A and B throws a a pair of dice alternatively.A wins if he throws 6 before B throws 5 and B wins if he throws 5 before A throws 6.Find B chance of winning if A makes the first throw.

• p(A winning) = 1/6 + 5/6*5/6*1/6 +.....so on.
  p(B winning) = 5/6*1/6 + 5/6*5/6*5/6*1/6 +....so on. infinite GP
  hence (5/36)/(1-25/36) = 5/11 ans.
• 9 years agoHelpfull: Yes(27) No(6)
• @ravi ranjan 5/11=45.45%
  it is aroun 50% which seems to be impossible(if we think practially)
  even if 1st chance was given to B then his chances to win would be 16.66%
  so definetly it must be less than 16.66% becoz there is a condition that A must lost first...(this is what i think,i may be wrong)
• 9 years agoHelpfull: Yes(5) No(4)
• event is that B should win
  for B to win A should loss and simultaneously B must win
  A loss=5/6 and B win=1/6
  p(Event)=(5/6)*(1/6)=5/36
• 9 years agoHelpfull: Yes(3) No(6)
• 5/9 am i correct
  
• 9 years agoHelpfull: Yes(2) No(9)
• it must be 5/11
  
• 9 years agoHelpfull: Yes(2) No(3)
• 31/36 x 4/36 + 31/36 x 32/36 x 31/36 x 4/36 + 31/36 x 32/36 x 31/36 x 32/36 x 31/36 x 4/36 + ....infinite times.
  
  31/36 x 4/36 [ 1 + 31/36 x 32/36 + ( 31/36 x 32/36 ) ^2 + ( 31/36 x 32/36 ) ^3 + ( 31/36 x 32/36 ) ^4 +...]
  
  Solving this infinite G.P series we get
  
  = 31/76 (Ans)
• 4 years agoHelpfull: Yes(1) No(0)
• P(A) winning=5/11
  P(B)winning=(5/36)/(1-5/6)=5/6
• 9 years agoHelpfull: Yes(0) No(3)