A game of two

Logic Problem #3

Problem

Let’s imagine that you and I are to play a game. We shall take it in turns to call out integers. The first person to call out '50' wins. The rules are as follows:
– The player who starts must call out an integer between 1 and 10, inclusive;
– A new number called out must exceed the most recent number called by at least one and by no more than 10.

Do you want to go first, and if so, what is your strategy?

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Clue

You will win if you shout out 50. So you need to think under what circumstances within the rules of the game will allow you to shout out 50, as in what numbers being called out by the opponent will permit this.

How can you force his opponent into that position, and work backwards to the beginning to find out how the game must start for you to win?

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Solution

Let’s backtrack our way up to the solution. Consider the players as W and L, with W representing the winner and L representing the Loser. Now,

We'll call the players 'W' and 'L' (can you guess which one wins?) The trick is to start at the end and work backwards.

W wins if he says 50, he can say this if -
L says any number in the range 40 - 49 and will have to say one of these if -
W says 39, he can say this if -
L says any number in the range 29 - 38 and will have to say one of these if -
W says 28, he can say this if -
L says any number in the range 18 - 27 and will have to say one of these if -
W says 17, he can say this if -
L says any number in the range 7 - 16 and will have to say one of these if -
W starts on 6.

So that is pretty much it!

W wins if he starts on 6 then simply follows the sequence 6, 17, 28, 39, 50. W forces each pair of numbers to go up in 11's regardless of what L says, hence he starts on 6 which is 50 - 4x11.

Thus, you can win the game if you start by saying 6!

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