I’ve been thinking about a card game/trick shared by a colleague. Take a pack of cards with JQK removed, and deal them out in a long line of 40 cards. Place a counter on one of the first six cards, and then move forward that number of cards – for example, if you place your counter on a 3, you’d move 3 cards forward. Continue until you can’t move any further without going off the end of the line.

Then place another counter on another of the first six cards, and do the same.

If you try it, you may be surprised that quite often, you end up on the same card. Trying to explain why is quite fun. (Analysing the probabilities would be horrendous, but when I set this as a problem with students I’m planning to make it an experimental rather than theoretical probability task.)

My unanswered question is this though; is it possible to order such a pack of cards in such a way that each of the first six cards will take you to a DIFFERENT finishing point? We think we have a convincing argument why this is impossible… but maybe we’ve missed something and it is possible, or maybe you have a really neat way of thinking about this – if so, please share it!

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September 6, 2010 at 12:09 |

Hm, interesting … will have a think! Mostly commenting to say: you might want to specify that you deal the cards out face up – I assumed face down which was slightly confusing for a minute.

Why the first six cards, rather than any other number of cards?

What you’re looking for really is entirely disjoint paths. Which makes me think of graph theory. Which may or may not be useful.

September 6, 2010 at 12:13 |

Good point – we put “face up” when we started writing this up for the NRICH site! We started with six and then looked at starting with other numbers, and also what happens if you had cards numbered 1-11 instead of 1-10 – there are lots of interesting variations and extensions. Graph theory is not my speciality, although if someone comes up with a nice graph-theory way of solving such problems, I’ll do my best to engage with it.

September 6, 2010 at 20:06 |

Intriguing indeed. I thought I had proved it impossible twice but each time realised I had overlooked something…..

September 6, 2010 at 22:32 |

The 6 card start limits the possibilities

September 7, 2010 at 13:50 |

A very rich and interesting problem Alison! It had our Maths teachers intrigued tonight. As often with problem solving we needed to clarify the problem. In your example, do you continue moving 3 cards forward (my assumption) or according to the new card you land on? This would make an interesting variation. Getting mental predictions of which card is landed on is powerful maths. One teacher used algebra to do this. Another suggestion was to create a “gambling” game out of this situation. The power of a group of teachers working together on a problem and asking “What if” questions was evident. Is this problem on the nrich site yet? Thanks for your great problems…

September 7, 2010 at 14:00 |

Hi webmaths – thanks for your comment. In our original problem, we intended that you would always move forward according to the new card you landed on, but of course making new problems from an original idea is a great part of working on rich mathematics. Really glad you and colleagues enjoyed working on the problem – it will appear on the NRICH site in November together with a variety of probability tasks.