Yes! by looking at the numbers of coins and Voltorbs, combined with simple math and logics! Here is a great way to play:

Always start with picking 3 or 4 cards from high value and low Voltorb rows.

Eliminate (ignore) any row or coloumn that has voltorbs and coins that add up to 5. That means that in a row with (example) 4 coins and 1 voltorb, there will only be 1x cards. 1+4=5 you have 4 other cards that aren't voltorbs, and to total 5, they must all be ones. you can now eliminate this and other 4and1 or 3and2 rows, because there will never be more than 1 coin per card here.

By just looking at a row with one exposed card can tell you alot. If I have a row that has 5 coins and I have seen one 1x coin, I'll know that there is a 2x card there.

Because: 1 card out, right? 4 left. One is a Voltorb. So this means I have 3 more cards with value. 4 coins in 3 cards means only one thing 2 1x cards, and a 2x card. You can use this simple algebra to easily figure out where the value cards are in most rows and coloumns. Then use the same math to figure out if there is any other good cards on that row, or if they're all ones or Voltorbs.

These two pro tips are excellent to use. Of course, you still need some luck to hit a good card.

- The not-so-smart finishing moves: Once you've used this technique succsessfully, you'll notice that you can eliminate tons of rows very quickly. So at the end you just use process of elimination, and pick off remaining squares.

You may be left with 4 squares you know are good, while you've eliminated 5 rows without even selecting more than 1 card in each one. The safe way to play!

Good Luck.