# Positioning cards labeled with numbers from 0-9

Ann and Bob play a game. On a table there are 10 cards which are labeled with number from 0 to 9 each. Bob is allowed to change the position of the cards with a sequence of his preference. When he is done, Ann takes the left card or the right card of this sequence. From the remaining cards, Bob takes the left card or the right card of the sequence. Continuing that way, Bob will finally take the last card.

Now Ann and Bob sum up the value of their cards. The person with the higher sum wins. Who can force to win the game? Whatever the order is, consider the card positions [1...10]. Ann should add up the values of the cards in odd positions, then add up the values of the cards in even positions. Since the sum of all 10 cards is odd, one set will always have a higher value.

Then,

If odds have a larger sum, Ann takes the leftmost card. This forces Bob to take an even-position card (card 2 or card 10). Whatever he chooses, Ann can take another card that was originally in an odd position, forcing Bob to choose an even position again. She can just keep doing this to force him to take all the evens, which we already know has a smaller sum than her odds. Similarly, if evens have a larger sum, Ann can just take the rightmost card and do the same thing.

If she wants, Ann can do even better by changing to evens or odds on each turn - she just recomputes the remaining sums and uses the same strategy. However, this will just be running up the score.

However

Bob puts the card in an order, Ann can always win the game by +1.

Why?

let's think the game with 1,3,4,5 where the sum is not even.

so whatever Bob can arrange these cards in the best case scenario,

Ann will win by +1, $$4,1,5,3$$, because Ann needs any two cards without 1 only and since Ann starts the game she can arrange it whatever the original arrangement. Ann will take $$4 + 5$$ or $$3+4$$ since Ann has advantage choosing at the end between two numbers.

Let's make it more complex;

the game with 0,1,2,3,4,5, and the sum is not even again, so only one draw is possible so, Ann will win the game if she takes 8 or more points with 3 cards. and after choosing her first two cards, she can choose between two cards and take the biggest at worst before ending the game so Ann will have advantage at least +1 at the end. Moreover, Ann needs to arrange her strategy by summing every number by passing the number next to it in the list, such as let's say order is $$1,4,2,5,3,0$$, so let's take sum from the beginning by passing one number at a time, so the sum becomes $$1+2+3$$, so we should not start from there, start from the other end to guarantee to win the game.

so even it is 0 to 100, Ann will

win the game if the sum is odd with the methodology above.