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Sunday Times Teaser 2720 – Better Tetra

by Michael Fletcher

I have four regular tetrahedral dice with one of the numbers 1, 2, 3 or 4 on each of the sixteen faces. The numbers on each die sum to ten and none of the dice have the same four numbers.

My friend and I play a game in which we each throw a die and win if our face down number is higher [1].

But, because I am always more likely to win, my friend has changed the rules so that we now throw our die twice and win if the sum of our two face down numbers is higher. He now knows that there is one dice that he can choose that will win more often than the others.

What are the four numbers on this dice?

[1] My wording follows that in the Sunday Times but it seems likely that this wording should have been: “My friend and I play a game in which he chooses a die and I then choose another, after which we each throw our die and win if our face down number is higher”. 

10 Comments Leave one →
  1. brian gladman permalink

    I am not sure about this one as I can only find a solution if I assume that (a) the friend always plays first in both types of games, and (b) they have to choose different dice.

  2. Peter Hayes permalink

    I am of the opinion that the ST original was poorly written; ie it didn’t make a lot of sense to me as it was, but I made the same assumptions as your two and arrived at an answer.

  3. brian gladman permalink

    Hi Peter,

    I agree that the wording is poor, which is a pity since only small changes are needed to provide a nice teaser. I had some difficulty in deciding how to re-write it since, as you say, it does not make sense as written. But I decided to follow the ST wording in case it was my interpretation that was in error.

  4. mike fletcher permalink

    Yes it should have said ‘My friend chooses a dice first and then I choose one of the remaining three dice’

    • brian gladman permalink

      Hi Mike, Thanks for confirming the wording. Its a nice teaser, one which I enjoyed after realising what the wording should have been. It would seem that the ST sub-editors have struck again!

  5. mike fletcher permalink

    Every year the village has a fete and the biggest attraction is the ‘Guess how many marbles there are in the jar’ competition. This year it was more difficult: there were two colours of marbles. Contestants had to guess how many red marbles and how many black marbles there were in the jar.
    ‘That’s difficult’, I commented.
    ‘Well, to make it easier,’ said the stallholder, ‘I’ll give you a clue.’
    ‘If I were to close my eyes and take out four marbles the chance they would all be red is 1 in 152099999999.’
    How many red marbles and how many black marbles are there in the jar?

    Thought you might like this!

  6. brian gladman permalink

    Red 626, Black 389375

  7. mike fletcher permalink

    Too easy!

  8. brian gladman permalink

    Hi Mike,

    Yes, 152099999999 = (390000)^2 – 1 is a pretty good clue. I didn’t even have to get a calculator out for this one, although I did need a pencil and paper.

    Are you aware of my other site for manual solvers of ST teasers at:

    Your teaser got some discussion there that you might be interested in. I also posted your correction and this puzzle there.

  9. mike fletcher permalink

    I guess I should have said ‘solve without calculator’

    152099999999 = 152100000000 -1 and 39 x 39 =1521 and a bit of manipulation getsit out

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