The Three-Card Puzzle

As promised I did my turn at the Brighton Science Festival yesterday. The Sallis Benney Theatre wasn’t quite full but there was a decent crowd, which was mildly surprising because the event I was involved in hadn’t really been advertised very well. If you want to know how my talk went then you should ask someone who was in the audience because I wasn’t really paying attention. However, I was preceded by John Haigh (seen below in mid-talk) whose presentation included a nice puzzle for the “Cute Problems” file:wp-1456735427335.jpeg

Imagine you have three cards. One is blue on both sides. One is pink om both sides. One is pink on one side and blue on the other. Other than the colours the cards are identical. For the demonstration John glued playing cards together, but they don’t have to be playing cards. Anyway, you put the three cards into a bag (seen on the stage in the picture), pull out one card “at random” and look at the colour of one side but not the other. If the colour you see is blue, what is the probability that the other side is also blue?

Try to answer this without googling. I’ll post the solution when there have been enough responses to the poll:


OK. Over a hundred people responded so I have now closed the voting.

As always seems to be the case with this sort of problem, the majority went for the “obvious” answer, which turns out to be wrong!

SOLUTION: If the card is blue on one side then it must be either the blue-blue or blue-pink one. I think most people voted for 1/2 because there are two possible cards. But the relevant consideration is that there are three possible sides: side 1 of the blue-blue card; side 2 of the blue-blue card; and the blue side of the blue-pink card. Each of these is equally likely and two of them result in the other side being blue. The correct answer is therefore 2/3; it is twice as likely for the other side to be blue as it is to be pink.

6 Responses to “The Three-Card Puzzle”

  1. 2/3. I didn’t have to Google it; Sleeping Beauty told me.

  2. 2/3, Bayes told me!

  3. There are three card “position” blue-pink blue1-blue2 blue2-blue1, so the probability that the other side is blue is 2/3.
    I ever think that the Peter Coles statistical problems have not a simple solution, and so I looked for a non-trivial solution; but is it the real solution?

  4. 2/3. this is a reformulation of the monty hall problem, isn’t it?

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