It is complicated by the fact that it's Scrabble tiles and not fair coins (it only works for spherical chickens in a vacuum)...
If you throw a fair coin, the probability of it landing heads-up is equal to the probability of it not. So the scenario of x fair coins landing heads-up is just as likely as that of x-1 or x-2 or x+i landing heads-up: that is, each of the possible scenarios has exactly the same probability, which means that said probability is 1 divided by the number of possible scenarios.
So if you have a hundred fair coins, the probability of exactly half of them landing heads-up is equal to the probability of none of them landing heads-up which is equal to the probability of all of them landing heads-up (and all possibilities in between) - ie 1/101
However, if you throw something which has, say a 2/3 chance of landing heads-up then not all of your scenarios have an equal chance.
This is where a tree diagram comes in. (I mean, you can do this for the fair-coin problem too, but when you do you'll see you don't need it.)
If you have two scrabble tiles your possibilities are
HH - 2/3 x 2/3 = 4/9
HT 2/3 x 1/3 = 2/9
TH 1/3 x 2/3 = 2/9
TT 1/3 x 1/3 = 1/9
(quick check - total adds up to one)
Here your probability of exactly half being face-up is 2/9 + 2/9 = 4/9
Add another tile and you get into 2/3 x 2/3 x 2/3 = 8/27 and so on and obviously then you can't get half of them face-up and I can't be arsed to do it for four 16/81