Mathematics Puzzle: Adding Random Numbers

Posted on February 28, 2008
Filed Under Math Puzzles |

I have no intention of turning this blog into a math blog, but every now and then I run into interesting mathematical puzzles. My two favorite fields are combinatorics and probability, and this puzzle belongs to one of them:

Suppose you start adding up random numbers chosen from the interval [0,1]. How many numbers, on the average, would you have to add before their sum becomes equal to or exceeds one? 

(Hint: the answer is NOT 2 … )

Comments

6 Responses to “Mathematics Puzzle: Adding Random Numbers”

  1. admin on March 1st, 2008 6:53 am

    Ok, here is a hint (stop reading if you want to think about it for yourself!).

    A naive answer would be two numbers. The expectation value of X, a random variable uniformly distributed between [0,1], is 1/2. So doesn’t it make sense that the expectation value of X+X be 1/2+1/2 = 1, meaning we’ll need to numbers on the average?

    The answer is no. The above reasoning is a sort of “backwards”-reasoning, and it omits a very important piece of information: no matter what, a SINGLE random variable X would NEVER give us 1.

    When you compute the average of X+X+X+X+…+X you’re actually looking at a large number of trials, and you’re not excluding any sort of trial - even the ones in which you have a single random variable.

    Try and looking at the cumulative probability distribution of X+X+X+…+X and see where that gets you.

  2. wandering.the.cosmos on March 2nd, 2008 9:07 am

    I believe the average number of random numbers picked before their sum exceeds unity, is Euler’s number e = 2.71828…

  3. admin on March 2nd, 2008 3:22 pm

    Yep!

    Care to share your solution with the readers?

  4. Shlomo on March 3rd, 2008 8:41 am

    {
    class Program
    {
    static void Main(string[] args)
    {
    int Numbers = 0;
    int one = 1;
    double sume = 0.0;
    int I ;
    double RR;
    int j;
    double Eo = 0.0;
    for (j = 0; j < 10; j++)
    {
    for (I = 0; I 1.0)
    {
    Numbers++;
    sume = 0.0;
    }
    RR = new Random().NextDouble();
    sume += RR;

    }

    Eo = Eo + 2000.0/Numbers;
    Numbers = 0;
    }
    Console.WriteLine(Eo/10);
    Console.ReadLine();

    }
    }
    }

  5. admin on March 3rd, 2008 12:32 pm

    Hey, that’s cheating :).
    I meant a mathematical argument.

  6. admin on March 5th, 2008 6:39 am

    Ok, final hint: prove that the expectation for the time for something to happen equals the sum of all probabilities of it not happening (in the 1st time, 2nd time, 3rd time, etc … ). In other words, prove:

    E(X) = F1 + F2 + F3 + …

    where Fi is the probability of the experiment failing to produce the event in i steps. Now use this to compute the expectation in this riddle. Good luck!

Leave a Reply