}}))\in W_{i}({\textbf{A}})\cap W_{j}({\textbf{A}})\) and, thus,
\(f(w({\textbf{A}}))\in W_{i}({\textbf{A}})\) holds.
We now show that \(w({\textbf{A}})\ne w^{\prime }({\textbf{A}})\) for \(w({\textbf{A}}),w^{\prime }({\textbf{A}})\in W_{j}({\textbf{A}})\) implies \(f(w({\textbf{A}} ))\ne f(w^{\prime }({\textbf{A}}))\).
If \(w_{m_{ij}^{x}}({\textbf{A}})\ne w_{m_{ij}^{x}}^{\prime }({\textbf{A}})\) or \( w_{m_{qr}^{x}}({\textbf{A}})\ne w_{m_{qr}^{x}}^{\prime }({\textbf{A}})\) holds for some \(q\in N{\setminus } \left\{ i,j\right\} \), \(r\in N{\setminus } \left\{ q\right\} \), and \(x\in \left\{ 1,\ldots ,k\right\} \), then \(f(w_{m_{ij}^{x}}( {\textbf{A}}))\ne f(w_{m_{ij}^{x}}^{\prime }({\textbf{A}}))\) and \( f(w_{m_{qr}^{x}}({\textbf{A}}))\ne f(w_{m_{qr}^{x}}^{\prime }({\textbf{A}}))\) clearly follows due to parts (1) and (3), respectively, of the above construction.
If
\(j=w_{m_{j\ell }^{x}}({\textbf{A}})\ne w_{m_{j\ell }^{x}}^{\prime }( {\textbf{A}})=\ell \) for some
\(x\in \left\{ 1,\ldots ,k\right\} \) and
\(\ell \in N\setminus \left\{ i,j\right\} \), then the following four cases are possible:
(a)
\(w_{m_{i\ell }^{x}}({\textbf{A}})=w_{m_{i\ell }^{x}}^{\prime }({\textbf{A}} )=i\). We have then \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=j\), \(f(w_{m_{i\ell }^{x}}({\textbf{A}}))=i\), \(f(w_{m_{j\ell }^{x}}^{\prime }({\textbf{A}}))=j\), \( f(w_{m_{i\ell }^{x}}^{\prime }({\textbf{A}}))=\ell \);
(b)
\(w_{m_{i\ell }^{x}}({\textbf{A}})=w_{m_{i\ell }^{x}}^{\prime }({\textbf{A}} )=\ell \). We have then \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=\ell \), \( f(w_{m_{i\ell }^{x}}({\textbf{A}}))=i\), \(f(w_{m_{j\ell }^{x}}^{\prime }( {\textbf{A}}))=\ell \), \(f(w_{m_{i\ell }^{x}}^{\prime }({\textbf{A}}))=\ell \);
(c)
\(w_{m_{i\ell }^{x}}({\textbf{A}})=i\) and \(w_{m_{i\ell }^{x}}^{\prime }( {\textbf{A}})=\ell \). We have then \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=j\), \( f(w_{m_{i\ell }^{x}}({\textbf{A}}))=i\), \(f(w_{m_{j\ell }^{x}}^{\prime }( {\textbf{A}}))=\ell \), \(f(w_{m_{i\ell }^{x}}^{\prime }({\textbf{A}}))=\ell \);
(d)
\(w_{m_{i\ell }^{x}}({\textbf{A}})=\ell \) and \(w_{m_{i\ell }^{x}}^{\prime }( {\textbf{A}})=i\). We have then \(f(w_{m_{j\ell }^{x}}({\textbf{A}}))=\ell \), \( f(w_{m_{i\ell }^{x}}({\textbf{A}}))=i\), \(f(w_{m_{j\ell }^{x}}^{\prime }( {\textbf{A}}))=j\), \(f(w_{m_{i\ell }^{x}}^{\prime }({\textbf{A}}))=\ell \).
