Probability

Question

I toss a coin repeatedly. The coin is unfair and P(H) = p. The game ends the first time that two consecutive heads (HH) or two consecutive tails (TT) are observed. I win if HH is observed and lose if TT is observed. For example if the outcome is HTHTT, I lose. On the other hand, if the outcome is THTHTHH, I win. Find the probability that I win.

Ans: $\frac{p^{2} (2-p)}{(1-p+p^{2} )}$

Comments

probability of win = p(HH)+p(HT)+p(TH)

p(HH)=p²

p(HT)=p(1-p)

p(TH)=(1-p)p

adding all three

p²+p(1-p)+(1-p)p=p²(1-p)

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I think your answer is not correct. Please read the question again. I think you misunderstood the question.

Answer 1

P(H) = p

P(T) = (1-p)

Winning sequence possible,

HH , THH, HTHH, THTHH, HTHTHH......

So respective Probabilities Sequence,

$p^2$, $(1-p)p^2$, $p(1-p)p^2$, $p(1-p)^2p^2$, $p^2(1-p)^2p^2$.....

So Probability to win =

$p^2$ + $(1-p)p^2$ + $p(1-p)p^2$ + $p(1-p)^2p^2$ + $p^2(1-p)^2p^2$ +.....

=  $p^2$[1 + $(1-p)$ + $p(1-p)$ + $p(1-p)^2$ + $p^2(1-p)^2$ +......]

= $p^2$[(1 + $(1-p)$) + $p(1-p)(1 + (1-p))$ + $p^2(1-p)^2(1 + (1-p))$ +......]

= $p^2$[$(2-p)$ + $p(1-p)(2-p)$ + $p^2(1-p)^2(2-p)$ +.......]

= $p^2(2-p)$[1 + $p(1-p)$ + $p^2(1-p)^2$ +.......]

= $p^2(2-p)$×$\frac{1}{1 - p(1-p)}$

[Since sum of infinite G.P is $\frac{a}{1 - r}$ , here $a$ is 1 and r is $p(1-p)$ ]

= $\frac{p^2(2-p)}{(1 - p + p^2)}$