How to solve below question of probability ?

Question

2 friends Alice and bob have found an unfair coin ,it has 72% chance of coming up heads .Alice and bob play a game with this coin ,If coin comes up with head and then tails,Alice wins and if its in the reverse way then bob wins .And if neither of the two things happen then the game restarts and  continues until there is a winner ,what is the probability that bob wins ?

 

In this how to deal with the case when we have HH or TT ?

Answer 1

P(H)=0.72           P(T)=0.28

P(HT)=0.72 * 0.28=0.2016 say x1    ,     P(TH)=0.28 * 0.72=0.2016 say x2

to repeat same process....probability=1-(x1+x2)=0.5968 say y1

ans=x1 + y1*x1 + y1* y1*x1+......

       =x1(1+y1+y1*y1+...)

       =x1(1/(1-y1))

       =0.2016(1/(1-0.5968))

       =0.5

so ans should be 0.5

Comments

"to repeat same process....probability=1-(x1+x2)=0.5968"

Can you explain this point ?What is the logic behind this?

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P(bob wins) = P(bob wins in Game one) + P(No result in 1st Game  and bob wins in 2nd Game)

                       + P(No result in 1st Game  and No result in 2nd Game and bob wins in 3rd Game)

                       + ........

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P(bob wins in game i, (irrespective of what i is)) = P(tail in first throw and head in 2nd throw)

                                                                               = (0.28)(0.72)

                                                                               = 0.2016

P(No result occurs in a Game i (irrespective of what game i is))   =  ??

       =P(head in first throw  and head in 2nd throw) +  P(tail in first throw  and tail in 2nd throw)

                                = (.72)(.72) + (.28)(.28)

                                = 0.5968

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P(bob wins) $= .2016 + (.5968)(.2016) + (.5968)(.5968)(.2016) + .........$

                            ^           ^         ^

                             |            |           |____ P(bob wins in 2nd Game))

                             |    P(No result in 1st Game))

               P(bob wins in Game one)

                    $= .2016 (1 + .5968 + .5968^{2}$ + $.5968^{3} + ........ )$

                    $\\ = .2016\ *\ \frac{1}{1-0.5968} \\ \\ = \frac{0.2016}{0.4032} \\ \\ = \frac{1}{2} \\$

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thank you :)