Thus,
\(f(w({\textbf{A}}))\ne f(w^{\prime }({\textbf{A}}))\) holds for each of the possible cases.
Finally, if \(i=w_{m_{i\ell }^{x}}({\textbf{A}})\ne w_{m_{i\ell }^{x}}^{\prime }({\textbf{A}})=\ell \) for some \(x\in \left\{ 1,\ldots ,k\right\} \) and \(\ell \in N{\setminus } \left\{ i,j\right\} \) the situation is completely analogous to the previous one, so it can also be derived in this last case that \(f(w( {\textbf{A}}))\ne f(w^{\prime }({\textbf{A}}))\). We conclude then that f is a bijection, so \(\left| W_{i}({\textbf{A}})\right| =\) \(\left| W_{j}( {\textbf{A}})\right| \) holds. Hence, the league competition \(\left( s,N\right) \) satisfies WET with respect to \({\textbf{A}}\). \(\square \)
Proof of Theorem 3
We first show that if \(N=\left\{ 1,2\right\} \), then (s, N) satisfies MS. Let \({\textbf{P}}\in {\mathcal {P}}_{R}\) be an arbitrary but fixed probability matrix. Let \({\textbf{A}}\in {\mathcal {A}} ^{(s,N)}\) and notice that, by \(N=\left\{ 1,2\right\} \), \(a_{12}=s\). Now assume that \(p_{12}>0.5\). It needs to be shown that \(\varphi _{1}({\textbf{A}}, {\textbf{P}})-\varphi _{2}({\textbf{A}},{\textbf{P}})\ge 0\) holds. If \(s=1\) then the inequality follows immediately. Assume then that \(s\ge 2\) and consider the following two cases.
Case 1 (
\(2s>\left| N\right| =2\) and
s is odd). Note that in this case we have
$$\begin{aligned} \varphi _{1}({\textbf{A}},{\textbf{P}})=p_{12}^{s}+p_{12}^{s-1}\cdot p_{21}+p_{12}^{s-2}\cdot p_{21}^{2}+\ldots +p_{12}^{(s+1)/2}\cdot p_{21}^{(s+1)/2-1} \end{aligned}$$
and
$$\begin{aligned} \varphi _{2}({\textbf{A}},{\textbf{P}})=p_{21}^{s}+p_{21}^{s-1}\cdot p_{12}+p_{21}^{s-2}\cdot p_{12}^{2}+\ldots +p_{21}^{(s+1)/2}\cdot p_{12}^{(s+1)/2-1}. \end{aligned}$$
Thus,
\(\varphi _{1}({\textbf{A}},{\textbf{P}})-\varphi _{2}({\textbf{A}},{\textbf{P}} )=\left( p_{12}^{s}-p_{21}^{s}\right) +p_{12}\cdot p_{21}\cdot \left( p_{12}^{s-2}-p_{21}^{s-2}\right) +\ldots +p_{12}^{(s+1)/2-1}\cdot p_{21}^{(s+1)/2-1}\cdot \left( p_{12}-p_{21}\right) >0\), where the inequality follows from
\(p_{12}>0.5\). We conclude that (
s,
N) satisfies MS.
Case 2 (\(2s>\left| N\right| =2\) and s is even). When expressing \(\varphi _{1}({\textbf{A}},{\textbf{P}})\) and \(\varphi _{2}( {\textbf{A}},{\textbf{P}})\) for this case there are two differences compared to the corresponding expressions in Case 1. First, \((s+1)/2\) must be replaced by \(s/2+1\), and \((s+1)/2--1\) by \(s/2-1\). Second, in both expressions the probabilities of the vectors of match winners where both players, 1 and 2, are winners must be considered; that is, the term \(p_{12}^{s/2}+p_{21}^{s/2}\) must be added in both expressions. Since these vectors for player 1 and player 2 do coincide (due to s being even), the added terms are the same for both players. Thus, they cancel out when the difference between \(\varphi _{1}({\textbf{A}},{\textbf{P}})\) and \(\varphi _{2}({\textbf{A}},{\textbf{P}})\) is taken. By reproducing the same reasoning as in Case 1, we can conclude that (s, N) satisfies MS also in this case.
