Steem Riddle of the Week #2: Flipping Cards [Prize Doubled - 2SBD]
Riddle of the Week #2
No one was able to solve last weeks riddle, despite some good discussion. So, I'm going to repost the same problem with a slightly different statement to make it easier / more clear and double the prize. The first person to upvote and comment with a valid solution (as decided by me and only me) prior to 12-11-2017 midnight UTC will get 2 SBD!
The problem:
There is a deck of 52 cards. Each card is black on one side and white on the other side. All of the cards start with the black side up. You are blindfolded. Your opponent randomly picks 10 of the cards and turns them white side up and hands you the deck.
You must split the deck into two piles such that each pile has the same number of cards with the white side facing up. Your strategy must work 100% of the time and cannot rely on luck.
You cannot see anything, nor can you feel the difference between the front and backs of the cards. How can this be done?
PS. If no one can solve it this week, I'll post the answer, enjoy! Be sure to follow if you want to be notified when the answer is posted :)
PPS. I am newish here so a follow and/or resteem are greatly appreciated!
Count out 21 cards and lay them down into 2 piles (21 cards in each pile), as they are already on the white side facing up.
I don't think that works, there are 52 cards total, your strategy needs to split the deck into 2 piles that together have all 52 cards. Of those 52 cards, at the start, 42 are black side up and 10 are white side up. Your strategy would leave 10 cards left over.
Furthermore, even if you just deal out 2 groups of 21 cards, since you don't know which cards are white side up, you might get more white side up cards in one of those two piles than in the other.
It can be helpful to imagine the state of the deck when it is handed to you as a list like
w
b
w
b
b
b
w
b
b
b
.
.
.
.
b
Where you know there are 42 b's and 10 w's but you don't know the order.
This post has received a 17.41 % upvote from @booster thanks to: @droopy.