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3 votes
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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?

5 Answers

Best answer
14 votes
14 votes

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

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

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
3 votes
3 votes
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 $= \frac{0.2016}{0.4032} = 0.5$
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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 head, head = 0.72 * 0.72

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 by
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Prob of Head = 0.72

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 

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