Just to replicate the progress, if I put the following into Mathematica, I get the same answers as @byorgey is getting.

Take P\left(t\;\middle|\;m,k\right) to be the probability of needing exactly t turns given m pairs still face down and k of those 2m cards already seen.

P\left(0\;\middle|\;0,k\right) = 1

P\left(1\;\middle|\;1,k\right) = 1

P\left(t\;\middle|\;m,k\right) = 0 if t\leq 0 or m \le 1

P\left(t\;\middle|\;m, 2m\right) = 0

P\left(t\;\middle|\;m, 2m-1\right) = 0

P{\left(t\;\middle|\;m, k\right)} = \frac{k}{2m-k} P\left(t-1\;\middle|\;m-1,k-1\right) \\\quad\quad\quad\quad\quad\quad+ \left(1-\frac{k}{2m-k}\right)\frac{k}{2m-k-1} P\left(t-2\;\middle|\;m-1,k\right) \\\quad\quad\quad\quad\quad\quad+ \left(1-\frac{k}{2m-k}\right)\frac{1}{2m-k-1} P\left(t-1\;\middle|\;m-1,k\right) \\\quad\quad\quad\quad\quad\quad+\left(1-\frac{k}{2m-k}\right)\left(1 - \frac{k+1}{2m-k-1}\right) P\left(t-1\;\middle|\;m,k+2\right)

Or in Wolframaic:

```
p[t_, m_, k_] := p[t, m, k] = Which[
t == 0 && m == 0, 1, (* no more pairs so no more turns *)
t == 1 && m == 1, 1, (* 1 more pair so 1 more turn *)
t <= 0 || m <= 1, 0, (* no other way to have 0 more turns or 1 more pair *)
k >= 2 m - 1, 0, (* prevents division by zero *)
True, k/(2m-k) p[t-1, m-1, k-1]
+ (1-k/(2m-k)) k/(2m-k-1) p[t-2, m-1, k]
+ (1-k/(2m-k)) 1/(2m-k-1) p[t-1, m-1, k]
+ (1-k/(2m-k)) (1-(k+1)/(2m-k-1)) p[t-1, m, k+2]
]
```

(I get 80.1732 turns in expectation for a 10x10 grid of 50 pairs.)

I don’t think Mathematica can actually solve the recurrence relation.

And, wait, the plot thickens! Or straightens? I made a plot of the expected number of turns as a function of the number of pairs and it turns out it’s pretty much linear. The function is extremely well-approximated by 1.6138329300699403p - 0.5206273290971469. What do we make of that?

PS: I refined that linear regression to try to focus on what the function is converging to in the limit and am getting this: 1.6137060532056546696p - 0.51154898513237434532.