A simple math puzzle about dice probability

I love reading about probability puzzles: even the easiest puzzles can take a moment to figure out exactly what is going on.

The book Luck, Logic, and White Lies starts out with a fun little puzzle about dice throwing:

With a pair of dice, one can throw the sum 10 either as the combination 5+5 or 6+4. The sum 5 can also be obtained in two ways, namely, by 1+4 or 2+3. However, in repeated throws, the sum 5 will appear more often than 10. Why?


photo by Ella’s Dad

Like many probability calculations, this puzzle has a counter-intuitive feel to it. There are two things that seem to have the same chance of occurring and yet one is more likely than the other.

Do you know why? Give it a try before reading the solution.

The solution

The trick is all about the wording of the puzzle which creates a mystery where there is none.

The sum 10 can be obtained in three ways by dice roll: namely (5,5), (4,6), and (6,4); the sum 5 in four ways: (1,4), (4,1), (2,3) and (3,2).

So the sum 10 is obtained with probability 3/36 versus the sum 5 with probability 4/36.

Pictorially:

1 3 4 5 6
1 5
5
3 5
4 5 10
5 10
6 10

The puzzle demonstrates that it’s always important to consider the events in probability. Sly wording, like this puzzle’s ways describing sums rather than pairs of rolls, can easily confuse.



Share this post:

| More

Previous post:

Next post:



  • Benvitale

    Yes, we must consider ‘Events’ in probability.
    Possible sums are: …2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
    distribution of sums: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1
    This would look much better in a table.
    This tells us that 2 appears in one way (as 1 + 1),
    3 appears in two ways (1+2 and 2+1)
    4 appears in 3 ways, 5 in 4 ways, 6 in 5 ways,
    7 in 6 ways, 8 in 5 ways, 9 in 4 ways,
    10 in 3 ways, 11 in 2 ways, 12 in one way.

    S is the sample space of the experiment and E is the event of interest.
    n(S) is the number of elements in the sample space S
    and n(E) is the number of elements in the event E

    The sample space S of two dice is 36
    For example, let E be the event “sum equal to 2″
    Only 1 possible outcome give a sum equal to 2;
    E ={(1,1)}, hence
    P(E) = n(E) / n(S) = 1/36
    And so forth.

  • Benvitale

    Here’s a puzzle:
    Could you find another way to put numbers on the dice with the same sum distribution?

  • Mike

    @Benvitale:

    One such way:

    Die 1:
    1/2, 3/2, 5/2, 7/2, 9/2, 11/2

    Die 2:
    3/2, 5/2, 7/2, 9/2, 11/2, 13/2

    Then I looked at it more and realized the general formula is this:

    Each side of Die 1 + K
    Each side of Die 2 – K

  • Benvitale

    @Mike.

    Yes. But if we only consider positive integers, then the only solution is: Sicherman dice.

    See the Mathematical justification:

    http://en.wikipedia.org/wiki/Sicherman_dice#Mathematical_justification





Previous post:

Next post:

Other posts you may enjoy reading: