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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 = \(4\).

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 = \(24\).

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 \(72\) 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 = \(4! \times 4\) = \(96\)

Hence total number of ways = \(196\)

Now, let us find no. of favourable ways. Note that now, in every case last turn will have only \(1\) 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. \(4\) ways.(every possible way will satisfy)

Case-2 : First 2 turns are different and third turn is same as first.

\(4 \times 3 \times 1 = 12\) ways

Case - 3: First 3 turns are different and fourth one is same as the first one :

\(4 \times 3 \times 2 \times 1 = 24\) ways.

Case-4 : First four turns are different and fifth turn is same as first turn :

\(4 \times 3 \times 2 \times 1 \times 1 = 24 \) ways.

Hence total number of favorable ways = \(64\).

Hence probability = \(\frac{64}{196} = \boxed{\frac{16}{49}} .\)

## Comments

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TopNewestEach way you list in Case 1 is four times more probable than each way you list in Case 2. Similar with other things. They are not equiprobable. – Ivan Koswara · 3 years, 9 months ago

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