Nice easy one for ya ......

[Gizmonster]

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 ???

[mike11111]

If you take 80 then the card at the high end each time you will always be at least 2 ahead of your opponent.

[Gizmonster]

Soz I think you misunderstood.
The cards' vlaues do not run from 1 - 80. The values on the cards is random and can change from game to game.
I was simply referring to the card's relative positions to one another when I referred to cards 79 and 80 etc.

[grasscarp]

I never really understand why people put puzzles that they already know the answers for.

[Gizmonster]

.... so other people can enjoy them.
I'm like that - I like to share :)

[Prudie]

If they're random than apart from always choosing the highest value of the 2 seems logical (unless you have time to go away and plan the possibilities out on paper).

[sunny-dave]

Take the one which leaves the lowest total sum on the two ends ...

[Gizmonster]

Prudie that wouldn't be very wise 'cos when you remove the highest one out of the 2 choices, you could leave an even higher for your opponent.
It's dead easy - you'll kick yourself when u find out :)

[Gizmonster]

No takers ??????

[factor-fiction]

I'm afraid I don't know Gizmonster- I've only glanced through it but if the order of the cards is random i can't see any strategy working.

[factor-fiction]

Agh sorry- I've just realised you can see the numbers

[factor-fiction]

always choose the even numbers?

[Jemisa]

My heads in a mucking fuddle!!!

jem

[Gizmonster]

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 :)

[Prudie]

Interesting Giz - I would hope that you have time to do this, with a calculator. Most people couldn't add up 2 sets of 40 numbers in their head too quickly!

[nescio]

gizmonster,
I do not understand your explanation at all.

[nescio]

Oh hang on.
You mean the cards' positions in the line, rather than their values ?
I get it now.

[mike11111]

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.

[factor-fiction]

The problem and solution might be more realistic if there were only 10 cards (numbered 1 to 10)- that way adding up the odds and evens quickly should be easy

[mike11111]

Actually you only need to add up one set of numbers, either odd or even as
Σ80 = (80² + 80)/2 =6480 so depending whether the total is more or less than 3240 you know which to choose.

[factor-fiction]

I agree , mike11111- except that you have forgotten to divide by 2. The sum of the numbers 1 to 80 = 81 x 80/2= 3240. If your set of alternate numbers add up to more than half of that- 1620- then you choose that set; otherwise you switch