投壶是从先秦延续至清末的中国传统礼仪和宴饮游戏，投壶礼来源于射礼.投壶的横截面是三个圆形，投掷者站在距离投壶一定距离的远处将箭羽投向三个圆形的壶口，若箭羽投进三个圆形壶口之一就算投中.为弘扬中华传统文化，某次文化活动进行了投壶比赛，比赛规定投进中间较大圆形壶口得$3$分，投进左右两个小圆形壶口得$1$分，没有投进壶口不得分.甲乙两人进行投壶比赛，比赛分为若干轮，每轮每人投一支箭羽，最后将各轮所得分数相加即为该人的比赛得分，比赛得分高的人获胜.已知甲每轮投一支箭羽进入中间大壶口的概率为$\dfrac{1}{3}$，投进入左右两个小壶口的概率都是$\dfrac{1}{6}$，乙每轮投一支箭羽进入中间大壶口的概率为$\dfrac{1}{4}$，投进入左右两个小壶口的概率分别是$\dfrac{1}{5}$和$\dfrac{1}{6}$，甲乙两人每轮是否投中相互独立，且两人各轮之间是否投中也互相独立.若在最后一轮比赛前，甲的总分落后乙$1$分，设甲最后一轮比赛的得分为$X$，乙最后一轮比赛的得分为$Y$.
$(1)$求甲最后一轮结束后赢得比赛的概率；
$(2)$求$|X-Y|$的数学期望.

(1) $\\dfrac{1}{4}$

(2) $\\dfrac{233}{180}$

$(1)$设甲一轮的得分为$\xi $，
则$P\left(\xi =1\right)=\dfrac{1}{6}×2=\dfrac{1}{3}$，$P\left(\xi =3\right)=\dfrac{1}{3}$，$P\left(\xi =0\right)=1-P\left(\xi =1\right)-P\left(\xi =3\right)=\dfrac{1}{3}$；
设乙一轮的得分为$\eta $，
则$P\left(\eta =1\right)=\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{11}{30}$，$P\left(\eta =3\right)=\dfrac{1}{4}$，$P\left(\eta =0\right)=1-P\left(\eta =1\right)-P\left(\eta =3\right)=\dfrac{23}{60}$；
则甲最后一轮反败为胜的概率$P=P\left(X=3,Y=0\right)+P\left(X=3,Y=1\right)=\dfrac{1}{3}×\dfrac{23}{60}+\dfrac{1}{3}×\dfrac{11}{30}=\dfrac{1}{4}$.
$(2)$由题意知：$|X-Y|$所有可能的取值为$0$，$1$，$2$，$3$，
$P\left(|X-Y|=0\right)=P\left(X=0,Y=0\right)+P\left(X=1,Y=1\right)+P\left(X=3,Y=3\right)=\dfrac{1}{3}×\dfrac{23}{60}+\dfrac{1}{3}×\dfrac{11}{30}+\dfrac{1}{3}×\dfrac{1}{4}=\dfrac{1}{3}$，
$P\left(|X-Y|=1\right)=P\left(X=0,Y=1\right)+P\left(X=1,Y=0\right)=\dfrac{1}{3}×\dfrac{11}{30}+\dfrac{1}{3}×\dfrac{23}{60}=\dfrac{1}{4}$，
$P\left(|X-Y|=2\right)=P\left(X=1,Y=3\right)+P\left(X=3,Y=1\right)=\dfrac{1}{3}×\dfrac{1}{4}+\dfrac{1}{3}×\dfrac{11}{30}=\dfrac{37}{180}$，
$P\left(|X-Y|=3\right)=P\left(X=0,Y=3\right)+P\left(X=3,Y=0\right)=\dfrac{1}{3}×\dfrac{1}{4}+\dfrac{1}{3}×\dfrac{23}{60}=\dfrac{19}{90}$，
$\therefore |X-Y|$的数学期望$E\left(|X-Y|\right)=0\times \dfrac{1}{3}+1\times \dfrac{1}{4}+2\times \dfrac{37}{180}+3\times \dfrac{19}{90}=\dfrac{233}{180}$.