long odds

Quote: nick

... how many 12s should be added to the equation to make it an even proposition?

#include  
#include 
#include 

main() {
 int s, d1, d2; 
 int noS = 0, noT = 0;
 int ctr, x = 0, y = 0;

 srand(time(NULL));

 x = 0;
 y = 0;
 
 for (ctr = 0; ctr < 100000000; ctr++) {
 if (ctr%1000000 == 0)
 fprintf(stderr, "%d\r", ctr);

 d1 = rand()%6+1;
 d2 = rand()%6+1;
 s = d1+d2;

 if (s == 7)
 x++;

 if (s == 12)
 y++;

 // input value of x 7's before y 12's
 if (x == 34) {
 noS++;
 x = 0;
 y = 0;
 }

 // input value of y 12's before x 7's
 if (y == 6) {
 noT++;
 x = 0;
 y = 0;
 }
 }

 printf("\n");
 printf("Number of times 7's completed first: %d\n", noS); 
 printf("Number of times 12's completed first: %d\n", noT); 
}

->OUTPUT->
Number of times 7's completed first: 294860
Number of times 12's completed first: 283584

Quote: nick

Basically you are agreeing that it is approximately even odds whether you get 6 consecutive 7s or 1 12 before those 6 7s are thrown

Quote: nick

what are the chances of getting 6 sevens before 1 twelve.

Quote: DJTeddyBear

The odds of hitting six 7s in a row, an event that requires all six throws to resolve, is ( 6 / 7 ) ^ 6 = 39.7%
Therefore the odds of NOT doing that, i.e. hitting AT LEAST one 12 in those six throws, is 100% - 39.7% = 60.3%

Quote: Wizard

...the probability that the first six times a 7 or 12 is rolled it is a 7 every time is (6/7)^6 = 39.66%

Quote: nick

My question is based on dice odds. I know that there are 6 ways to get 7 and 1 way to get 12 but what are the chances of getting 6 sevens before 1 twelve. Are they even and if not how many 12s should be added to the equation to make it an even proposition?

Thank you
NIck