Now assume that (
s,
N) satisfies MS. We show that
\(\left| N\right| >2\) leads to a contradiction. Consider first the case where
\(N=\left\{ 1,2,3\right\} \) and the probability matrix
\({\textbf{P}}\in {\mathcal {P}}_{R}\) is as defined below:
$$\begin{aligned} {\textbf{P}}=\left( \begin{array}{ccc} 0.5 &{} 0.5+\varepsilon &{} 1-\varepsilon \\ &{} 0.5 &{} 1-2\varepsilon \\ &{} &{} 0.5 \end{array} \right) \end{aligned}$$
Take the assignment
\({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\) defined by
\(a_{12}=1\) and
\(a_{23}=s-1\). If
\(s=2\), then
\(\varphi _{1}({\textbf{A}},{\textbf{P}})\approx 0.5\) and
\(\varphi _{2}({\textbf{A}},{\textbf{P}})\approx 1\) in contradiction to
\( p_{12}>0.5\) and (
s,
N) satisfying MS.
If \(s>2\), given \({\textbf{A}}\) and \({\textbf{P}}\), we have that player 2 has a probability arbitrarily close to 1 of winning all matches except one and therefore a probability close to one of winning most of her matches. Thus, \( \varphi _{2}({\textbf{A}},{\textbf{P}})\approx 1\) and \(\varphi _{1}({\textbf{A}}, {\textbf{P}})\approx 0\), again reaching a contradiction to \(p_{12}>0.5\) and (s, N) satisfying MS.
Next assume that \(\left| N\right| >3\) holds and consider a probability matrix \({\textbf{P}}\in {\mathcal {P}}_{R}\) such that \(p_{23}>0.5\) and \(p_{3k}\approx 1\) for all \(k\in N{\setminus } \left\{ 1,2\right\} \). We distinguish the following two cases.
Case 1 (\(2s=\left| N\right| \)). In this case each player participates in exactly one match according to every assignment. Take \( {\textbf{A}}\in {\mathcal {A}}^{(s,N)}\) to be such that \(a_{12}=a_{34}=1\). Note that, for any vector of match winners, a player is a final winner of the entire competition only if she wins her single match according to \({\textbf{A}} \). Thus, \(\varphi _{3}({\textbf{A}},{\textbf{P}})\approx 1\) and \(\varphi _{2}( {\textbf{A}},{\textbf{P}})\le 0.5\) in contradiction to \(p_{23}>0.5\) and (s, N) satisfying MS.
Case 2 (\(2s>\left| N\right| \)). Consider the assignment \({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\) defined as follows: \(a_{12}=1\), \(a_{1j}=a_{2j}=0\) for all \(j\in N{\setminus } \left\{ 1,2\right\} \), and \( a_{3j}=1\) for each \(j\in N{\setminus } \left\{ 1,2\right\} \). Note that, according to \({\textbf{A}}\) and \({\textbf{P}}\), player 3 has a probability arbitrarily close to one of winning all her matches, so she has a probability close to one of winning every match in the competition except one (the one played between players 1 and 2). Hence, \(\varphi _{3}({\textbf{A}},{\textbf{P}})\approx 1\) and \(\varphi _{2}({\textbf{A}},{\textbf{P}})\approx 0\) in contradiction to \(p_{23}>0.5\) and (s, N) satisfying MS. \(\square \)
Proof of Theorem 4
We start with the case \(2\,s=\left| N\right| \). Let \({\textbf{P}}\in {\mathcal {P}}_{R}\) be an arbitrary but fixed probability matrix and, recalling that \(\left| N\right| \) is even, consider the assignment \({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\) defined for all \(i,j\in N\) as follows: \(a_{ij}=1\) if and only if \(i+j=\left| N\right| +1\). That is, \({\textbf{A}}\) matches the best player with the worst one, the second best with the second worst, and so on. Clearly, given \( {\textbf{A}}\), a player is a final winner in a vector of match winners if and only if she is the winner of the single match that she plays. Thus, for \( k\in N\), we have \(\varphi _{k}({\textbf{A}},{\textbf{P}})=p_{kk^{\prime }}\) with \(k^{\prime }\in N\) being the player such that \(a_{kk^{\prime }}=1\).
