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2 friends Alice and Bob have found an unfair coin,It has 72% chance of coming up heads.Alice and Bob plays a game with this coin.If coin comes up head then tails,Alice wins.If it's reverse(tails,then heas),Bob wins.And if neither of those two things happens,the game restarts and continues untill there is a winner

What is Bob's probability of winning?
ans given is 0.5...
I think it should be TH  out of sample space { TT, TH, HT, HH }

But , not getting the answer.

Write the whole procedure,no problem if it is wrong,need to see the arppraoch!!!

$P\left ( Alice\ Wins \right ) = P\left ( HT \right ) = 0.72 *0.28 = 0.2016$

$P\left ( Bob\ Wins \right ) = P\left ( TH \right ) = 0.28 *0.72 = 0.2016$

Probability to restart the game = Probability if $HT$ and $TH$ do not occur = $P'=\left ( 1-(0.2016+0.2016) \right ) = 0.5968$

When this process continues, then

• $P\left ( Bob\ Wins \right ) = P(TH) + P'P(TH) + P'P'P(TH)+P'P'P'P(TH).......$
• $P\left ( Bob\ Wins \right )= P(TH) * \frac{1}{1-P'}$
• $P\left ( Bob\ Wins \right )= 0.2016 * 2.4801 = 0.5$

selected by
@Kapil

How u took :

P′P(TH)+P′P′P(TH)+P′P′P′P(TH)+.....=$\frac{1}{1−P′}$

Couldn't understand
@Prajwal

Infinite GP

P(TH) + P′P(TH)+P′P′P(TH)+P′P′P′P(TH)

P(TH) (1+ P' + P'P' +..............)
@Kapil Sir, please verify is'nt my answer equivalent to your only ? Please do check it.. i am saying after checking 2 -3 times !!

Did you check your calculation here TH/(1-((HH + TT))  ?

Yes sir i checked the answer, it is having difference as in calculation i took all values upto 2 decimal i.e why final answer have some difference in points.
Bob can win in the first chance or third chance or fifth chance or ...

Let $p$ deonte the probability of getting a head. $p = 0.72$.

P(First chance win) $= p(1-p)$

P(Third chance win) $= [p^2 + (1-p)^2]^2 * p(1-p)$

P(Fifth chance win) $= [p^2 + (1-p)^2]^4 * p(1-p)$

and so on ...

Let $p(1-p) = K$ and $[p^2 + (1-p)^2] = Z$

Answer $= K + Z^2K + Z^4K + ...$

$= K(1 + Z^2 + Z^4 + ...)$
$= K(\frac{1}{1-Z^2})$
Substitute the values and you'll get answer $= \frac{0.2016}{0.4032} = 0.5$
didnt refresh the page before posting so missed the already accepted answer.
No problem, nicely explained :)

I did this. Please see why I am getting the answer in a strange way!!

Let Probability of getting tail = p, Probability of getting head = (1-p)

Bob wins if the outcome is (TH) or(TT+HH)TH or (TT+HH)(TT+HH)TH and so on.

So, P(Bob winning) = p(1-p) + [p2 + (1-p)2]p(1-p) + [p+ (1-p)2]2p(1-p) + ...........

Let [p+ (1-p)2] be x.

So, Probaility = p(1-p)[1 + x + x2 . .....] = p(1-p)($\frac{1}{1-x}$) = $\frac{p-p^{2}}{1-[p^{2}+1+p^{2}-2p]}$ = $\frac{p-p^{2}}{2p-2p^{2}}$ = 0.5

Will this question give 0.5 for every cases?? I have not even substituted the values.

Probability of head then tails = 0.72 * (1-0.72) = 0.72 * 028 = 0.4032

Probability of tail then head = 0.28 * 0.72 = 0.4032

Probabilty of tail ,tail = 0.28 * 0.28

Probababilty of Bob's winning =  0.72 * 028 + (0.72 * 0.72 + 0.28 * 0.28) * 0.72 * 0.28 +  (0.72 * 0.72 + 0.28 * 0.28)^2 * * 0.72 * 0.28 + ...... = 0.72 * 0.28 / (1 - (0.72 * 0.72 + 0.28 * 0.28)) = 0.5
edited
Mandeep can you explain last point little clearly,

couldn't understand what you have taken properly
look Bob will win only when head will follow tail, so take the probability of this case . Now if head, head or tail, tail comes in these two cases, game restarts so now compute the probability of bob's wining again and again until either of one wins so it will become a geometric progression where Ist term is bob's chance and common multiple is probability of head and tail tail.
@Mandeep you solved wrong !! We need to find Prob for Bob not ALICE your end result is correct as both winning prob is 0.5 here.
OOps my mistake. thanks for pointing it out. Correct it now

Prob of Tail = 0.28

Lets see how Bob can win,

In first step : "TH"

In second step : "HH TH" or "TT TH"

In third step : " HH HH TH " or " HH TT TH" or " TT TT TH " or "TT HH TH"

Now the pattern is easy to see, we even can write a regular expression for this : (HH + TT)* TH  :)

P(Bob wins)

= TH + (HH + TT)TH + (HH + TT)2TH+..................

= TH [1 + (HH + TT) +  (HH + TT)2 + ................... ]

=TH/(1-((HH + TT))   [Applied Infinite GP sum = a/1-r where a is first term and r is common ration]

Now subtitute values,

P(TH) = 0.28 * 0.72

P(HH) = (0.72)2

P(TT) = (0.28)2

On solving P(Bob wins) = 0.4897

The second step......HHTH

But when T appears, Alice already wins.
Favourable event is TH out of sample space {TT, TH, HT, HH}

So, P(Bob Wins) = 0.28 * 0.72 = 0.2016

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