Challenge – Equal Probability between 1 and 7

              Write a method to generate a random number between 1 and 7, given a method that generates a random number between 1 and 5. The distribution between each of the numbers must be uniform.

              Let’s think of this like a decision tree. Each rand5() will be a decision. After 2 tries, we have 25 possible solutions. We try to get maximum bunches of 7 as we can (1 – 21, 3 sets of 7). If we get any of the other values, we just try again. Since the probability of getting each of 21 values are the same every time, trying again won’t affect their probabilities.

int rand7() {
    while (1) {
        int num = 5*(rand5()-1) + rand5();
        if (num < 22) return ((num % 7) + 1);
That was fun, right? Anyone up for another challenge? Watch out for it next tuesday (March 1st).


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