Bucky Roberts dice challenge. LUL

bucky_challenge_2.c

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

#define DICE_COUNT 3
#define SIDE_COUNT 6

unsigned int roll(unsigned int side_count) {
    int n;
    
    /* When RAND_MAX isn't evenly divisible by side_count, we'll have a bias problem.
     * Lets analyse the problem: If rand() only returns a value from [ 0, 1, 2, 3, 4, 5 ]
     * for a six-sided dice, then each side has equal distribution. However if you tack
     * another number onto the end of that, and we have to use modulo to reduce the set of
     * seven numbers to six, then one side will occur twice as often as the any of others.
     * We can eliminate the bias by discarding any numbers greater than or equal to the
     * largest multiple of six that's less than RAND_MAX. */
    
    do {
        n = rand();
    } while (RAND_MAX - n <= RAND_MAX % side_count);
    return n % side_count + 1;
}

int main(void) {
    srand(time(NULL));

    unsigned long int last_total = 0, next_total = 0;
    for (unsigned int x = DICE_COUNT; x > 0; x--) {
        last_total += roll(SIDE_COUNT);
        next_total += roll(SIDE_COUNT);
    }

    printf("The last %u rolls totaled to %lu\n", DICE_COUNT, last_total);
    printf("Will the next %u rolls total [h]igher, [l]ower or [s]ame?\n", DICE_COUNT);

    int c = getchar();
    if (c == EOF) {
        puts("EOF encountered :(");
        exit(EXIT_FAILURE);
    }

    if (strchr("hls", c) == NULL) {
        puts("Invalid selection :(");
        exit(EXIT_FAILURE);
    }

    printf((c == 'h' && next_total > last_total)
        || (c == 'l' && next_total < last_total)
        || (c == 's' && next_total == last_total)
         ? "Aww, crikey!\n"
         : "Nah, m8. I reckon ur full of shit, m8. The next rolls have totaled %lu\n", next_total);

    return 0;
}