There are 100 coins in a jar. Two players take turns removing anywhere from 1-10 coins...

There are 100 coins in a jar. Two players take turns removing anywhere from 1-10 coins from the jar. The player who empties the jar by removing the remaining coin(s) wins the game. To guarantee that you win the game, would you choose to move first or second, and what strategy would you follow?

Expert Solution

To win, we must first play this game and can arrange the game

Let's think backwards of this game

We wish to ensure that we make an adversary move at points if jar has the next number of coins

89,78,67,56,45,34,23,12,1

That means that when we make the first move, we should remove 1 coin in the second step and then in stage 2 adversary can pick any between 1 and 10 so that we can assume he takes 10 coins and then the whole coin is taken away 11 Coins in the third step, then we must ensure that 12 Coins are removed by removing 1 Coin again.

Again in the fourth step, we will make sure that the total coins are removed by 23 at the end of our turn by the opponent (1 to 10).

We are repeating the strategy to remove coins at round 2 in such a way that we remove total coins equal to 89 and then we are the last person to empty the jar in the end, in the case of every single nunner of coins removed by an opponent.

For instance,

If the opponent takes 1 coin in n-2 step, we have removed 90 total coins and 10 coins to empty the jar.

If opponent removes 2 purses in step n-2 we have removed 91 of the total coins and 09 purses for emptying the jar.

If the opponent deletes three coins in n-2 step, then the total number of coins has been removed, 92 and 08 coins are removed.

If the adversary takes 10 coins in n-2 step, then total coins have taken 90 of us away and 1 coins can be taken to empty the jar. Simialrly.

Rahul Sunny answered 2 years ago

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