I want nobody to tell the correct answer here.I already bought the discussion and the answer seems wrong to me. I just wanted to know what is wrong with this approach:
First, let us calculate total number of such possible ways ,
Case-1 : We get the same numbers in first and second turns,
For the first turn, we have 4 options and for the second turn , we have only one option, hence total number of ways = .
Case-2 : First 2 are different and the third is same as either first or second.
Hence, first turn has 4 options, second turn has 3 options and third turn has 2 options(either same as first turn or same as second turn). Hence no. of ways = .
Case-3 : First three numbers are different and fourth one is same as either of them.
First turn has 4, second turn has 3, third turn has 2 and fourth turn has 3 options, hence ways.
Case-4 : First four turns are different and the fifth turn is either of them.(Note that fifth turn has 4 options).
Number of ways = =
Hence total number of ways =
Now, let us find no. of favourable ways. Note that now, in every case last turn will have only option, as it is same as the first one. Hence everytime i will be multiplying by 1 showing last turn has 1 option only.
Case - 1: First 2 turns are same. ways.(every possible way will satisfy)
Case-2 : First 2 turns are different and third turn is same as first.
Case - 3: First 3 turns are different and fourth one is same as the first one :
Case-4 : First four turns are different and fifth turn is same as first turn :
Hence total number of favorable ways = .
Hence probability =