每日一题[1880]概率分布

已知正三角形被线段划分成 $9$ 个小正三角形，$1$ 个小球从 $P$ 区域出发，每 $1$ 秒后等可能的到达相邻的区域，求 $n$ 秒后小球到达 $Q$ 区域的概率．

答案    $\begin{cases} 0,&n=2k-1,\\ \dfrac13\left(1-\dfrac{1}{2^k}\right),&n=2k.\end{cases}$．

解析    如图染色，则奇数秒后小球位于白色区域，偶数秒后小球位于蓝色区域．

设 $n$ 秒后到到达 $P,Q,R$ 区域的概率分别为 $P_n,Q_n,R_n$，则当 $n=2k-1$ 时，$P_n=Q_n=R_n=0$；当 $n=2k$ 时，有\[\begin{cases} P_{2(k+1)}=P_{2k}\left(\dfrac 13\cdot 1+\dfrac 13\cdot \dfrac 12+\dfrac 13\cdot \dfrac 12\right)+Q_{2k}\left(\dfrac13\cdot \dfrac 12\right)+R_{2k}\left(\dfrac 13\cdot \dfrac 12\right),\\ Q_{2(k+1)}=P_{2k}\left(\dfrac 13\cdot\dfrac 12\right)+Q_{2k}\left(\dfrac13\cdot \dfrac 12+\dfrac 13\cdot 1+\dfrac 13\cdot \dfrac 12\right)+R_{2k}\left(\dfrac 13\cdot \dfrac 12\right), \\ R_{2(k+1)}=P_{2k}\left(\dfrac 13\cdot \dfrac 12\right)+Q_{2k}\left(\dfrac13\cdot \dfrac 12\right)+R_{2k}\left(\dfrac13\cdot \dfrac 12+\dfrac 13\cdot 1+\dfrac 13\cdot \dfrac 12\right), \end{cases}\]而事实上 $Q_n=R_n$，因此\[\begin{cases} P_{2(k+1)}=\dfrac 23P_{2k}+\dfrac 13Q_{2k},\\ Q_{2(k+1)}=\dfrac 16P_{2k}+\dfrac 56Q_{2k},\end{cases}\]于是\[P_{2(k+1)}+\alpha\cdot Q_{2(k+1)}=\left(\dfrac 23+\dfrac 16\alpha\right)P_{2k}+\left(\dfrac 13+\dfrac 56\alpha\right)Q_{2k},\]且 $P_0=1$，$Q_0=0$，令\[\dfrac{\dfrac 13+\dfrac 56\alpha}{\dfrac 23+\dfrac 16\alpha}=\alpha\iff x=-1\lor x=2,\]于是\[\begin{cases} P_{2(k+1)}-Q_{2(k+1)}=\dfrac 12\left(P_{2k}-Q_{2k}\right),\\ P_{2(k+1)}+2Q_{2(k+1)}=P_{2(k+1)}+2Q_{2(k+1)},\end{cases}\implies \begin{cases} P_{2k}-Q_{2k}=\dfrac{1}{2^k},\\ P_{2k}+2Q_{2k}=1,\end{cases}\]解得 $Q_{2k}=\dfrac 13\left(1-\dfrac{1}{2^k}\right)$． 综上所述，$n$ 秒后小球到达 $Q$ 区域的概率 $Q_n=\begin{cases} 0,&n=2k-1,\\ \dfrac13\left(1-\dfrac{1}{2^k}\right),&n=2k.\end{cases}$