Two students play a game based on the total roll of two standard dice. Student A says that a 12 will be rolled first. Student B says that two consecutive 7s will be rolled first. The students keep rolling until one of them wins. What is the probability that A will win?

• Let p be the probability that student A wins. We consider the possible outcomes of the first two rolls. (Recall that each roll consists of the throw of two dice.) Consider the following mutually exclusive cases, which encompass all possibilities.
  
  If the first roll is a 12 (probability 1/36), A wins immediately.
  If the first roll is a 7 and the second roll is a 12 (probability 1/6 · 1/36 = 1/216), A wins immediately.
  If the first and second rolls are both 7 (probability 1/6 · 1/6 = 1/36), A cannot win. (That is, B wins immediately.)
  If the first roll is a 7 and the second roll is neither a 7 nor a 12 (probability 1/6 · 29/36 = 29/216), A wins with probability p.
  If the first roll is neither a 7 nor a 12 (probability 29/36), A wins with probability p.
  Note that in the last two cases we are effectively back at square one; hence the probability that A subsequently wins is p.
  
  Probability p is the weighted mean of all of the above possibilities.
  
  Hence p = 1/36 + 1/216 + (29/216)p + (29/36)p.
  
  Therefore p = 7/13.
• 12 years agoHelpfull: Yes(8) No(9)
• seems correct.
  
  check Copy paste solution from
  
  http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-cat08-t-27454/p-919607/r-1220540?page=473
• 12 years agoHelpfull: Yes(5) No(3)
• please explain if the first roll is neither a 7 nor a 12 how does A wins because in que it is given that if 12is rolled first then A wins
• 12 years agoHelpfull: Yes(0) No(1)