Number Game: 31 - Winning Strategy?

My Maths teacher taught us how to play a game called 31 on Friday. Not once did my Maths teacher lose. I want to know why.

I'll explain the game...

31 is a game between two people.

Let's say you've got a grid of numbers - 6 rows, 4 columns. Each column contains the numbers 1-6, such that the first row will contain only 1's, the second will contain only 2's and so on.

Player one begins by choosing an entry from the grid, crossing it off and recording the number chosen. Player two then chooses another entry from the grid (could be the same number), crosses it off and adds this to the number previously recorded. Once an entry has been crossed off, it cannot be reused. The objective is to make the total reach 31 on your turn.

I've established that if you make the total reach 24 on your turn, the next turn a number would have to be chosen which would not allow the total to exceed 30, and hence you would win on your next turn, providing that the entry required to reach 31 had not been crossed off. Furthermore, if you make the total reach 17 on your turn, the next turn a number would have to be chosen which would not allow the total to exceed 23, and hence you would reach 24 on your next turn, providing that the entry required to reach 24 had not been crossed off. Further "magic numbers" are hence 10 and 3.

Using this "magic number" logic, I decided to attempt to beat my Maths teacher. He still beat me - I reached 24 no problem, but ran out of entries in a row (sorry - cannot remember which, think it was row 3), and so, he was crowned the winner to yet another student.

I'm really struggling to devise a fail-proof method. Can you please assist me?

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You can get a lot of information on this game from online sources. This one is quite entertaining. –  André Nicolas May 27 '12 at 5:24

Most likely it was row $3$: you probably started by taking a $3$, and then he took nothing but $4$’s, to which you responded with $3$’s to hit the ‘magic numbers’. Unfortunately, after four moves apiece the total was at $28$, and you’d used all the $3$’s. Your strategy would have been perfect had there been at least five of each number instead of just four.

You can make the same idea work for you as first player if you start by taking $5$. If the second player takes another $5$ to bring the total to $10$, you take $2$. If he then takes another $5$ to make $17$, you take $2$ again. In order to make $24$, he has to take the last $5$. You then take another $2$; there are no $5$’s left, so he can’t hit $31$. No matter whether he takes $1,2,3$, or $4$, you can reach $31$; the closest call is when he takes a $3$, but there’s still one $2$ left, so you still win.

If the second player doesn’t take $5$ on his first turn, there are two cases.

Case 1: He takes $1,2,3$, or $4$, making a total less than $10$. In this case you bring the total to $10$ on your next turn by taking $4,3,2$, or $1$, respectively. At this point there are still at least three of every number left, and only three more numbers to hit ($17,24,31$), so he can’t force you to use up what you need.

Case 2: He takes a $6$, making a total of $11$. Go ahead and take a $6$, making $17$. The worst he can do to you is take a $1$, forcing you to take the third $6$, but that brings the total to $24$, and even if he takes $1$ again, there’s still a $6$ that you can take to reach $31$.

In other words, if he doesn’t respond to your opening $5$ by taking a $5$ himself, you just play the ‘magic number’ strategy.

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That is a dirty plan. There's only one thing that I do not like about it. If he starts the game by taking a 5, I know I'm screwed. ;-) –  clookid May 27 '12 at 5:44