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

What's Bob's probability of winning?

My answer:0.1689

but correct ans is given 0.5?

  • 1
    $\begingroup$ How did you get 0.1689, by the way? $\endgroup$ – Clement C. Jan 22 '16 at 14:26
  • $\begingroup$ I made an infinite GP, then did something, I don't know where it went wrong. $\endgroup$ – abc Jan 22 '16 at 14:32

Hint: Since the two consecutive coin tosses are independent, $\Pr[HT] = \Pr[H]\cdot\Pr[T]$, and $\Pr[TH] = \Pr[T]\cdot\Pr[H]$. Does the winner depend on the bias $p$ at all?

(Note that the probability that someone wins in the first round does depend on the bias; but since the game is repeated until either HT or TH happens, it does not. For more, you may want to read about von Neumann's trick; essentially, Alice and Bob have equal chances of winning, but the (expected) duration of the game will depend on the bias $p=0.72$))

  • $\begingroup$ Suppose the question asks "what is the probability of bob winning in the 1st move"? then what is the answer. Is it 0.72*0.28 or 0.72*0.28/2 ? $\endgroup$ – abc Jan 22 '16 at 14:29
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    $\begingroup$ It would be $\Pr[TH] = (1-p)p = 0.72\cdot0.28$. This is also equal to the probability Alice wins in the first round (and the probability no one wins in the first round is $p^2 + (1-p)^2$). Fortunately,as a sanity check we do have that $p(1-p) + p(1-p) + p^2 + (1-p)^2 = (p+(1-p))^2 = 1$ -- the probabilities do sum to $1$. $\endgroup$ – Clement C. Jan 22 '16 at 14:33

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