# A coin flip's long tail

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 $\mathbb P(HH) = \mathbb P(TT) = \frac13$?

• Is this a trick question? Your statement "the tosses are independent and identically distributed" makes it mathematically impossible to have both probabilities equal 1/3. Commented Dec 2, 2015 at 15:19
• @Gamow: I believe my statement is accurate. This wouldn't be an interesting puzzle if it weren't for the apparent contradiction :) Commented Dec 2, 2015 at 15:26
• I'll think about how to give subtle hints if no-one has got closer to the answer by tomorrow. Commented Dec 2, 2015 at 16:00
• Must the coin have two faces, one with a single head and the other with a single tail? Commented Dec 2, 2015 at 16:24
• Yes. (Actually, until you mentioned it I didn't deeply think about how you'd even go about weighting a coin to change the outcome of its probabilities. Probably the actual easiest thing would be to put a bunch of heads and tailses on a die and weight that appropriately. But that's irrelevant to the puzzle: from the puzzle's perspective the only thing that matters is that there are exactly two possible outcomes of each individual coin toss.) Commented Dec 2, 2015 at 16:35

$p$ is a random variable, chosen with uniform probability over the interval [0,1]. The probability of two heads is $\int_0^1 p^2=\frac{1}{3}$, and by symmetry so is the probability of two tails.

• Now that I think about it this makes perfect sense. There was definitely a reason why Ben Millwood said the flips were identically distributed.
– orp
Commented Dec 2, 2015 at 19:39
• Also explains why they are possibly unfair. Commented Dec 2, 2015 at 19:46
• Of course, this destroys independence, though. In this case, the flips are only conditionally independent, conditional on the value of $p$. Commented Dec 2, 2015 at 20:10
• @orp: No, it's about whether one event tells you something about another event. Also, you can't just suddenly fix $p$ like that. Let me put it this way. Let $H?$ be the event that the first coin lands heads, and let $?H$ be the event the second coin lands heads. In the sample space of all possible values of $p$, flip 1, and flip 2, $p(H?) = 1/2$ and $P(HH) = 1/3$, so $P(?H|H?) = P(HH)/P(H?) = (1/3)/(1/2) = 2/3 \ne P(?H)$. Thus, by the definition of independence, $?H$ and $H?$ are not independent. Commented Dec 2, 2015 at 21:28
• A generalization to n coins.
– xnor
Commented Dec 3, 2015 at 7:05

This seems to be mathematically impossible. The probability of two heads is $1/3$ and the probability of two tails is $1/3$.

Since they're equal, this gives us $$p^2 = (1-p)^2$$ $$p = 1-p$$ $$p = 1/2$$

But $(1/2)^2 = 1/4$, not $1/3$.

If there really is a solution to this I'd be really interested to see it.

• I agree that it seems impossible, but I insist that it is not! Commented Dec 2, 2015 at 15:54
• @Ben Millwood Just to clarify that this isn't a trick question, are you insisting that it is mathematically possible to answer the question, or that it is mathematically possible for P(HH) = P(TT) = 1/3? If this is the dirty trick I'm suspecting, "Under no circumstances" is a potentially valid answer. Commented Dec 2, 2015 at 18:08