## Wednesday, June 16, 2010

This is borrowed from folj.com and I don't know the solution.  Submit ideas in the comments section - maybe you'll get it before I do.

There are twenty coins sitting on the table, ten are currently heads and ten are currently tails. You are sitting at the table with a blindfold and gloves on. You are able to feel where the coins are, but are unable to see or feel if they are heads or tails. You must create two sets of coins. Each set must have the same number of heads and tails as the other group. You can only move or flip the coins, you are unable to determine their current state. How do you create two even groups of coins with the same number of heads and tails in each group?  [Note: I assume you need a guaranteed method of getting even groups, rather than method that will likely work.]

Solution:

Rob got the solution faster than I could, and here it is:
Split into two groups of 10. Group A will have X heads, and 10-X tails. Group B will have X tails, and 10-X heads. Flip every coin in group B. Both groups now have X heads and 10-X tails.

1. Split into two groups of 10. Group A will have X heads, and 10-X tails. Group B will have X tails, and 10-X heads. Flip every coin in group B. Both groups now have X heads and 10-X tails.

2. Does that work for the general case of N coins, where N is even? No, it doesn't. Does it work just for groups that are multiples of 10 coins?
Why does it work?

3. Yes, it does work for the general case obviously. I missed that the first time around. The reason it works is simply the even number and the change of state.

4. We don't know value of X here, however we know for sure that they have equal heads and tails on each table

can we stretch it a bit further?
will it be possible to divide it into groups of 10 heads and 10 tails?
If yes how?
If no then can we prove it that it is impossible