Question 737203
<pre>
To be successful you either have to get

HH or HTHH or HTHTHH or HTHTHTHH

or get

TT or THTT or THTHTT or THTHTHTT


So it's the probability of starting with a head and
ending with two heads + the probability of starting 
with a tail and ending with two tails.

The desired probability is the sum of sums of these 
two infinite geometric series:

1.   P(HH)+P(HTHH)+P(HTHTHH)+··· =  
{{{1/2^2}}}{{{""+""}}}{{{1/2^4}}}{{{""+""}}}{{{1/2^6}}}+··· =
{{{1/4}}}{{{""+""}}}{{{1/16}}}{{{""+""}}}{{{1/64}}}+···   


2.   P(TT)+P(THTT)+P(THTHTT)+··· =  
 {{{1/2^2}}}{{{""+""}}}{{{1/2^4}}}{{{""+""}}}{{{1/2^6}}}+··· =
{{{1/4}}}{{{""+""}}}{{{1/16}}}{{{""+""}}}{{{1/64}}}+···   

Each of those has a<sub>1</sub>={{{1/4}}} and r ={{{1/4}}}

S<sub>&#8734;</sub> = {{{a[1]/(1-r)}}} = {{{(1/4)/(1-1/4)}}} = {{{(1/4)/(3/4)}}} = {{{(1/4)*(4/3)}}} = {{{1/3}}}

So the desired probability is twice that or {{{2/3}}}

Edwin</pre>