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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?

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  • $\begingroup$ 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. $\endgroup$ – Matti P. Dec 28 '18 at 10:58
  • $\begingroup$ 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. $\endgroup$ – saisanjeev Dec 28 '18 at 11:04
  • $\begingroup$ 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. $\endgroup$ – badjohn Dec 28 '18 at 12:25
  • $\begingroup$ 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)" $\endgroup$ – Mark S. Dec 30 '18 at 18:50
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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).

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

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  • $\begingroup$ 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? $\endgroup$ – saisanjeev Dec 28 '18 at 11:13
  • $\begingroup$ 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. $\endgroup$ – Akiva Weinberger Dec 31 '18 at 13:24

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