# Ask Uncle Colin: A Coin Toss Conundrum

Dear Uncle Colin,

Please can you settle an argument? I say, if you toss a coin three times, the probability of getting all heads is one in four, because the only possibilities are HHH, HHT, HTT and TTT. My friend says it’s one in eight, being $\frac{1}{2}\times \frac{1}{2} \times \frac{1}{2}$. Who’s right, and why?

Tiffing Over Some Sums

Hi, TOSS, and thanks for your message!

I’m afraid it’s your friend who’s correct. Their interpretation is spot on: the probability of getting a head is a half, and the probability of tossing three in a row is $\br{\frac{1}{2}}^3$.

The more interesting question is, why is your interpretation wrong?

### Some are more equal than others

The problem is that your four possibilities are not all equally likely. Imagine you have three distinct coins - a penny, a 10p piece and a pound coin ((Other currencies are available)).

The possibility HHT is really three possibilities: it could be any one of the three coins that’s the tail. The same goes for HTT - any of the three could be the head.

Meanwhile, there’s only one way to get all heads, and one way to get all tails.

So, rather than four possibilities, there are, in fact, eight.

Hope that helps!

- Uncle Colin