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Using % with rand is wrong.

Assuming your RAND_MAX is the same as mine 2147483647 then the probabilities for each number are:

dice1 = rand() % 6 + 1;
1: 357913942/2147483647 Notice a slightly higher probability for a 1.
2: 357913941/2147483647
3: 357913941/2147483647
4: 357913941/2147483647
5: 357913941/2147483647
6: 357913941/2147483647

The solution use C++11 random functionality.
Correct for the skew in C++03's rand()

Unfortunately I can't find a correct answer on SO for using rand().

int dieRoll() // return a number between 1 and 6
{
 static int maxRange = RAND_MAX / 6 * 6; // note static so calculated once.
 int result;
 do
 {
 result = rand();
 }
 while(result > maxRange); // Anything outside the range will skew the result
 return result % 6 + 1; // So throw away the answer and try again.
}

Note:

int result = rand() * 1.0 / range; // does not help with distribution

Using % with rand is wrong.

Assuming your RAND_MAX is the same as mine 2147483647 then the probabilities for each number are:

dice1 = rand() % 6 + 1;
1: 357913942/2147483647 Notice a slightly higher probability for a 1.
2: 357913941/2147483647
3: 357913941/2147483647
4: 357913941/2147483647
5: 357913941/2147483647
6: 357913941/2147483647

The solution use C++11 random functionality.
Correct for the skew in C++03's rand()

Unfortunately I can't find a correct answer on SO for using rand().

int dieRoll() // return a number between 1 and 6
{
 static int maxRange = RAND_MAX / 6 * 6; // note static so calculated once.
 int result;
 do
 {
 result = rand();
 }
 while(result > maxRange); // Anything outside the range will skew the result
 return result % 6 + 1; // So throw away the answer and try again.
}

Note:

int result = rand() * 1.0 / range; // does not help with distribution