# Thread: mathematical game

1. ## mathematical game

This is a challange problem.

you play a game against a smart partner. the game goes like this. each person, in his turn, chooses a number from 1-9 that hasn't been chosen before. game ends when all numbers have been chosen, or when someone wins. someone wins if he has chosen any three numbers which sum up to 15, so for example, if i have 2,4,8,9, i win, because 2+4+9=15.

find a strategy for this game; can the first player win? can the second? which one has a strategy?

pmed Jameson the Admin the solution(the only mod i could find, sorry >_<)

Moderator edit: Approved.

2. If I am the first player and choose 9 as my first pick, then the other player will be forced to pick 6 because if not, I win. Then I pick 5 (sum is now 14) and the other player will be forced to pick 1, since if he does not, I win. His sum is now 7 (if he picks 8 I'm toast). Now I pick 10 and get a sum evaluating to 15 and I win.

So yes, the first player can win every time with the exact same strategy! If the first player implements this strategy, the second player cannot win, so in that case the second player cannot, in general, win. So yea, this is my answer.

Now, are there any additional restrictions?

3. no, this is not what i meant you have to have EXACTLY three numbers which sum up to 15; if you have 6,7,8,9 you don't win - perhaps i didn't explain it right. 6 and 9 are only 2 numbers, so having them does not win also, the only numbers you can choose aer 1-9, you cannot pick 10...

4. Oh, never mind then!

5. It's way easier than it seems, I suggest you keep thinking, you'll figure it out in a few minutes or so

6. Well, the combinations to win are:
186, 195, 276, 285, 294, 384, 375.

Since 2 and 5 are the numbers that repeat themselves the most, I would pick either one. Say I pick 5.
Now, my opponent, knowing that I picked 5 can pick 1, 2, 3, 7, 8, 9, the numbers which get to 15 using 5 to prevent me from winning. But if my opponent is the smartest, he'll pick 2, forbidding me from 3 combinations in one play. I'm left with only two solutions: 195, 186, 284 or 375.
If I pick 1, he'll pick 9 (to prevent me from winning 195), then I pick 4 (to prevent him from winning 294), he'll pick 7 (to try getting 276), I pick 6 (to prevent him from winning) and he picks 8. Then, there's draw.

So, I can't find some way to really win if your opponent doesn't makes a mistake...

7. you've given me an example for an almost-best-moves game, but not a strategy. you also forgot the 456 combination
hint: it is essential to have the numbers 1-9 as options. if we had the numbers, say, 5-13, the goal sum would have to be 27..

8. yes, I know I forgot something, but what, it's just now that I realise it

But, hey, if I changed nothing, then it means using what I did above gives me the win

Ok, let me try again, hmm...

186, 195, 276, 285, 294, 384, 375, 456.

5 is the most common, so, first pick for me.
2 and 4 are the next most common, so, there goes he. Say, 2.
4 is left, so this is what I pick.
He picks 6 to prevent 456.
I pick 7 to prevent 276.
He picks 3 to prevent 375.
I pick 1, blocking 168 and taking an attempt at 195.
But, he picks 9.
I'm left with 8.

Draw again

I might need to change things... there may be some way to trap which I don't see yet . Looking more into it.

9. If you get a draw, it isn't necessarily wrong. but i'm expecting a strategy, or some kind of uh...description of the game in a simpler way; I wanna see that you understood I'll also be happy if others try thinking this one out too but whatever your findings are, share them

10. Ah, this problem is simpler than it looks at first sight, if I'm correct. The first person starts with the number five. There are exactly ten numbers in the set. All you need to do next is choose two numbers that average to five. So say your opponent picks a 1. That eliminates a nine for you. You choose 2. Your opponent is forced to choose 8. You choose 6 ( To block your opponent ). He is forced to choose 4. You choose 3 (To block). Your opponent chooses 7. You are forced to choose 9, the only remaining number, and WIN.

11. Never mind, that last post has some seriously wrong logical, and is wrong outright. Ignore it sorry. I thought I had something but I didn't.

12. Originally Posted by shos
If you get a draw, it isn't necessarily wrong. but i'm expecting a strategy, or some kind of uh...description of the game in a simpler way; I wanna see that you understood I'll also be happy if others try thinking this one out too but whatever your findings are, share them
this is going to be drabble, but I'm thinking on your point on exactly 3 numbers adding to 15.
(1,2,3,4) 5 (6,7,8,9)

You need at least one number on each "side" of 5 to add to exactly 15. ((3+4+5 = 12, 5+6+7 = 18)). so if opponent chooses a number from one side, you start by choosing a number from the other side to block him. (thinking... blocking sums of 9, 10, 11???)

13. Well, the only thing that I found using your last 'hint' was that:

1+2+3+4+5+6+7+8+9 = 45
Then, divide by 3 to get 15.

5+6+7+8+9+10+11+12+13 = 81
Then divide by 3 to get 27.

So, it's like some kind of pattern... except that I'm not seeing what hides behind this...

14. that is correct. I can give you another hint if you'd like, but this one should be quite helpful, so I'll put it in spoiler tags, in case others don't want to see it:
Spoiler:
what relation is there between that 3 and the number of numbers you have?

15. This game is isomorphic to noughts and crosses (or tic-tac-toe as it's called in America). The connection between the two games involves a magic square, and is explained here (scroll down to "Variations" and look at the seventh variation).

Page 1 of 2 12 Last