91影视

A coin that comes up heads with probability p is flipped until either a total of n heads or of m tails is amassed.

From the given information we need to calculate the results,

$min(X,Y)+max(X,Y)=X+Y$

$min(X,Y)=X+Y鈭�max(X,Y)$

$E[min(X,Y\left)\right]=E\left[X\right]+E\left[Y\right]鈭�E[max(X,Y\left)\right]$

The expectation values of X and Y are given as follows:

$E\left[X\right]=\frac{n}{p}$

$E\left[Y\right]=\frac{m}{1鈭�p}$

conditioning on N, E[M] is given below,

$E\left[M\right]=鈭�E[M鈭�N=i]P\{N=i\}$

If $i<n$then $n-i$heads need to be obtained,

If $i>n$then $m鈭�(n+m鈭�1鈭�i)$tails need to be obtained.

So,

$E\left[M\right]=\underset{i}{鈭�}鈥�E[M鈭�N=i]P\{N=i\}$

$=\underset{i=0}{\overset{n鈭�1}{鈭�}}鈥�E[M鈭�N=i]P\{N=i\}+\underset{i=n}{\overset{n+m鈭�1}{鈭�}}鈥�E[M鈭�N=i]P\{N=i\}$

$=\underset{i=0}{\overset{n鈭�1}{鈭�}}鈥�\left(n+m鈭�1+\frac{n鈭�i}{p}\right)P\{N=i\}+\underset{i=n}{\overset{n+m鈭�1}{鈭�}}鈥�\left(n+m鈭�1+\frac{i+1鈭�n}{1鈭�p}\right)P\{N=i\}$

Now simplify the equation we get,

$E\left[M\right]=n+m鈭�1+\underset{i=0}{\overset{n鈭�1}{鈭�}}鈥�\frac{n鈭�i}{p}\left(\begin{array}{c}n+m鈭�1\\ i\end{array}\right)p^i(1鈭�p{)}^{n+m鈭�1鈭�i}+\underset{i=n}{\overset{n+m鈭�1}{鈭�}}鈥�\frac{i+1鈭�n}{1鈭�p}{p}^i(1鈭�p{)}^{n+m鈭�1鈭�i}$

now we need to solve by using the above results,

$E[min(X,Y\left)\right]=E\left[X\right]+E\left[Y\right]鈭�E[max(X,Y\left)\right]$

$=\left(\begin{array}{l}\frac{n}{p}+\frac{m}{1鈭�p}鈭�n鈭�m+1鈭�\underset{i=0}{\overset{n鈭�1}{鈭�}}鈥�\frac{n鈭�i}{p}\left(\begin{array}{c}n+m鈭�1\\ i\end{array}\right)p^i(1鈭�p{)}^{n+m鈭�1鈭�i}\\ 鈭�\underset{i=n}{\overset{n+m鈭�1}{鈭�}}鈥�\frac{i+1鈭�n}{1鈭�p}{p}^i(1鈭�p{)}^{n+m鈭�1鈭�i}\end{array}\right)$

$=\frac{n}{p}+\frac{m}{1鈭�p}鈭�E\left[M\right]$

Thus the result is

$E[min(X,Y\left)\right]=\frac{n}{p}+\frac{m}{1鈭�p}鈭�E\left[M\right]$

The expected number of flips will be$E[min(X,Y\left)\right]=\frac{n}{p}+\frac{m}{1鈭�p}鈭�E\left[M\right]$.