Given 10 cards with numbers 1,2,3...10, 2 distinct cards are randomly chosen and duplicated, and then the 12 cards are shuffled.
What is the probability that the numbers 1,2,3,4 are the numbers on the first 4 cards (in that order)?
Given that the above is true, what is the probability that 1 was duplicated?
$\color{Blue}{\textbf{Part 1 }}$
Heya don't beat yourself up this question is definitely very fiddly. We are going to solve it together. We are going to solve the problem using conditioning. Here are the four steps we will take:
$\textbf{Step 1:}$
Define $A_0,A_1$ and $A_2$ the events that respectively when duplicating we get $0,1$ or $2$ of the "helpful cards", the $1,2,3 $ or $4$. I.e if we duplicate a 3 and a 7 then $A_1$ has occured. Instead if we duplicated the $1$ and $4$ then $A_2$ occurred. There is a term for these three events and mathematicians call them a "disjoint partition". This is because 1 of them and exactly 1 of them must happen.
$\textbf{Step 2:}$
We know wish to work out the probability of the $A_i$ events.
$\mathbb{P}(A_2)=\frac{4}{10}\cdot\frac{3}{9} = \frac{12}{90}$
$\mathbb{P}(A_0)=\frac{6}{10}\cdot\frac{5}{9} = \frac{30}{90}$
$\mathbb{P}(A_1)=1-\mathbb{P}(A_2)-\mathbb{P}(A_0)= \frac{48}{90} $
$\textbf{Step 3:}$
We now wish to work out the probability of success in 3 separate circumstances. Let us call $S$ success. We will first calculate $\mathbb{P}(S|A_0)$, this is the probability of success given we didn't manage to duplicate any helpful cards. And as such there is exactly one and only one combination of the initial first four cards for success. We must draw the $1$ followed by the $2$ followed by the $3$ followed by the $4$. And as such we have only $1$ choice for the first 4 cards and after this we then have $8!$ ways to order the remaining cards and hence the number of successful combinations is $\frac{8!}{12!} = \mathbb{P}(S|A_0)$
Now onto $\mathbb{P}(S|A_1)$.
Instead of having only $1$ successful combination for our initial four cards we now have $2$. Let us say the duplicated card was the $3$. And call the original three $ 3_O$ and the duplicated three $3_D$. Our two successful first four card combinations are $1$ followed by the $2$ followed by the $3_O$ followed by the $4$ and $1$ followed by the $2$ followed by the $3_D$ followed by the $4$. Hence we have twice as many successful combinations and as such $\mathbb{P}(S|A_1) = 2\cdot \mathbb{P}(S|A_0) = \frac{2\cdot8!}{12!}$
And finally $\mathbb{P}(S|A_2)$.
You might have guessed it but we know have four successful combinations for our initial four cards. There are now two cards that we have 2 different choices for. For sake of a nice example let us say the 2 and 4 have been duplicated. The four successful initial combinations are: $(1,2_O,3,4_O),(1,2_O,3,4_D),(1,2_D,3,4_O),(1,2_D,3,4_D)$
And hence $\mathbb{P}(S|A_2)= 4\cdot \mathbb{P}(S|A_0) = \frac{4\cdot8!}{12!}$
$\textbf{Step 4:}$
The law of total probability states that if $A_1, A_2 , A_3, ..., A_n$ are a disjoint partition then for any event $X$ we have $\mathbb{P}(X) = \sum\limits_{i=1}^n\mathbb{P}(X|A_i)\cdot\mathbb{P}(A_i)$
We have our disjoint partition and their probabilities and we have our probability of success conditioned on this partition so let us put this all together:
$\mathbb{P}(S) = \sum\limits_{i=1}^n\mathbb{P}(S|A_i)\cdot\mathbb{P}(A_i) = \mathbb{P}(S|A_0)\cdot\mathbb{P}(A_0) + \mathbb{P}(S|A_1)\cdot\mathbb{P}(A_1) + \mathbb{P}(S|A_2)\cdot\mathbb{P}(A_2) =$
$\frac{8!}{12!} \cdot \frac{30}{90} + \frac{2\cdot8!}{12!} \cdot \frac{48}{90} + \frac{4\cdot8!}{12!} \cdot \frac{12}{90} =\frac{8!}{90\cdot 8!}(30+96+48) = \frac{174\cdot8!}{90\cdot 12!} \approx 0.0162 \% $ This is the answer to your first question.
$\color{Blue}{\textbf{Part 2 }}$
Now we will move onto the next part. The probability that a 1 was duplicated given we succeeded. We will do this in three steps:
$\textbf{Step 1:}$
Bayes formula gives us $\mathbb{P}(A_i|S) = \frac{\mathbb{P}(S|A_i)\cdot\mathbb{P}(A_i)}{\mathbb{P}(S)}$ We have already worked out each of these values so I will just list them here for you, all the factorials cancel out so the answers are nice and simple:
$\mathbb{P}(A_0|S) = \frac{5}{29}$
$\mathbb{P}(A_1|S) = \frac{16}{29}$
$\mathbb{P}(A_2|S) = \frac{8}{29}$
$\textbf{Step 2:}$
Call $D$ the event that a $1$ was duplicated. We know would like to find out the probability a $1$ was duplicated under each of the $A_i$
$\mathbb{P}(D|A_0)$ is clearly $0$ as $A_0$ means none of $1,2,3$ or $4$ were duplicated.
$\mathbb{P}(D|A_1)$ is clearly $\frac{1}{4}$ as under $A_1$ only one of $1,2,3,4$ were duplicated each with equal likelihood
$\mathbb{P}(D|A_2)$ is clearly $\frac{1}{2}$ as under $A_1$ two of $1,2,3,4$ were duplicated each with equal likelihood
$\textbf{Step 3:}$
Again using the law of total probability we have: $\mathbb{P}(D|S) = \mathbb{P}(D|A_0)\cdot\mathbb{P}(S|A_0) + \mathbb{P}(D|A_1)\cdot\mathbb{P}(S|A_1) + \mathbb{P}(D|A_2)\cdot\mathbb{P}(S|A_2) = $
$0 \cdot \frac{5}{29} + \frac{1}{4} \cdot \frac{16}{29} + \frac{1}{2} \cdot \frac{8}{29} = \frac{24}{116} =\frac{6}{29} \approx 21\% $
I hope this helped. This was a fiddly question so no worries for having found it hard. If you would like anything made more clear let me know.