\frac{M^3-4M}{24} & \mbox{if $2\mid M$}\end{array}\right.$. If ${{\cal T}}$ is irreducible, the $3$-cycles are at least $M-2$.

#### Number of tournaments

The number of tournaments with $M$ nodes (up to relabeling) is $T(M)=\displaystyle{\sum_{(d)}\frac{2^D}{N}}$, where

-   $(d)=(d_1,d_2,\dotsc,d_M)$ is multi-index with $d_{2i}=0$, $d_{2i+1}\geq0$ and $\sum_{i=1}^{M}i\cdot d_i=M$,

-   $D$ is $\frac12\left(\sum_{i,j=1}^{M}d_id_j\gcd(i,j)-\sum_{i=1}^{M}d_i\right)$,

-   $N$ is $\prod_{i=1}^M i^{d_i}d_i!$.

The values of $T(M)$ for $M\leqslant12$ are given in Table \[tab:num\_tournaments\].

  ----- -------- -- ----- -------- -- ----- --------------
    $M$   $T(M)$      $M$   $T(M)$      $M$         $T(M)$
      1        1        5       12        9         191536
      2        1        6       56       10        9733056
      3        2        7      456       11      903753248
      4        4        8     6880       12   154108311168
  ----- -------- -- ----- -------- -- ----- --------------

  : The number of tournaments.[]{data-label="tab:num_tournaments"}

As $M$ tends to infinity, $T(M)\rightarrow\infty$ and $T(M)\sim\displaystyle{\frac{2^{\binom{M}{2}}}{M!}}$ hold.

The probability $P(M)$ that a tournament with $M$ nodes is irreducible can be computed recursively by the formula $$P(M)=1-\sum_{i=1}^{M-1}\binom{M}{i}\frac{P(i)}{2^{t(M-t)}}.$$ The values of $P(M)$ for $M\leqslant16$ are given in Table \[tab:prob\_irred\_tournament\].

  ----- -------- -- ----- ---------- -- ----- ---------- -- ----- ----------
    $M$ $P(M)$        $M$ $P(M)$          $M$ $P(M)$          $M$ $P(M)$
      1 1               5 0.53125           9 0.931702         13 0.993671
      2 0               6 0.681152         10 0.961589         14 0.996587
      3 0.25            7 0.799889         11 0.978720         15 0.998171
      4 0.375           8 0.881115         