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So if we have a Bernoulli trial with the property that the accumulated number of successes is twice the accumulated number of failures and the ratio has never exceeded 2. Is the probability of this happening at the $3k$th trial:

$$P_{3k}=3^{k-1}(p^2q)^k ?$$

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Can you explain where your conjectured answer comes from? Also, your wording is slightly unclear. Do you mean to ask, "Assume that we are running Bernoulli trials with probability of success $p$, and $q=1-p$. What is the probability that, after $3k$ trials, the ratio of successes to failures is exactly $2$ and that at no point prior had the accumulated ratio of successes to failures been strictly greater than $2$?" If this is not what you mean to ask, then you are asking for some sort of related conditional probability, but I'm not sure which. –  Aaron Sep 26 '12 at 6:38
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Taking Aaron's interpretation, when $k=3$, i.e. after $9$ steps, there are twelve possible patterns while your expression would suggest there are nine.

The expression should be $$\frac{1}{2k+1}{3k \choose k} (p^2 q)^k$$ related to OEIS A001764.

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and since the patterns can be reversed, this is also the probability that the ratio of successes to failures is exactly 2 and that at no prior point had the number of successes been strictly less than double the number of failures. –  Henry Sep 26 '12 at 7:18
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