I have a (possibly-) unfair coin, which lands heads with probability $p$ and lands tails with probability $1-p$. I toss it twice; the tosses are independent and identically distributed.
The probabilities of the various outcomes will vary depending on $p$. For example, if I know that $p = \frac12$, then the probability of tossing two heads, which I'll abbreviate $\mathbb P(HH)$, can be calculated as $p^2 = \frac14$. Symmetrically, the probability of tossing two tails $\mathbb P(TT) = \frac14$.
On the other hand, if I know that $p > \frac12$, e.g. $p = \frac1{\sqrt 3}$, then $HH$ becomes more likely: $\mathbb P(HH) = \frac13$. Correspondingly, $TT$ becomes less likely: $\mathbb P(TT) = (1 - \frac1{\sqrt 3})^2 \approx 0.18$.
It seems like I can push $HH$ up above $\frac14$, but only by pushing $TT$ down below that same number. With that in mind, the following question might seem surprising:
Under what circumstances does $\mathbb P(HH) = \mathbb P(TT) = \frac13$?