Turn 6 cards upside down

Question

Six identical cards are placed on a table. Each card has number '1' marked on one side and '2' on the other. All cards are placed with '1' facing upward on a table. In one try, exactly four cards (neither more nor less) can be turned upside down. In how many least number of tries can the cards be turned so that all six cards show '2' on the upper side? Please explain how.

Answer 1

Start: 6 cards show '1'.

1 1 1 1 1 1

First turn: Turn over four of the '1' cards. Now 4 cards show '2', and 2 cards show '1'.

2 2 2 2 1 1

Second turn: Turn over three of the '2' cards and one of the '1' cards. Now 2 cards show '2' and four cards show '1'.

1 1 1 2 2 1

Third turn: turn over the four cards showing '1'. Now all six cards show '2'.

2 2 2 2 2 2

How do you get to this solution? Well, there are only three moves you can make, really. You can turn over four cards of one type and none of the other, or two of both, or one of one kind and three of the other. It's obvious that the first two won't work, so it must be the last option that yields the solution.

Answer 2

They start like this:

1 1 1 1 1 1

and here is the three required steps:

2 2 2 2 1 1

2 1 1 1 2 1

2 2 2 2 2 2

I do not know how to phrase this 'mathematically', but this is the answer. Easy to see two turns is not enough, since you end up with at least two 1s.

Answer 3

From a state with $k$ cards showing 2, by turning $a$ cards showing 2 and $4-a$ cards showing 1, we reach a state with $k-a+(4-a)=k+4-2a$ cards showing 2. Of course, we must obey the restrictions that $0\le a\le k$ and $0\le 4-a\le 6-k$, i.e. $\max\{k-2,0\}\le a\le\min\{k,4\}$.

Thus we have the following allowed moves: $$0\stackrel{a=0}\longrightarrow 4 \begin{cases}\stackrel{a=2}\longrightarrow4\\\stackrel{a=3}\longrightarrow2\stackrel{a=0}\longrightarrow 6\\ \stackrel{a=4}\longrightarrow0\\\end{cases}$$ where the second moves leading to $k=4$ or $k=0$ are obviously no good, hence the three moves shown are the shortest possibility.