Notes from an Inspector's Notebook: March 2006

In March 2006 I received an email from John which had an air of excitement. It is reproduced below. The problem exists in our collection as Task 46, Duelling Dice, and as a Maths300 lesson with the same name which extends the investigation even further with companion software.

The pages from John's diary are also reproduced below. They may be a little unclear but are worth the small amount of time it will take to dig into how a mathematician thinks. Click on a photo to enlarge it to the original size for clearer viewing.

For us the delight in reproducing this work is in its demonstration of the principle that a mathematician is never finished with a problem. They are only finished for now.

Greedy Pig (Task 200) re-awakened my interest in probability and other bits of mathematics which seem to me to be counter intuitive.

While trying to support a trainee teacher last week I was introduced to a beauty. She had given out the following investigation sheet, told the class to work in pairs and provided each pair with three dice marked up as required.

For this investigation you need 3 dice and a partner.

  • Dice A is labelled 2 2 2 2 6 6
  • Dice B is labelled 1 1 5 5 5 5
  • Dice C is labelled 3 3 3 4 4 4
With your partner, play a game with the following rules:
  1. The first player chooses a dice.
  2. The second player chooses a die from the remaining two dice.
  3. Each player rolls his or her die and the player with the higher number wins a point.
  4. Repeat this with the same dice 12 times. The player with the most points wins the game.
With these rules is it possible for one player to win nearly all the time?
If this is so how is it possible?

After 20 minutes or so each pair came up with the result that in order to win you must be the second player, ie: the one that chooses from two, not three, dice.

But this made no sense to me.
If Dice A beats Dice B and Dice B beats Dice C, how is it possible for Dice C to beat Dice A? But it does!! And I can prove it!!!

All this reminded me of a similar problem that I had had in my notebook for a number of years and to which I had no satisfactory solution.

The problem was to select a dice from a set of four with the best possible chance of winning over any of the others. Of course the dice are marked up in odd ways:

Dice A: 3 3 3 3 3 3
Dice B: 2 2 2 2 6 6
Dice C: 1 1 1 5 5 5
Dice D: 4 4 4 4 6 6
At last I had a good method, obtained from my exploration of the three dice game above, and a good use for adding and multiplying fractions! And not only could I select the right dice, but I could rank them in order of their winning chances.

Better perhaps to play the games (A v B, A v C, A v D, B v C, B v D and C v D) but good to be able to prove the theoretical results and disprove all those who said all I needed to do was find the average possible score for each dice in order to rank them.

And why do my diagrams remind me of the handshake problem?

Notebook Page 1

200604 Page 1

Notebook Page 2

200604 Page 2

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