# Why does A always win in this game?

I have the following question with me:

"A and B start with p = 1. Then they alternately multiply p by one of the numbers 2 to 9. The winner is the one who first reaches 1000. Who wins : A or B?"

My book tells that A always wins. However, I give steps for B to win in the following way:

Step 1: A multiplies by 2 to give 2.

Step 2: B multiplies by 9 to give 18.

Step 3: A again multiplies by 9 to give 162

Step 4: B again multiplies by 9 to give 1458

Thus crossing 1000 first, B wins the game right. Is this not a possible way in which B can proceed? Is there any problem with my interpretation of my question?

• Maybe it means "A always has a sure winning strategy" or something along those lines. A can choose their second-to-last number so that B cannot cross the line. And then on the last turn, he takes the winning step. Something like that. – Matti P. Dec 28 '18 at 10:58
• So the question is basically to find a strategy for A to win rather than who should win? By the way the question is from the book Problem Solving Strategies by Arthur Engel. – saisanjeev Dec 28 '18 at 11:04
• I guess that it means that A can always win (if he wants to and does not make a mistake). What you have shown is that he could lose if he wanted to or played badly. – badjohn Dec 28 '18 at 12:25
• If the book literally says "A always wins", then it's a confusing shorthand for "A has a way to ensure they always win (but they are only guaranteed to win if they follow that winning strategy)" – Mark S. Dec 30 '18 at 18:50

## 2 Answers

The OP is correct that A has a winning strategy and that there are hypothetical lines of play in which B wins.

"Who wins : A or B?" is a common, if confusing, shorthand for "Who wins in a situation where the players play perfectly?", or "Who has a winning strategy?" (for these types of games those are equivalent).

• By "these types of games" you mean games with no hidden information between the players and nothing left to chance, right? (Like, not poker.) – Akiva Weinberger Dec 31 '18 at 13:26
• @AkivaWeinberger That, and also specifically two-players. – Mark S. Dec 31 '18 at 13:27

Winning strategy for $$A$$:

Start by multiplying by $$6$$.

$$B$$ must then return one of $$\{12,18,24,30,36,42,48,54\}$$

No matter which of those $$B$$ returns, $$A$$ can win. To see this note that all $$A$$ has to do is to hand $$B$$ a number $$N$$ with $$56≤N≤111$$. If $$B$$ is handed such an $$N$$, all the possible responses lie between $$112$$ and $$999$$ and $$A$$ can just multiply by $$9$$ for the win.

It is easy for here. If $$B$$ returns $$12$$, say, then $$A$$ returns $$60$$ and wins. If $$B$$ returns $$54$$ then $$A$$ returns $$108$$ for the win, and so on. To be specific, if $$B$$ returns $$\{12,18,24,30,36,42,48,54\}$$ then $$A$$ returns $$\{60,72,72,60,72,84,96,108\}$$ respectively.

• Yeah I was able to give a winning strategy for A. However, I was worried that I was able to give a sequence of moves as mentioned in the question by which B was winning. Is there any problem with my interpretation of the question? – saisanjeev Dec 28 '18 at 11:13
• Sure, if A plays poorly, they can lose. The point is, if A knows what they're doing, they can win, no matter what B does. – Akiva Weinberger Dec 31 '18 at 13:24