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If you were listening on Thursday 16 April, you might remember Sarah Montague and John Humphrys talking about a maths logic puzzle which has gone viral over the internet because it is just so baffling.

It was a logic problem called Cheryl's birthday that was posed to high achieving teenagers in Singapore - which has topped maths leagues tables since the 90s.

We then set you our own maths challenge about three prisoners and coloured hats. Unfortunately, despite having the answer written in front of her, Sarah ended up giving you the WRONG answer – so we turned to school children for help.

Jess Fisk, 10, Rory Clarke, 11, and Jimmy Allan, 11 all attend Primary Schools in Hampshire, but know each other as they street dance with the Integr8 dance school.

The prisioner hat problem:

There were three prisoners, one of whom was blind. They were offered their freedom if they could succeed in the following game. Their jailer produced three white hats and two red hats and, in the dark, placed a hat on each prisoner. The prisoners were then taken into the light where, except for the blind man, they could see one another (but they could not see the hat on their own head).

The game was for any prisoner to state correctly the colour of the hat he himself was wearing. The jailer asked one of the men who could see if he knew, and the man said 'no'. Then the jailer asked the other man who could see if he knew, and his answer was 'no'.

The blind man at this point correctly stated the colour hat he was wearing, winning the game and freedom for all three. What colour hat was he wearing, and how did he know?

The answer:
The answer...

There were seven possible combination of hats possible. The three options below hold the key. If we can account for these, we have the answer.

The three options were:

Blind 1st seeing 2nd seeing

R W R

R R W

R W W

The blind prisoner listened to what the other two prisoners had to say and in his mind, he deduced the following:

• The first option is eliminated when the 1st seeing prisoner, if he had seen the other two wearing red hats, would have known he was wearing white, but he was unable to say this so he could not see two red hats.

• The second option could be similarly eliminated when the 2nd seeing prisoner could not state that he was wearing white and so could not see two red hats.

• But before making his decision, the 2nd seeing prisoner would have heard the 1st seeing prisoner answer. From this the 2nd seeing prisoner would have followed the same argument as the 1st and realised that he, the 2nd seeing prisoner, could not have been wearing red.

• Therefore, if the blind prisoner was wearing red, as in the third option, and the 2nd seeing prisoner knew he could not be wearing red, then he, the, would have to be wearing white.

• However, the 2nd seeing prisoner was unable to state that he was wearing white; therefore the third option was also not possible and could be eliminated.

As all the other possible combinations had the blind prisoner wearing white, the blind prisoner, having reasoned through the arguments above, could deduce that he must be wearing white.

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