I really need help on this statement/question about probability theory?

Question

the packet state-

Activity:

Roll Two dice to show that the experimental probability of rolling a sum of 8 and a sum of 6 before rolling two of 7 is approximately 55 percent.If you're familiar with probability theory,prove that the theoretical probability is 54.6 percent.

my question is how they arrive at the percentage of 54.6 percent ?please explain

Answer 1

this is probably confusing, but it gets the answer.

so p(6) = 5/36, p(7) = 6/36, p(8) = 5/36.

if any other sum comes besides 6,7,8, we don't care.  so let's ignore those rolls...the probabilities would become:

so p(6) = 5/16, p(7) = 6/16, p(8) = 5/16.   we get the 16 = 5 + 6 + 5.

now break it down into cases, only considering our successful rolls:

case 1: a 7 comes up before a 6 or 8.  thus are successful rolls would be 768 and 786. p(7) = 6/16 from above. then p(6or8) = p(6) + p(8)=10/16.  next be careful that 768 is not our only possible win, but also 7668, 76668, and so on.

assume we got a 7 then a 6.  now we dont care if a 6 comes down.

thus p(7)= 6/11 and p(8) = 5/11.   again 11= 5 + 6 like above.

thus any successful case one will occur with probability:

p(7) * p(6or8) * p(8) = 6/16 * 10/16 * 5/11 = *******************

case 2: either a 6 or a 8 comes before a 7. then we want 68 or 678

again, if it was a 6 comes...p(7)= 6/11 and p(8) = 5/11.   again 11= 5 + 6 like above.

         if it was a 8, p(7)= 6/11 and p(6) = 5/11.

so then the probability for case 2 is

p(6or8) * (p(8) + p(7)*p(8)) = 10/16 (5/11 +6/11*5/11) = ............

p(case 1 + case2) = ************ + ..................

                           = .54558 = .546

done.  sorry I couldn't explain it better, hope you understand.