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Two people start flipping coins. The probability of heads is 0.5 (bonus if you can do it for $p_1$ and $p_2$). What is the probability that both will hit two consecutive heads simultaneously (as opposed to one of them doing so before the other)?

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  • $\begingroup$ I don't see why the probability isn't 1. If both are flipping coins forever, at some time it will happen that both get heads. Are there other rules to the flipping? $\endgroup$ – Jasper Sep 26 '18 at 6:03
  • $\begingroup$ Sorry, I should have mentioned, this is before any one of them flips two heads before the other. $\endgroup$ – Rohit Pandey Sep 26 '18 at 6:10
  • $\begingroup$ Sorry, it was partly my own mistake, I missed "consecutive" in the title. $\endgroup$ – Jasper Sep 26 '18 at 16:28
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Ok, figured it out. Thought I'd post the answer instead of deleting (BTW, I really don't understand the downvote - especially without having the courtesy to add a comment).

Let A be the event the two sequences reach HH simultaneously. Let's condition on the result of the first toss for each of them.

$P(A) = \frac{P(A|HH)+2*P(A|HT)+P(A|TT)}{4}$ Also,

$P(A|TT)=P(A)$

since it just resets.

$P(A|HH)=1/4+P(A)/4$

(If we get another HH, we're done. If we get TT, we reset. All other cases, it doesn't happen).

$P(A|HT) = 1/2(P(A|HT)/2+P(A)/2)$

Solving these, we get $P(A) = 3/47$

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    $\begingroup$ The downvote wasn't me, but it may have been for not mentioning what efforts you'd made to solve the problem. $\endgroup$ – Geoffrey Brent Sep 26 '18 at 5:43

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