2 friends decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. They continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.
2 friends decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. They continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.
2 friends decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. They continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.
Always collect coins with at least as much value on every turn, or cumulatively?
Either way the answer seems to be too simple to me, they go first so they set the strategy before their first turn
Always collect coins with at least as much value on every turn, or cumulatively?
Either way the answer seems to be too simple to me, they go first so they set the strategy before their first turn
Obviously cumulatively. But first player can't tell which coins 2nd player will choose. So, how is it too simple?
Not that it's very difficult, but lemme hear the strategy....
2 friends decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. They continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.
Hidden:
Since 50 is an even number, it can’t be a pattern of more, less, more less etc. so he’ll always have an option of taking the coin with more or equal value first.
Since 50 is an even number, it can’t be a pattern of more, less, more less etc. so he’ll always have an option of taking the coin with more or equal value first.
2 friends decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. They continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.
Hidden:
It’s all about not exposing the higher denominator coins to the opponent? Looking at the highest values of the first 2 coins on each side instead of only the first
It’s all about not exposing the higher denominator coins to the opponent? Looking at the highest values of the first 2 coins on each side instead of only the first
Are you expecting an answer in an equation?
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Might or might not work. You might have to choose between two pennies, exposing a quarter. There's a more definite strategy that will work.
