Shopping & Style0 min ago
Nice easy one for ya ......
22 Answers
80 cards are laid out on a table in a single file.
All of the cards are blank on one side and have a number on the other.
All of the cards are face up - i.e. the value of the number is shown.
You now play a game against one other person. You each take it in turns to remove 1 card at a time.
You can only remove a card from either end of the line of cards - you cannot remove a card from somewhere in the middle. So, for example, if you remove card no 1, your opponent can only remove card 2 or 80. Or, if you remove card no 80, your opponent can only remove card 1 or 79.
At the end of the game, you'll both have 40 cards (obviously!!) - the winner is the one who's sum of all the values on the cards is the greatest.
You get to choose first, so what strategy should you employ to ensure that you never lose ???
All of the cards are blank on one side and have a number on the other.
All of the cards are face up - i.e. the value of the number is shown.
You now play a game against one other person. You each take it in turns to remove 1 card at a time.
You can only remove a card from either end of the line of cards - you cannot remove a card from somewhere in the middle. So, for example, if you remove card no 1, your opponent can only remove card 2 or 80. Or, if you remove card no 80, your opponent can only remove card 1 or 79.
At the end of the game, you'll both have 40 cards (obviously!!) - the winner is the one who's sum of all the values on the cards is the greatest.
You get to choose first, so what strategy should you employ to ensure that you never lose ???
Answers
Best Answer
No best answer has yet been selected by Gizmonster. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.
Very close Factor30 - in fact it's so close I might as well tell ya.
First add up all the odd numbered cards and all the even numbered cards. Whichever is the highest - make sure you always choose them. So, say for example, the odds were the greatest, you would start by removing the card at postition 1. Your opponent will always be left with a choice of 2 even cards. Whichever one they take, they will leave an odd card for you. You remove that odd card on your turn and again they are only left wth a choice of 2 even cards - this way you will end up with all the odd cards and your opponent will end up with all the even cards ....... tada :)
First add up all the odd numbered cards and all the even numbered cards. Whichever is the highest - make sure you always choose them. So, say for example, the odds were the greatest, you would start by removing the card at postition 1. Your opponent will always be left with a choice of 2 even cards. Whichever one they take, they will leave an odd card for you. You remove that odd card on your turn and again they are only left wth a choice of 2 even cards - this way you will end up with all the odd cards and your opponent will end up with all the even cards ....... tada :)
I think he means find the total of the cards in the odd positions, 1 to 79. Then find the total of the cards in the even positions, 2 to 80. Whichever is the greater, take a card from that position, i.e. at the start of the game if the cards in the odd positions have the greater total you will choose the first card whereas if the cards in the even positions total more you will choose card 80 and so on. This would take some time, mentally totalling two sets of 40 cards.
