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# Number picking I. Even or odd?

### #1

Posted 31 May 2014 - 06:04 PM

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?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #2

Posted 31 May 2014 - 09:54 PM

### #3

Posted 31 May 2014 - 10:31 PM

opps...

### #4

Posted 01 June 2014 - 05:58 AM

opps...

Spoiler for correction

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.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #5

Posted 01 June 2014 - 06:01 PM Best Answer

### #6

Posted 02 June 2014 - 02:51 AM

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.

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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