## Thursday, June 12, 2014

### Problem

Variant 1-----------
There is a table on which a number of coins are placed. You also know that there are as many coins with Head up as many coins with Tail up. Now you have to divide the coins (number of coins is even) into two equal piles such that number of coins with Heads up and Tails up in either piles be the same. The catch is you are blind folded and you cannot determine the sides (for sure) if you are blinded

Variant 2-----------
Your friend pulls out a perfectly circular table and a sack of quarters, and proposes a game.
"We'll take turns putting a quarter on the table," he says. "Each quarter must lay flat on the table, and cannot sit on top of any other quarters. The last person to successfully put a quarter on the table wins."
He gives you the choice to go first or second. What should you do, and what should your strategy be to win?

### Solution

Variant 2-
You should go first, and put a quarter at the exact center of the table.

Then, each time your opponent places a quarter down, you should place your next quarter in the symmetric position on the opposite side of the table.

This will ensure that you always have a place to set down our quarter, and eventually your oppponent will run out of space.

References

1. Variant 1 -

Suppose we have 10 coins with Heads faced up and 10 more with Tails.
Now we randomly divide the total(20) coins into two pipes of 10 coins each.
pipe1 pipe2
1 0
1 0
1 0
1 0
0 1
0 1
0 1
0 1
0 1
0 1

since there are 10 heads(1) and 10 tails(0) , if there are 4 Heads in pipe1, then remaining 6 should be in pipe2.

Now to get the Solution, flip all the coins in either of the pipes.
Result will be the identical number of Heads and Tails faced coins in both pipes.

1. Thanks for the solution Amith. :)