here only $1$ coin.
we tossed several times and observed the result
i.e. first time$==>\left\{H,T \right \}$
Second time $==>\left \{HH,TT,HT,TH \right \}$
third time $==>\left \{HHH,TTH,HTH,THH,HHT,TTT,HTT,THT \right \}$
all are independent
We want both Head or Tail in even number of tosses.
lets take both head case(even number of tosses):-
Even number of tosses means$:2nd$ time,$4th$ time$,6th$ time ,........and so on
$2nd$ $time==>\left \{HH\right \}==>P(HH)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}=(\frac{1}{2})^{2}$
$4th$ $time==>\left \{HTHH\right \}==>P(HTHH)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}=(\frac{1}{2})^{4}$
$6th$ $time==>\left \{HTHTHH\right \}==>P(HTHTHH)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{64}=(\frac{1}{2})^{6}$
$8th$ $time==>\left \{HTHTHTHH\right \}==>P(HTHTHTHH)=(\frac{1}{2})^{8}$
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$(OR)$
lets take both tail case(even number of tosses):-
Even number of tosses means$:2nd$ time,$4th$ time$,6th$ time ,........and so on
$2nd$ $time==>\left \{TT\right \}==>P(TT)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}=(\frac{1}{2})^{2}$
$4th$ $time==>\left \{THTT\right \}==>P(THTT)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}=(\frac{1}{2})^{4}$
$6th$ $time==>\left \{THTHTT\right \}==>P(THTHTT)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{64}=(\frac{1}{2})^{6}$
$8th$ $time==>\left \{THTHTHTT\right \}==>P(THTHTHTT)=(\frac{1}{2})^{8}$
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Now even number of tosses until both heads are coming $=(\frac{1}{2})^{2}+(\frac{1}{2})^{4}+(\frac{1}{2})^{6}+(\frac{1}{2})^{8}+........$
This is infinite Geometric Series $S=\frac{a}{1-r}$ When $r<1$ and $S=\infty $ When $r>1$
Here $a=(\frac{1}{2})^{2}$ and $r=(\frac{1}{2})^{2}$
So,$S_{1}=\left [\frac{(\frac{1}{2})^{2}}{1-(\frac{1}{2})^{2}} \right ]$
$S_{1}=\left [ \frac{\frac{1}{4}}{1-\frac{1}{4}}\right ]$
$S_{1}=\left [ \frac{\frac{1}{4}}{\frac{3}{4}}\right ]$
$S_{1}=\left [{ \frac{1}{4} }\times { \frac{4}{3} }\right ]$
$S_{1}=\frac{1}{3}=0.33$
Similarly for both tails comming in even number of tosses $S_{2}=\frac{1}{3}=0.33$
Now,both heads or tails comming in even number of tosses $S=S_{1}+S_{2}$
$S=0.33+0.33=0.66$