Pearls II

And when he starts and goes for the last, fourth row, he's an idiot, because he's going to lose either way. If he takes one (which he never does), I’ll make 3 3 5 5. If he takes two, I’ll make 3 4 3 4. If he takes three, I’ll make 3 4 4 3. If he takes four, I’ll make 2 3 4 5—the foundation of a future winning combination. If he takes five or six, I’ll make 1 4 5, and he’s done for.

If he starts and takes the 3 from the third row (he won’t take 1, 2, or 5, because he’d lose—I’d again form a winning combination from the rest), then there are several possibilities, which I’ll get to right away. Well, he won’t take all 4, because the computer gives us a chance to win :) But back to those scenarios: from the first row, where there are 3 beads, I can’t take one or two, because I’d lose. The Magician would create winning combinations. If I took 1, it would create 2 4 2 4 (the same number of beads in the even rows). If I took 2, it would create 2 4 6—also a winning combination. So I can only take 3, and by doing so, I’ll create the winning combination 2 4 6 myself. Now the second row, where there are 4 beads. Here, I don’t stand a chance. If I take 4, he’ll make 1 2 3. If I take 3, he’ll make 1 2 3 again. If I take 2 beads, he’ll make 3 2 2 3—again, the same number of beads in the even rows, which is a winning combination. And if I take one, it’s the same thing all over again. He’ll make 3 3 2 2—again, the same number of beads in the even rows. Now the third row—the one the magician was taking from. If I take both, I lose. The magician makes 2 4 6. But if I take just one, I’ll beat him. He’ll then have to make a move, so he’ll have to take a bead. And as soon as he takes one, I’ll make a winning combination from what’s left (2 4 6—the same number of beads in the even rows—or 2 3 4 5, whichever suits me best). And what about the last, fourth row, where there are 6 beads? If I take one, I’ll create 2, 3, 4, 5—again, the basis for a future winning combination—because the magician, whether he likes it or not, has to take a bead. If I take 2, he’ll create 2 4 2 4, so don’t do that. If I take 3 beads, he’ll create the combination 3 2 2 3, so don’t do that either. If I take 4 beads, he’ll create 3 3 2 2, so again, no. If I take 5, he’ll make 1 2 3—also no. If I take 6, he’ll make 1 2 3 again—don’t do that either.

When it starts and takes 3 from the second row (it doesn't take 1 or 2, because I would form winning combinations. If it took one, I would form 3 3 5 5—the same number in the even rows. If he took two, I’d create 2 3 4 5. I just don’t understand why he doesn’t take all 4—because then I’d lose. I guess the computer always gives us a chance to win), there are again several possibilities. First row. If I take 1, I’ll put him in a situation where he has to take something, and I’ll form a winning combination, so yes. If I take 2, he’ll form 1 1 5 5, so no. If I take 3, he’ll form 1 4 5, so no again. Second row: if I take the remaining one, I win. He’ll have to take something, and I’ll set up a winning combination of 1 2 3 or 2 4 6 or 1 4 5 or the same number in two rows. So basically, he’s in trouble if he takes from the second row :) Third row: if I take 1, I’ll set myself up. He’ll take something, and then I’ll form a winning combination, so yes. If I take a 2, he’ll form the combination 3, 1, 3, 1—so no. If I take a 3, he’ll form 1, 2, 3—so no again. If I take 4, he’ll make 3 1 1 3—again, no. And if I take all 5, he’ll make 1 2 3 again, so also no. Fourth row: if I take 1, he’ll make the combination 1 1 5 5, so don’t take it. If I take 2 beads, it makes 1 4 5, so no. If I take 3, the magician makes 3 1 1 3, so no again. If I take 4 or 6, it makes 1 2 3, so no again. If I take 5, the magician creates 3 1 3 1—again, no. So don’t touch the bottom fourth row at all!

I've already posted the solution here. There are several possibilities. When the game starts and the magician takes the 1 from the first row (he never takes the 2 or 3, because he would lose), I would use the remaining beads to form the winning combination 1 4 5), there are several possibilities. From the first row, where the magician took a bead, I can’t take anything. If I took one, he would form the winning combination 1 4 5. If I took two pearls, he’d again form 1, 4, 5. So, thanks, but no to the top row. The second row—this looks more promising. If I take 1, he’ll form 2, 3, 4, 5, so don’t take it. If I take two, he’ll form 2 2 5 5, so don’t take those either. But if I take 3, he has to take something, and then I’ll be able to form one of the winning combinations. And if I take 4, he’ll form 2 4 6, so no to that one, too. Third row—if I take 1, he’ll form the winning combination 2 4 4 2 (the same number of beads in the even rows), or 2 4 6, so no. If I take 2, he’ll make 2 4 6—so that won’t work either. If I take 3, he’ll make 2 4 2 4 or, again, 2 4 6—so that won’t work either. And if I take 4, it again forms 2 4 6. Only if I take all 5 will I form the combination 2 4 6 and win. And the fourth row? No chance. I take 1 bead; he makes 2, 2, 5, 5 (again, the same number of beads in the even rows). If I take 2, it creates 2 4 2 4; if I take 3 beads, he creates 2 3 4 5, which sets up the next winning combination, because it’s my turn and I have to take a bead—and no matter which one I take, it won’t help me. If I take 4 beads, he’ll form 2 4 4 2 again. If I take 5, he’ll form 1 4 5, and if I take all 6, he’ll form 1 4 5 again. So, the bottom row is out.

great game

Hey, ReadyTeddy, Tutorial 1 is great—when I start, I always get it right, but once he starts, I’ve actually lost before. Where am I going wrong? I’m trying to make a winning combination like you described, but he sometimes does it differently! For example, I have 1, 4, 6, but I can’t force it to be 1, 1, 1 or 2, 2, 2. When there are a lot of balls, I just can’t keep track :-(

zormba - if you weren't such a dummy, you'd read the walkthrough that Ready Teddy posted here; I actually finished the game

YOU CAN'T WIN!!!