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# Number picking I. Even or odd?

## Question

You play a betting game with a friend.

You each place \$1 on the table. A computer then prints a random string of digits, from 0 to 9.

Players take turns erasing a digit from either end of the string and adding that number to their score.

When the last digit has been erased, the player with the higher score wins the \$2.

Players take turns moving first and choosing whether an odd or even number of digits are printed.

You're moving first on the next game. Which do you choose: even or odd?

## 5 answers to this question

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taking the sum of the even numbers in even positions and odd numbers in odd positons comparing them and taking the larger of the two, each turn.

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its even. then you can employ the following.

if the sum of all numbers in the odd postions > the sum of all numbers in the even postions,

then select from the left for the first move. else select form the right.

then simply select the oppisite of whatever sde your opponent picks from.

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opps...

right idea wrong finish.

you need to implement the sum each turn.

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opps...

right idea wrong finish.

you need to implement the sum each turn.

Are you sure this is best?

I'm assuming that "implement the sum" means "take (erase) the larger" of the two available numbers.

Might that not expose a much larger number for your opponent's next turn?

But maybe I misunderstood.

The reasoning in your first post is on the right track.

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If you play first and you choose to have an even number of numbers, you can erase all the even numbers or all the odd numbers, whichever is larger, thus guaranteeing you will not lose. Worst case is if the even and odd sums are equal. Your approach seems better, I think, possibly guaranteeing a win in every case.

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