It is easy to work out how likely it is that you will throw six sixes in six throws of a die. There is 1/6 probability that you will throw a six, therefore there is a  (1/6)⁶ probability that you will throw 6 in a row.

However, if we want to know the probability that we throw at least one six things get a bit more complicated.

We could get it on the first throw – probability 1/6. We could get it on the second throw, however, we must not have thrown it on the first go or else we will double count, so the probability is 5/6 x 1/6. We end up with a lot of possible combinations which will result with a six appearing. Likewise, we could work these out and add them together, but there is an easier way to approach the problem.

The way to do this is to work out the probability that we do not throw any sixes at all. Since the probability of not throwing a six is 5/6 then for six throws there are (5/6)⁶ ways of doing so.

The probability of not throwing a six is then 1-(5/6)⁶ since we either throw a six or do not.

When we tackle problems such as this, it is often a lot easier to work out the probability of something not happening and take it away from one.

I have seen several people get this wrong – even ending up with a probability of 3.6 (360%) due to the double counting.