In Laplace’s rule of succession (Example 5e), show that if the first $n$ flips all result in heads, then the conditional probability that the next m flips also result in all heads is$(n+1)/(n+m+1)$.

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

$F_n$ - the first $n$ flips of a chosen coin are heads.

$H_{n+1}-n+1$-st flip is heads.

Conditioning on which coin is chosen, we get the formula:

$\left.F_n\right)=\frac{{\sum }_{i=0}^k{\left(\frac{i}{k}\right)}^{n+1}}{{\sum }_{j=0}^k{\left(\frac{j}{k}\right)}^n}$

Also for large $k$:

$\frac{1}{k}\sum _{i=0}^k{\left(\frac{i}{k}\right)}^n\approx {\int }_0^1{x}^{n}dx=\frac{1}{n+1}$

Thus, for large $k$:

$P\left(H_{n+1}\mid F_n\right)\approx \frac{n+1}{n+2}$

Calculation of $P\left(H_{n+m}{H}_{n+m-1}\dots H_{n+1}\mid F_n\right)$

Use the formula of conditional probability (first row) and multiplication rule (second row):

$P\left(H_{n+m}{H}_{n+m-1}\dots H_{n+1}\mid F_n\right)$

$=\frac{P\left(H_{n+m}{H}_{n+m-1}\dots H_{n+1}{F}_n\right)}{P\left(F_n\right)}$

$=\frac{P\left(H_{n+m}\mid H_{n+m-1}\dots H_{n+1}{F}_n\right)·P\left(H_{n+m-1}\mid H_{n+m-2}\dots H_{n+1}{F}_n\right)·\dots ·P\left(F_n\right)}{P\left(F_n\right)}$

$=P\left(H_{n+m}\mid H_{n+m-1}\dots H_{n+1}{F}_n\right)·\dots ·P\left(H_{n+1}\mid F_n\right)$

$=P\left(H_{n+m}\mid F_{n+m-1}\right)·P\left(H_{n+m-1}\mid F_{n+m-2}\right)·\dots ·P\left(H_{n+1}\mid F_n\right)$

For as little approximation as possible, substitute formula (1) into this equation directly:

$P\left(H_{n+m}{H}_{n+m-1}\dots H_{n+1}\mid F_n\right)$

$=\frac{{\sum }_{i=0}^k{\left(\frac{i}{k}\right)}^{n+m}}{{\sum }_{j=0}^k{\left(\frac{j}{k}\right)}^{n+m-1}}·\frac{{\sum }_{i=0}^k{\left(\frac{i}{k}\right)}^{n+m-1}}{{\sum }_{j=0}^k{\left(\frac{j}{k}\right)}^{n+m-2}}·\cdots ·\frac{{\sum }_{i=0}^k{\left(\frac{i}{k}\right)}^{n+1}}{{\sum }_{j=0}^k{\left(\frac{j}{k}\right)}^n}$

$=\frac{{\sum }_{i=0}^k{\left(\frac{i}{k}\right)}^{n+m}}{{\sum }_{j=0}^k{\left(\frac{j}{k}\right)}^n}$

All the fractions are reduced because by changing the name of the iterator the sums become identical.

Now use approximations:

$\frac{1}{k}\sum _{i=0}^k{\left(\frac{i}{k}\right)}^n\approx {\int }_0^1{x}^{n}dx=\frac{1}{n+1}$

Therefore:

The product of probability is$P\left(H_{n+m}{H}_{n+m-1}\dots H_{n+1}\mid F_n\right)\approx \frac{n+1}{n+m+1}$.