Now assume that \(p_{ij}>0.5\) holds for some \(i,j\in N\). It needs to be shown that \(\varphi _{i}({\textbf{A}},{\textbf{P}})\ge \varphi _{j}({\textbf{A}}, {\textbf{P}})\) follows in such a case. To that end, let \(i^{\prime },j^{\prime }\in N\) be such that \(a_{ii^{\prime }}=a_{jj^{\prime }}=1\). If \(i^{\prime }=j \), then \(\varphi _{i}({\textbf{A}},{\textbf{P}})=p_{ij}>p_{ji}=\varphi _{j}( {\textbf{A}},{\textbf{P}})\) follows. If \(i^{\prime }\ne j\), we have from \( p_{ij}>0.5\) that iRj holds, while \(j^{\prime }Ri^{\prime }\) holds due to the construction of \({\textbf{A}}\). Thus, \(p_{ii^{\prime }}\ge p_{jj^{\prime }}\) follows by (3). We then have \(\varphi _{i}({\textbf{A}},{\textbf{P}} )=p_{ii^{\prime }}\ge p_{jj^{\prime }}=\varphi _{j}({\textbf{A}},{\textbf{P}})\) as required for showing that (s, N) satisfies WMS with respect to \({\textbf{A}}\).
Now consider a league (
s,
N) such that
\(s\ge \left| N\right| -1\) holds. Fix a probability matrix
\({\textbf{P}}\in {\mathcal {P}}_{R}\) and consider the assignment
\({\textbf{A}}\) defined as follows:
-
\(a_{12}=s-\left| N\right| +2\),
-
\(a_{ij}=1\), if \(i=1\) and \(j\ne 2\),
-
\(a_{ij}=0\), if \(i\ne 1\). According to the above assignment,
-
player 1 plays all s matches (due to \(a_{12}+\left( \left| N\right| -2\right) \cdot 1=s\)),
-
each of the remaining players (except player 2) participates in exactly one match (which is, of course, against player 1), and
-
player 2 meets player 1 in each of the remaining (\(s-\left| N\right| +2\)) matches of the competition.
Clearly,
\({\textbf{A}}\in {\mathcal {A}}^{(s,N)}\) follows. We show that (
s,
N) satisfies WMS with respect to
\({\textbf{A}}\) in three steps. Below,
\(M^{12}( {\textbf{A}})\) stands for the set of
\(\left( s-\left| N\right| +2\right) \) matches played between 1 and 2 according to
\({\textbf{A}}\).
Step 1 If \(p_{12}>0.5\), then \(\varphi _{1}({\textbf{A}}, {\textbf{P}})\ge \varphi _{2}({\textbf{A}},{\textbf{P}})\).
Proof. Recall that \(\varphi _{1}({\textbf{A}},{\textbf{P}} )=\sum \limits _{w({\textbf{A}})\in W_{1}({\textbf{A}})}pr(w({\textbf{A}}))\) and \( \varphi _{2}({\textbf{A}},{\textbf{P}})=\sum \limits _{w({\textbf{A}})\in W_{2}( {\textbf{A}})}pr(w({\textbf{A}}))\). Given a vector \(w({\textbf{A}})\) of match winners, let \(M_{w_{m}({\textbf{A}})=1}^{12}(w({\textbf{A}}))\) be the set of matches in \(M^{12}({\textbf{A}})\) which player 1 wins, and let \(M_{w_{m}( {\textbf{A}})=2}^{12}(w({\textbf{A}}))\) be the set of matches in \(M^{12}({\textbf{A}})\) which player 2 wins.
