Probability that final score is six 
A child throws 2 fair dice. If the numbers showing are unequal, he adds them together to get his final
  score. On the other hand, if the numbers showing are equal, he throws 2 more dice & adds all 4 numbers
  showing to get his final score. The probability that his final score is 6 is


Cases when they are unequal : (1,5)(2,4)(4,2)(5,1)
cases when they are equal :(1,1,2,2)(1,1,1,3)(1,1,3,1)(2,2,1,1)
Hence the probability should be 4/30 + 4/216
but the answer is given as 148/1296
 A: You have a few mistakes.  AugSB hinted at one of them.  Another is the $216$; this is $6^3$, how does that come in?  Note that the answer has $1296$ which is $6^4$ and makes more sense as 4 dice are involved. 
The first 2 dice can land in 36 ways with equal probability.  You and I know that in 6 cases, we will need to throw 2 more but the dice don't know that.  All 36 possibilities will be equally likely.  
So, the possibilities after the first two throws are:
$4/36$ we have a total of 6 without a double and we stop.
$26/36$ we don't have a total of 6 or a double and we stop.
$6/36$ we have a double and must continue.  We need to look more carefully.  
$1/36$ we have a double 1, we throw 2 more dice and have a $3/36$ chance of getting a total of 6.  
$1/36$ we have double 2, we throw 2 more and have a $1/36$ chance of getting 6.  
$4/36$ we have a double 3 or more, we throw again but we can't get 6.
So, overall a successful outcome has probability of $4/36 + 1/36\times3/36 + 1/36\times1/36 = 148/1296$.
