91影视

By choosig coin condition,

$P\left(H鈭�F_{n,m}\right)=\underset{i=0}{\overset{k}{鈭�}}鈥�P\left(H鈭�C_i{F}_{n,m}\right)P\left(C_i鈭�F_{n,m}\right)$

Probability head of coin,

$$\begin{gathered} P\left(H鈭�C_i{F}_{n,m}\right)=P\left(H鈭�C_i\right)=\frac{i}{k} \\ \begin{array}{l}P\left(C_i鈭�F_{n,m}\right)=\frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^n{\left(1鈭�\frac{i}{k}\right)}^m}{\underset{j=0}{\overset{k}{鈭�}}鈥�\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1鈭�\frac{j}{k}\right)}^m}\\ P\left(H鈭�F_{n,m}\right)=\underset{i=0}{\overset{k}{鈭�}}鈥�\frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^{n+1}{\left(1鈭�\frac{i}{k}\right)}^m}{\underset{j=0}{\overset{k}{鈭�}}鈥�\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1鈭�\frac{j}{k}\right)}^m}\\ P\left(H鈭�F_{n,m}\right)=\frac{\underset{i=0}{\overset{k}{鈭�}}鈥�{\left(\frac{i}{k}\right)}^{n+1}{\left(1鈭�\frac{i}{k}\right)}^m}{\underset{j=0}{\overset{k}{鈭�}}鈥�{\left(\frac{j}{k}\right)}^n{\left(1鈭�\frac{j}{k}\right)}^m}\end{array} \end{gathered}$$

By integral approximation of expression,

$\frac{1}{k}\underset{i=0}{\overset{k}{鈭�}}鈥�{\left(\frac{i}{k}\right)}^n{\left(1鈭�\frac{i}{k}\right)}^m鈮�{鈭�}_0^1鈥�y^n(1鈭�y{)}^{m}dy=:C_{n,m}$

The wanted probability approximation as,

$P\left(H鈭�F_{n,m}\right)鈮�\frac{C_{n+1,m}}{C_{n,m}}.$

By using partial integration,

$\begin{array}{l}{C}_{n,m}={鈭�}_0^1鈥�y^n(1鈭�y{)}^{m}dy=\left[\begin{array}{cc}{y}^{n+1}=u\left(y\right)& u^{\mathrm{鈥�}}\left(y\right)=(n+1)y^n\\ (1鈭�y{)}^m鈭�v(y)& v^{\mathrm{鈥�}}\left(y\right)=鈭�m(1鈭�y{)}^{m鈭�1}\end{array}\right]\\ C_{n,m}={鈭�}_0^1鈥�\frac{1}{n+1}{u}^{\mathrm{鈥�}}\left(y\right)v\left(y\right)dy\\ C_{n,m}=\frac{1}{n+1}\left[{\left.\left(u\right(y\left)v\right(y\left)\right)\right|}_0^1鈭�{鈭�}_0^1鈥�u\left(y\right)v^{\mathrm{鈥�}}\left(y\right)dy\right]\\ C_{n,m}=\frac{1}{n+1}\left[0鈭�{鈭�}_0^1鈥�y^{n+1}脳(鈭�m)(1鈭�y{)}^{m鈭�1}dy\right]\\ C_{n,m}=\frac{m}{n+1}{鈭�}_0^1鈥�y^{n+1}(1鈭�y{)}^{m鈭�1}dy\\ C_{n,m}=\frac{m}{n+1}{C}_{n+1,m鈭�1}.\end{array}$

The wanted recursion is

$\begin{array}{c}{C}_{n,m}=\frac{m}{n+1}{C}_{n+1,m鈭�1}\\ \end{array}$

and the integration as

$C_{n,0}={鈭�}_0^1鈥�y^{n}dy=\frac{1}{n+1}$

By repeating recursion as $m$times until second index reaches at $0$becomes as

$\begin{array}{l}{C}_{n,m}=\frac{m}{n+1}脳\frac{m鈭�1}{n+2}脳\frac{m鈭�2}{n+3}脳鈥�脳\frac{1}{n+1+m鈭�1}脳C_{n+m,0}\\ C_{n,m}=\frac{m!}{\frac{(n+m)!}{n!}脳\frac{1}{n+m+1}}\\ C_{n,m}=\frac{n!m!}{(n+m+1)!}.\end{array}$

The wanted probability as

$$\begin{gathered} \left.P\left(H鈭�F_{n,m}\right)鈮�\frac{C_{n+1,m}}{C_{n,m}}\right)=\frac{\frac{(n+1)!m!}{(n+m+2)!}}{\frac{n!m!}{(n+m+1)!}} \\ \left.P\left(H鈭�F_{n,m}\right)鈮�\frac{{\displaystyle C_{n+1,m}}}{{\displaystyle C_{n,m}}}\right)=\frac{n+1}{n+m+2}. \end{gathered}$$