Define the mapping \(f:W_{2}({\textbf{A}})\rightarrow W_{1}({\textbf{A}})\) by just interchanging the match winners for each match in \(M^{12}(w({\textbf{A}} )) \). Notice that for \(w({\textbf{A}}),w^{\prime }({\textbf{A}})\in W_{2}( {\textbf{A}})\) with \(w({\textbf{A}})\ne w^{\prime }({\textbf{A}})\), \(f(w({\textbf{A}}))\ne f(w^{\prime }({\textbf{A}}))\) follows and thus, f is a bijection between \(W_{2}({\textbf{A}})\) and a subset of \(W_{1}({\textbf{A}})\).
Note also that, for any vector \(w({\textbf{A}})\) of match winners, we have: \( pr(w({\textbf{A}}))=\prod \limits _{m\in M({\textbf{A}}){\setminus } M^{12}({\textbf{A}} )}\varphi _{w_{m}({\textbf{A}})}({\textbf{A}},{\textbf{P}})\cdot \left( p_{12}\right) ^{\left| M_{w_{m}({\textbf{A}})=1}^{12}(w({\textbf{A}} ))\right| }\cdot \left( p_{21}\right) ^{\left| M_{w_{m}({\textbf{A}} )=2}^{12}(w({\textbf{A}}))\right| }\).
Finally, note that, by \(w({\textbf{A}})\in W_{2}({\textbf{A}})\), in \(w({\textbf{A}} )\) player 2 wins at least as many matches against player 1 as player 1 against player 2; that is, \(\left| M_{w_{m}({\textbf{A}})=2}^{12}(w( {\textbf{A}}))\right| \ge \left| M_{w_{m}({\textbf{A}})=1}^{12}(w( {\textbf{A}}))\right| \). Thus, \(\left| M_{f(w({\textbf{A}} ))_{m}=1}^{12}(w({\textbf{A}}))\right| \ge \left| M_{f(w({\textbf{A}} ))_{m}=2}^{12}(w({\textbf{A}}))\right| \). We then have from \(p_{12}>p_{21}\) that \(\left( p_{12}\right) ^{\left| M_{w_{m}({\textbf{A}})=1}^{12}(w( {\textbf{A}}))\right| }\cdot \left( p_{21}\right) ^{\left| M_{w_{m}( {\textbf{A}})=2}^{12}(w({\textbf{A}}))\right| }\le \left( p_{12}\right) ^{\left| M_{f(w({\textbf{A}}))_{m}=1}^{12}(w({\textbf{A}}))\right| }\cdot \left( p_{21}\right) ^{\left| M_{f(w({\textbf{A}}))_{m}=2}^{12}(w( {\textbf{A}}))\right| }\) holds. Therefore, \(pr(w({\textbf{A}}))\le pr(f(w( {\textbf{A}})))\) holds for each \(w({\textbf{A}})\in W_{2}({\textbf{A}})\), which enables us to conclude that \(\varphi _{1}({\textbf{A}},{\textbf{P}})\ge \varphi _{2}({\textbf{A}},{\textbf{P}})\).
Step 2 If \(p_{ij}>0.5\) for \(i\in \left\{ 1,2\right\} \) and \(j\in N{\setminus } \left\{ 1,2\right\} \), then \(\varphi _{i}({\textbf{A}}, {\textbf{P}})\ge \varphi _{j}({\textbf{A}},{\textbf{P}})\).
