You might have noticed that one of our contributors is quite keen on Rugby. Well, this puzzle is inspired by something read in a match report, but don't worry, it is not actually about Rugby. All that is required is logic and arithmetic.

The match report in question said that in the first half a team had three players sent to the sin bin in quick succession, and had to play three players short for four minutes.

Our contributors are special people, and this one started to think "From that information can I work out how many minutes the team played with one player short and two players short?".

So, can you work it out, and what are the times?

For those less familiar with Rugby, if a player is sin binned he or she takes no part in the game for ten minutes.

As this happened in the first half, we can discount this happening close to the end of the game, which would complicate things. Most of you discounted this anyway.

Let the time that Player 1 leaves the field be 0 minutes, so they will return when the time is 10 minutes.

As the team were three players short for four minutes, Player 3 must be sin binned when the time is 6 minutes. And thus they return when the time is 16 minutes.

Let the time that Player 2 goes off be t (which must be between 0 and 6) and the time they return be t+10.

The period of time that the team is one player short is the time between Player 1 going off and Player 2 going off (t - 0) added to the time between Player 2 returning and Player 3 returning (16 - (t+10)).

$$t - 0 + 16 - (t + 10) = t + 16 - t - 10 = 16 - 10 = 6$$

So the team is a player short for 6 minutes.

We know that the team is three players short for 4 minutes, and the whole time that the team is any number of players short is 16 minutes (the time Player 3 returns), which leaves 6 minutes when the team is two players short.