For an arbitrary $p$, this cannot be accomplished in a single toss. If we toss twice, we observe that the probability of HT and TH is the same, i.e. the probability of heads in the first toss and tails in the second is the same as its reverse, i.e. $p \cdot (1-p) = (1-p) \cdot p$

Hence, we can declare these two as the two events. If we observe HH or TT, we retry.

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Follow-up question

What's the probability of an infinite repetition?

Follow-up Answer

Let us calculate the probability of the first retry

Probability of HH or TT = $p^2 + (1-p)^2$

$P(r)= 2p^2 -2p + 1$

This is a quadratic equation, with a minimum value of $P(r) = 0.5$ at $p=0.5$, and $0.5 < P(r)< 1$ for the range $0.5 < p < 1$

Thus, the probability of an infinite repetition is $P(r) \cdot P(r) \cdot ... = 0$ (because $P(r) < 1$)

Note that if $r$ is close to 1 then $P(r)$ is close to 1, and it might take a very long sequence before event 1 or event 2 occurs.