由题意，设甲获胜的局数为${X}$，则$X\sim B\left( 2n,\dfrac{1}{2} \right)$，$P(X=k)=\text{\rm {C}}_{2n}^{k}{{\left( \dfrac{1}{2} \right)}^{k}}{{\left( 1-\dfrac{1}{2} \right)}^{2n-k}}=\text{\rm {C}}_{2n}^{k}{{\left( \dfrac{1}{2} \right)}^{2n}}$，
故甲赢得比赛的概率为：
$P\left( n \right)=P\left( X\gt n \right)=\sum\limits_{k=n+1}^{2n}{\text{\rm {C}}_{2n}^{k}{{\left( \dfrac{1}{2} \right)}^{2n}}}={{\left( \dfrac{1}{2} \right)}^{2n}}\cdot \sum\limits_{k=n+1}^{2n}{\text{\rm {C}}_{2n}^{k}}$，
又因$\sum\limits_{k=n+1}^{2n}{\text{\rm {C}}_{2n}^{k}}=\sum\limits_{k=0}^{n-1}{\text{\rm {C}}_{2n}^{k}}$，$\sum\limits_{k=n+1}^{2n}{\text{\rm {C}}_{2n}^{k}}+\sum\limits_{k=0}^{n-1}{\text{\rm {C}}_{2n}^{k}}+\text{\rm {C}}_{2n}^{n}={{2}^{2n}}$，
$\therefore \sum\limits_{k=n+1}^{2n}{\text{\rm {C}}_{2n}^{k}}=\dfrac{{{2}^{2n}}}{2}-\dfrac{1}{2}\text{\rm {C}}_{2n}^{n}={{2}^{2n-1}}-\dfrac{1}{2}\text{\rm {C}}_{2n}^{n}$，
故$P\left( n \right)={{\left( \dfrac{1}{2} \right)}^{2n}}\cdot \left( {{2}^{2n-1}}-\dfrac{1}{2}\complement _{2n}^{n} \right)=\dfrac{1}{2}-\dfrac{\text{\rm {C}}_{2n}^{n}}{{{2}^{2n+1}}}$，故$\rm C$正确；
$P(2)=\dfrac{1}{2}-\dfrac{\text{\rm {C}}_{4}^{2}}{{{2}^{5}}}=\dfrac{5}{16}$，故$\rm A$错误；$P(3)=\dfrac{1}{2}-\dfrac{\text{\rm {C}}_{6}^{3}}{{{2}^{7}}}=\dfrac{11}{32}$，故$\rm B$错误；
$\because P\left( n \right)=\dfrac{1}{2}-\dfrac{\text{\rm {C}}_{2n}^{n}}{{{2}^{2n1}}}=\dfrac{1}{2}\left( 1-\dfrac{\text{\rm {C}}_{2n}^{n}}{{{2}^{2n}}} \right)$，
$\therefore P\left( n+1 \right)=\dfrac{1}{2}\left( 1-\dfrac{\text{\rm {C}}_{2n+2}^{n+1}}{{{2}^{2n+2}}} \right)$，
又$\because \dfrac{\dfrac{\text{\rm {C}}_{2n}^{n}}{{{2}^{2n}}}}{\dfrac{\text{\rm {C}}_{2n+2}^{n+1}}{{{2}^{2n+2}}}}=\dfrac{4\text{\rm {C}}_{2n}^{n}}{\text{\rm {C}}_{2n+2}^{n+1}}=\dfrac{4\cdot \dfrac{\left( 2n \right)!}{n!n!}}{\dfrac{\left( 2n+2 \right)!}{\left( n+1 \right)!\left( n+1 \right)!}}=\dfrac{4{{\left( n+1 \right)}^{2}}}{\left( 2n+2 \right)\left( 2n+1 \right)}=\dfrac{2\left( n+1 \right)}{2n+1}\gt 1$，
$\therefore \dfrac{\text{\rm {C}}_{2n}^{n}}{{{2}^{2n}}}\gt \dfrac{\text{\rm {C}}_{2n+2}^{n+1}}{{{2}^{2n+2}}}$，
$\therefore P\left( n \right)\lt P\left( n+1 \right)$，即$P(n)$单调递增，故$\rm D$正确$.$
故选：$\rm CD$