In Laplace’s rule of succession (Example 5e), suppose that the first $n$flips resulted in r heads and $n-r$tails. Show that the probability that the$(n+1)$flip turns up heads is $(r+1)/(n+2)$. To do so, you will have to prove and use the identity

${\int }_0^1{y}^n(1-y{)}^{m}dy=\frac{n!m!}{(n+m+1)!}$

Hint: To prove the identity, let $C(n,m)={\int }_0^1{y}^n(1-y{)}^{m}dy$. Integrating by parts yields

$C(n,m)=\frac{m}{n+1}C(n+1,m-1)$

Starting with $C(n,0)=1/(n+1)$, prove the identity by induction on $m$.

$C_i$ - the coin with probability $\frac{i}{k}$ of flipping heads is chosen, $i=0,1,\dots ,k$.

$F_{n,m}$ - the first $n+m$ flips resulted in $n$ heads and $m$ tails.

$H-n+m+1$. flip is heads.

Calculation of $P\left(H\mid F_{n,m}\right)$

Same logic as in Example 5e: Conditioning on which coin is chosen, we get the formula:

$P\left(H\mid F_{n,m}\right)=\sum _{i=0}^{k}P\left(H\mid C_i{F}_{n,m}\right)P\left(C_i\mid F_{n,m}\right)$

After applying formula for conditional probability and formula for total probability, as in the Example 5e:

$P\left(C_i\mid F_{n,m}\right)=\frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^n{\left(1-\frac{i}{k}\right)}^m}{{\sum }_{j=0}^k\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1-\frac{j}{k}\right)}^m}$

Substituting this in formula (1) this is obtained:

$P\left(H\mid F_{n,m}\right)=\sum _{i=0}^k\frac{\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{i}{k}\right)}^{n+1}{\left(1-\frac{i}{k}\right)}^m}{{\sum }_{j=0}^k\left(\begin{array}{c}n+m\\ n\end{array}\right){\left(\frac{j}{k}\right)}^n{\left(1-\frac{j}{k}\right)}^m}$

More nicely put:

Let $C_{n,m}$denote the integral approximation of the following expression:

$\frac{1}{k}\sum _{i=0}^k{\left(\frac{i}{k}\right)}^n{\left(1-\frac{i}{k}\right)}^m\approx {\int }_0^1{y}^n(1-y{)}^{m}dy=:C_{n,m}$

Then the wanted probability would be approximately:

$P\left(H\mid F_{n,m}\right)\approx \frac{C_{n+1,m}}{C_{n,m}}$

As hinted, first use partial integration :

$={\int }_0^1\frac{1}{n+1}{u}^{\text{'}}\left(y\right)v\left(y\right)dy$

$=\frac{1}{n+1}\left[{\left.\left(u\right(y\left)v\right(y\left)\right)\right|}_0^1-{\int }_0^{1}u\left(y\right)v^{\text{'}}\left(y\right)dy\right]$

$=\frac{1}{n+1}\left[0-{\int }_0^1{y}^{n+1}·(-m)(1-y{)}^{m-1}dy\right]$

$=\frac{m}{n+1}{\int }_0^1{y}^{n+1}(1-y{)}^{m-1}dy$

$=\frac{m}{n+1}{C}_{n+1,m-1}$

And integration shows that:

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

By repeating recursion $m$times, until $0$is reached as the second index the formula becomes explicit:

$C_{n,m}=\frac{m}{n+1}·\frac{m-1}{n+2}·\frac{m-2}{n+3}·\cdots ·\frac{1}{n+1+m-1}·C_{n+m,0}$

$=\frac{m!}{\frac{(n+m)!}{n!}}·\frac{1}{n+m+1}$

$=\frac{n!m!}{(n+m+1)!}$

Now returning to the wanted probabiltiy $P\left(H\mid F_{n,m}\right)$

The required probability is$\left.P\left(H\mid F_{n,m}\right)\approx \frac{C_{n+1,m}}{C_{n,m}}\right)=\frac{n+1}{n+m+2}$.