Little help finding a bug in this program

[Neotriz]

Here's the question:

This program is intended to calculate the probability that if a person
rolls two fair sixed-sided dice, that they will get a total of 8. This
is done by looping through all the possibilities and counting those that
do not have a sum that exceeds 9. However, there are some bugs in this
file. Can you find them?

#include <stdio.h>

int main() {

  int count, total;

  for (i=1; i<6; i++)
    for (j=0; j<=6; j++)
      sum = i+j;
      if (sum = 8)
        count++;
      total++;

  printf("The probability of rolling an 8 is %lf.\n", count/total);

  return 0;
}

So obviosuly, there are some undeclared variables so this is what I wrote:

int main() {

  int count, total;
  int i;
  int j;
  double sum=0;
  
  for (i=1; i<6; i++) {
    for (j=0; j<=6; j++)
      sum = i+j;
      if (sum = 8)
        count++;
      total++;
      }

  printf("The probability of rolling an 8 is %lf.\n", count/total);
  
  system("PAUSE");
  return 0;
}

From there, I'm not too clear of how to make it work

[Adak]

The loop counters i and j both need to go from 1 to 6. No 0's, and no < 6. < 7 is OK though (preferred IMHO).

After sum you need two = signs. sum == 8. Also, you need to set total and count variables = 0, before the for loops.

[Neotriz]

The loop counters i and j both need to go from 1 to 6. No 0's, and no < 6. < 7 is OK though (preferred IMHO).

After sum you need two = signs. sum == 8. Also, you need to set total and count variables = 0, before the for loops.

For the first comment, did you mean like this:

int main() {

  int count, total;
  int i=0;
  int j=0;
  double sum=0;
  
  for (i=1; i<6; i++) {
    for (j=1; j<6; j++)
      sum = i+j;
      if (sum == 8){
        count++;
        }
      total++;
      }

  printf("The probability of rolling an 8 is %lf.\n", count/total);
  system("PAUSE");
  return 0;
}

[abachler]

again, i<6 is incorrect, that will get you 1-5 only, you need i<7 in both for loops

other than that it looks good

[MK27]

Ummm....the chance is 1 in 11. Exactly.

Whatever else you are doing is pointless.

[abachler]

Ummm....the chance is 1 in 11. Exactly.

Whatever else you are doing is pointless.

you are wrong mk27, so horribly wrong.

The True Odds of rolling two dice

[bithub]

      if (sum == 8)

[MK27]

you are wrong mk27, so horribly wrong.

The True Odds of rolling two dice

Okay I see the point now. Sorry.

I always just bet on the red squares

[Neotriz]

again, i<6 is incorrect, that will get you 1-5 only, you need i<7 in both for loops

other than that it looks good

Something else is wrong. The output says the prob. is zero.

I'm pretty sure it has something to do with increments

[Adak]

Ummm....the chance is 1 in 11. Exactly.

Whatever else you are doing is pointless.

You couldn't be more wrong, MK27. He's learning how to code up a C program that finds the solution.

Since the answer is already known, he can check his program's answer.

The variables i and j need to go from 1 to < 7, not 1 to < 6

[abachler]

nop, he forgot ot enclose his inner loop in curly braces

int main() {

  int count, total;
  int i=0;
  int j=0;
  double sum=0;
  
  for (i=1; i<6; i++) {
    for (j=1; j<6; j++){
      if (i+j == 8) count++;
      total++;
      }
    }

  printf("The probability of rolling an 8 is %lf.\n", count/total);
  system("PAUSE");
  return 0;
}

[Adak]

Good catch, Abachler.