Proof
Note first that if \(\left| M^{12}({\textbf{A}} )\right| >2\) then \(W_{j}({\textbf{A}})=\emptyset \) and \(\varphi _{i}( {\textbf{A}},{\textbf{P}})\ge 0=\varphi _{j}({\textbf{A}},{\textbf{P}})\) follow immediately.
If \(\left| M^{12}({\textbf{A}})\right| =2\), then \(w({\textbf{A}})\in W_{j}({\textbf{A}})\) implies that every player wins exactly one match, i.e., all players are winners of the entire competition. Thus, \(w({\textbf{A}})\in W_{j}({\textbf{A}})\) implies \(w({\textbf{A}})\in W_{i}({\textbf{A}})\); that is, \( W_{j}({\textbf{A}})\subseteq W_{i}({\textbf{A}})\) holds and hence, \(\varphi _{i}( {\textbf{A}},{\textbf{P}})\ge \varphi _{j}({\textbf{A}},{\textbf{P}})\) follows.
Now assume that \(\left| M^{12}({\textbf{A}})\right| =1\). Recall that only vectors of match winners in \(W_{i}({\textbf{A}})\setminus W_{j}({\textbf{A}} )\) and in \(W_{j}({\textbf{A}})\setminus W_{i}({\textbf{A}})\) do matter for the comparison of \(\varphi _{i}({\textbf{A}},{\textbf{P}})\) and \(\varphi _{j}( {\textbf{A}},{\textbf{P}})\). If \(i=1\), then \(W_{j}({\textbf{A}}){\setminus } W_{1}( {\textbf{A}})=\left\{ w({\textbf{A}})\right\} \) with \(w({\textbf{A}})\) consisting of player 1 losing all her matches and thus \(pr(w({\textbf{A}}))=p_{21}\cdot p_{31}\cdot \ldots \cdot p_{n1}\). Meanwhile, the vector \(w^{\prime }({\textbf{A}})\) of match winners where player 1 wins all her matches definitely belongs to \(W_{1}({\textbf{A}}){\setminus } W_{j}({\textbf{A}})\), with \(pr(w^{\prime }( {\textbf{A}}))=p_{12}\cdot p_{13}\cdot \ldots \cdot p_{1n}\). Thus, given that \( p_{1k}\ge p_{k1}\) for each \(k\in N{\setminus } \left\{ 1\right\} \) (with strict inequality holding for \(k=j\)), we have that \(\varphi _{1}({\textbf{A}}, {\textbf{P}})>\varphi _{j}({\textbf{A}},{\textbf{P}})\).
In contrast, \(i=2\) implies \(W_{j}({\textbf{A}}){\setminus } W_{2}({\textbf{A}} )=\left\{ w({\textbf{A}})\right\} \) with \(w({\textbf{A}})\) consisting of 1 only winning her match against 2 and \(pr(w({\textbf{A}}))=p_{12}\cdot p_{31}\cdot p_{41}\cdot \ldots \cdot p_{n1}\). Consider now the vector \(w^{\prime }( {\textbf{A}})\) of match winners which differs from \(w({\textbf{A}})\) only in that at \(w^{\prime }({\textbf{A}})\) player 2 wins her single match against player 1 and player j loses her single match against player 1. This involves that \(w^{\prime }({\textbf{A}})\in W_{2}({\textbf{A}}){\setminus } W_{j}( {\textbf{A}})\) with \(pr(w^{\prime }({\textbf{A}}))\ge pr(w({\textbf{A}}))\) due to \(p_{21}\ge p_{j1}\) following from \(p_{2j}>0.5\) and condition (2). We conclude then that \(\varphi _{2}({\textbf{A}},{\textbf{P}})\ge \varphi _{j}( {\textbf{A}},{\textbf{P}})\).
Step 3 If \(p_{ij}>0.5\) for some \(i,j\in N\setminus \left\{ 1,2\right\} \), then \(\varphi _{i}({\textbf{A}},{\textbf{P}})\ge \varphi _{j}( {\textbf{A}},{\textbf{P}})\).