An alternative way to crack this is to treat the dot counts purely as a numeric sequence $a_1, a_2, \ldots, a_6$ and test a recurrence relation instead of just eyeballing the pattern.
Reading off the counts from the six visible tiles gives $a_1=0$, $a_2=1$, $a_3=1$, $a_4=2$, $a_5=3$, $a_6=5$.
Testing the hypothesis $a_n = a_{n-1} + a_{n-2}$ for $n \geq 3$: $a_3 = a_2+a_1 = 1+0=1$, correct. $a_4=a_3+a_2=1+1=2$, correct. $a_5=a_4+a_3=2+1=3$, correct. $a_6=a_5+a_4=3+2=5$, correct. Since the recurrence holds for every term for which it can be checked, it is reasonable to extend it one step further.
Computing $a_7 = a_6+a_5 = 5+3 = 8$.
Now compare $a_7=8$ against the dot totals in the four candidate tiles: 4, 6, 8, and 9 dots respectively. Only the tile with 8 dots satisfies the recurrence.
