How do you win the simple math game 20-19?

Question

My friend (I think)made up this game called 20-19. The objective is to start at 20 and be the person to say one. You take turns, being allowed to say either 1 or 2 numbers. Ex: P1: 20, 19; P2: 18, and so on. Based on how fast the people who figured out the "trick" at school, they won every time without counting it down to see who'd win. This is probably a really easy question I for some reason don't understand or am overthinking, but how do you win this game? What's the trick? Is it that you follow a pattern in your 1's and 2's, or only land on odds? I'm stumped.

Thanks!

Answer 1

This is very similar to NIM but much easier.

The trick is if your opponent says x, you say 3 - x. Thus each time your opponent and you take full turns the result goes down by 3.

So if you go first say 2 and the result goes from 20 => 18. Then whatever x, the other person says, you say 3 - x and the result goes from 20 => 18 => {17,16} => 15. Keep going. The end result will be that in your second to last turn the result goes to 3 and you win no matter what.

Our more completely:

P1: 20 => 18; P2: => {17, 16}; P1 => 15; P2 => {14,13}; P1 => 12; P2 => {11,10}; P1 => 9; P2 => {8, 7}; P1 => 6; P2 => {5,4}; P1 => 3; P2 => {2,1}; P1 => 0. Player 1 will always will if she follows this pattern. If she ever screws up Player 2 can take advantage and win.

Whoever makes a multiple of 3 first, can keep ending up with multiples of 3 and will win. As you start at 20, Player 1 can always take 2 and make a multiple of 3. If you started with 21 then Player 2 can always win as player 1 can't make a multiple of 3 but whatever she does Player 2 can.

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Oops! I see it is the player who say's "1" who wins; not the player who says "0".

Oh, well. It's still the same idea. Instead of trying to land on 3k (so you can say 0) you try to land on 3k+1 (so you can say 1).

So Player 1 say "19" then player 2 says "..." and player 1 say "16" (".." "13" "..." "10" ... "7" ... "4" ... "1! I WIN!").

Answer 2

If I say "1" I win. If I say "4" and stop then you can stop at "3" or "2", and in either case I say "1" and win. If I say "7" and stop then you can stop at "6" or "5", and in either case I can say "4" and stop and win. Same if I say 10, 13, 16 or 19.

If I start then I say 20, 19. Next round I stop at 16, then at 13, then at 10, then at 7, then at 4, then at 1 and I win.

Now with three players it is quite interesting.

If you don't have any clever mathematical idea, then with this kind of game you can start at the end as I did: If you say "1" you are the winner. Then you figure out who wins if you stop at "2". Then you figure out who wins if you stop at "3" and so on. If you are lucky you find a pattern. Sometimes there are difficult patterns. Sometimes there are no known patterns. In this case, the pattern is quite simple, but even without finding the pattern you could write down the winning strategy.

Answer 3

Please explain better the rules of the game. The numbers that are said by the players must always be consecutive, each the predecessor of the previous one to be uttered?

If so, consider the following analysis:

If in your turn, it's 1 or 2, you win, thus

if it's 3, you lose; if it's 4 you must say 4;

if it's 5 you must say 5,4;

if it's 6 no matter what you say your opponent is gonna be at 5 or 4, so he has a winning strategy; so you lose no matter what; thus

if it's 7, and you play 7, then you win; if it's 8, and you play 8,7 then you win; therefore, if it's 9 your opponent is gonna be at 8 or 7 no matter, and he's gonna have a winning strategy, so you lose;

You may begin to see a pattern here: whenever in your turn the number is a multiple of 3 you lose,if it's not you have a winning strategy by uttering one or two numbers according to whether your number leaves remained 1 or 2 upon division by 3, thus forcing your opponent into a losing position.

Thus, in particular, if you begin the game you have to say 20-19, and your opponent is then sure to lose provided you play the game correctly.