Same number of heads

[bonanova]

You are seated at a table on which lie ten coins, each capable of showing a head or tail. You are wearing gloves and the room is dark. Your task is to create two non-empty groups that show the same number h of heads. h may have any value from 0 to 5. You may use any or all of the coins, exactly three of which currently show heads. Proceed. The clock is ticking.

[bubbled]

A general solution:

 

Let S = the starting number of heads face up, where S = any number from 1 to 9, inclusive. Move S coins into a separate pile and flip them all. Let H = the actual number of the coins in the side pile you created that happened to be heads. After the flip, you will have two piles with exactly h = S - H of the coins showing heads. No matter what S is, h could be as little as 0. h can never be more than 5, and is limited by 10 - S when S > 5.

 

 

[bubbled]

A general solution:

 

Hidden Content

 

 

Clarification:

 

To be clear, H is the number of heads in the your side pile BEFORE you flip those coins.