expected number of moves to match pairs in matching game There are 6 tiles made up of 3 matching pairs (e.g. 1,1,2,2,3,3). They are mixed up and flipped over so you can't see the numbers. Assume you have perfect recall (i.e. if you flip over a tile, you remember what was under that tile forever).
For each move, you flip over two tiles sequentially. If they match, they remain flipped. If not, they are flipped back (but you still remember what was on them).
What are the expected number of moves that you will make in this game?
To clarify - if there were only 4 tiles, then the game could go two ways:
1) In the first turn, you flip two tiles with the same number. The game ends in two moves (probability is 1/3)
2) In the first turn, you flip two tiles with different numbers. You finish the game in three turns (probability is 2/3).
The overall expected number of moves is 1/3 * 2 + 2/3 * 3 = 8/3.
So - what would this be for 6 tiles?
 A: Outline: With $6$ tiles, the game cannot go on long, so we basically trace out all possibilities. For the expectation, we condition on the result of the first turning  of a pair. Please note that we are assuming that on each turn two tiles are turned over simultaneously. 
Suppose that on the first turn we get a match. This has probability $\frac{1}{5}$. In that case the expected number of turns is $\frac{8}{3}+1$.  That will make a contribution of $\left(\frac{1}{5}\right)\left(\frac{8}{3}+1\right)$ to the expectation. 
With probability $\frac{4}{5}$, we don't get a match on the first turn. Without loss of generality we may assume we got a $1$ and a $2$. Now on the next turn we turn over two new tiles.  Three things can happen: (i) we get a double $3$; (ii) we get a $1$ and a $2$; (iii) we get a $1$ or $2$, and a $3$.
For each case, we can compute the associated probability, and the expected total number of moves. 
A: At the start of any turn $t$:
Let $0\le n=2p$ be the number of cards left.
Let $0\le k\le p$ be the number of pairs that have previously had one of their members exposed. Once a pair has been identified but not removed, the next move will remove it - this move has to happen at some point and it makes the analysis easier if it is done immediately.
There are therefore $n-k=2p-k\ge 0$ cards that have not been exposed, of which $k$ belong to a previously exposed pair and $2(p-k)\ge 0$ don't.
There are 4 possibilities with each move:


*

*The first card of a move exposes a known pair - the second card will remove that pair.
$$p_1=\frac{k}{2p-k}$$
$$p\to p-1$$
$$k\to k-1$$
$$t\to t+1$$

*The first card doesn't match and the second card matches the first card - this pair is removed.
$$p_2=\frac{2p-2k}{2p-k}\frac{1}{2p-k-1}$$
$$p\to p-1$$
$$k\to k$$
$$t\to t+1$$

*The first card doesn't match and the second card matches a previously exposed card - this pair is then removed.
$$p_3=\frac{2p-2k}{2p-k}\frac{k}{2p-k-1}$$
$$p\to p-1$$
$$k\to k-1$$
$$t\to t+2$$

*The first card doesn't match and the second card doesn't match anything either.
$$p_4=\frac{2p-2k}{2p-k}\frac{2p-2k-2}{2p-k-1}$$
$$p\to p$$
$$k\to k+2$$
$$t\to t+1$$


$$p=3;k=0\begin{cases}p_1=0&, \text{not possible}\\
p_2=\frac{1}{5}&, p\to 2,k\to 0, t\to 1\\
p_3=0&, \text{not possible}\\
p_4=\frac{4}{5}&, p\to 3,k\to 2, t\to 1\\
\end{cases}$$
So after 1 turn we have


*

*a $\frac{1}{5}$ chance of going to $p=2, k=0$ (this is the 4 card state for which the expected number of moves is $\frac{8}{3}$ by your reasoning). So expected turns on this path is $\frac{11}{3}$.

*a $\frac{4}{5}$ chance of going to $p=3, k=2$, from here the probabilities are:
$$p=3;k=2\begin{cases}p_1=\frac{1}{2}&, p\to 2,k\to 1, t\to 2\\
p_2=\frac{1}{6}&, p\to 2,k\to 2, t\to 2\\
p_3=\frac{1}{3}&, p\to 2,k\to 1, t\to 3\\
p_4=0&, \text{not possible}\\
\end{cases}$$


It can be shown that each of these cases will be completed in 2 turns - a total of 4 turns on this path.
Summing over all options gives $E(t)=3\frac{14}{15}$.
