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Another annoying little dice puzzle

July 29, 2008 am31 9:26 am

Yesterday I asked:

Which is more likely, rolling three normal dice, and having one be the sum of the other two? or rolling three normal dice, and having one be the product of the other two?

In addition to your regular work, try to find a (good!) explanation that appeals to intuition.

And we got nice answers, and addition won. Fairly easily, actually (just read the comments)

So, new puzzle. Change the old puzzle to make it a close race.
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5 Comments leave one →
  1. July 29, 2008 pm31 1:56 pm 1:56 pm

    Thanks for an interesting question, which has stopped me going to the gym this morning!

    First generalisation:

    Change the number of dice. Make the condition “one die is the sum (or product) of all the others”.

    3 dice: 45 sum, 25 product
    4 dice: 80 sum, 67 product <– nice and close.
    5 dice: 75 sum, 141 product
    6 dice: 36 sum, 256 product
    7 dice: 7 sum, 421 product

    Second generalisation:

    Keep three dice, and change the maximum number of ‘spots’ on each die.

    1 to 2: 3 sum, 4 product
    1 to 3: 9 sum, 7 product <– closest race.
    1 to 4: 18 sum, 13 product
    1 to 5: 30 sum, 16 product
    1 to 6: 45 sum, 25 product
    1 to 7: 63 sum, 28 product

    1 to 20: 571 sum, 139 product

    It seems that ‘sum’ is now beating ‘product’, and will continue to do so. Is this true? Does the ratio of ‘sum’ over ‘product’ tend to a fixed number?

    Third generalisation:

    We can combine the above, and look at which maximum number of spots makes the race closest for a given number of dice.

    3 dice, 1 to 2: 3 sum, 4 product
    4 dice, 1 to 5: 40 sum, 37 product
    5 dice, 1 to 7: 175 sum, 151 product
    6 dice, 1 to 8: 336 sum, 466 product

    There’s nothing that great there, although I quite like the 4 dice one.

    Fourth generalisation:

    We can keep three dice, but change the spots on each die. For example, make the spots ‘0 … n’ rather than ‘1 … n+1’. This seems to alter the balance between sum and product:

    0 to 1: 4 sum, 8 product
    0 to 2: 10 sum, 23 product
    0 to 3: 19 sum, 44 product
    0 to 4: 31 sum, 74 product
    0 to 5: 67 sum, 107 product

    0 to 19: 571 sum, 1265 product

    On reflection you can see why adding ‘0’ to the mix immediately improves the chances of getting the multiplications to work out.

    It would be interesting to investigate other spot patterns (the primes, for example, would not be good for the ‘product’ count!), and other generalisations of the success rule, but that’s all I have time for at the moment!

    (All the numbers in this post were cooked up by a quickly generated program written while my daughter and wife prodded and talked to me, so please don’t take them as true without verifying them yourself!)

  2. July 29, 2008 pm31 5:07 pm 5:07 pm

    Neat. I was thinking of dragging 0 in. Changes the game dramatically, doesn’t it? But I’m not following your calculation on the last group. Eg, 3 dice, 0-3: I count 12 products and 10 sums. How are you getting your numbers?

    (I’ve assumed that you are reducing the number of sides to correspond with the number of numbers. Is that where I went off?)

    The 4 dice, 1 – 6 one is very close. It comes very close to being a fair game.

  3. July 29, 2008 pm31 5:22 pm 5:22 pm

    I’ll double-check my numbers now that my family aren’t poking me with sticks… I confidently expect at least one set of them to be wrong :).

    In any case, it’s an interesting problem with lots of extensions, and I may well try to work it into one of my lessons next year.

  4. July 29, 2008 pm31 5:39 pm 5:39 pm

    I teach a six week unit on expected value and probability as part of a “combinatorics” class, an elective for 12th year students (combining those who want extra math with those few who want me with those who need an extra math course because the regular courses are hard.

    Mix of kids means having a few harder problems around for some is always a good idea. And with these, sometimes you ask the question, play, and come up with nothing, and sometimes you come up with winners. I think, in the end, that this is complicated for the kiddies. But I will keep playing and asking.

    Last year I created some dice games with unequal payouts and convoluted rules and made them calculate expected values. They worked ok, but I found them, for the most part, boring. So I keep exploring.

    Thanks for playing along!

  5. July 29, 2008 pm31 9:02 pm 9:02 pm

    After having checked my numbers, I can confirm that at least some of them are wrong :). I won’t retype them all, but may write a post on this topic myself if I find anything interesting.

    I’ve asked something similar, but less numerically demanding, to my students taking a statistics A-level module (equivalent to your year 11 or 12): imagine you have two players, and two dice, one black and one white. The dice are rolled, and player one wins if the product is 6 or less. Who is more likely to win? The more interesting question is this: can we find a simple winning condition which will make the game fair for both players?

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