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Maths Puzzle: Coins

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Uploaded by on Jan 31, 2011

Take a random pile of coins. If I tell you how many coins are heads and how many coins are tails, can you separate the pile into two piles with the same number of heads in each. And can you do this with your eyes shut?

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Education

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Uploader Comments (singingbanana)

  • I'm glad you put this one online! It tekes me back to Cambridge when I found the solution with those pieces. I bet there are still my fingerprints on them...

    Did you know Martin Gardner had a card magic trick using the same principle?

  • @TyYann I forgot we did this in the restaurant. Yay for our YouTube get together. I disinfected the pieces after you touched them.

  • hmmm 11 coins u throw them and tell me there are 5 white ones i will flip over 3 of them and line then up in 2 lines... then i seperate them and i think it should be a high chance to get it right (3 blacks switched 8 white hmmm maybe?)

  • @Namezzzzzzz This time we're not looking for a high chance but 100%.

  • This is a really clever one that I'm familiar with, but yes you can. However, I don't know why it works everytime, and I'm assuming there's some math behind it...

  • @EighteenCharacters That's your challenge then. It can be proved mathematically.

Top Comments

  • Yes, if you have X white coins, you take any X coins aside and flip them.

    Problem solved.

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All Comments (108)

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  • @singingbanana but we live in only this universe. so there is no way we can superimpose any other universes even if we had them.

  • 2 people thought this video was racist

  • I got it....split into 2 piles.... 5 and 6...flip all d coins frm d one dat has 5.... But I dnt understand y....!!! Guess I'll c ur video

  • Get Chuck Norris to do it.

  • uhm did i miss something?

    *puts blindfold on*

    switches one white piece to the black pile, turns over one black stone.

    *looks unsure*

    again, did i miss something?

  • imposible it a odd number 11

  • Thanks for the great puzzle singingbanana! It took me a couple of days to figure this one out. Here's my solution

    Let there be n W(hite) pieces and m B(lack) pieces.

    You start off with a mix of nW and mB on the table.

    Since you know n, you remove n number of random pieces to one pile, leaving the remainder in the other pile. Let a+b=n.

    aW + bB : (n-a)W + (m-b)B

    Now you flip the pile with n pieces

    bW + aB : (n-a)W + (m-b)B

    Since n=a+b, b=n-a, so both sides have an equal number of W!

  • @nelsyeung Here's another run: There are a total of 52 coins, HHTTHTHHHTHHHHTTHTHHTHTTHTHHHH­TTHTHTHHHTTHHTTTHTHTHH 30 of which are heads, and 22 of which are tails. We will take a random selection of 30 coins: TTHHHTHHHTTTHTHTHHHTHHHTHHTHTH <-- My Pile HHTHHHHTTHHTTHHTTHTTTH <-- Remaining Pile Now we flip everything we took out HHTTTHTTTHHHTHTHTTTHTTTHTTHTHT The pile we chose now has 12 heads The other pile has 12 heads. SUCCESS!
  • @nelsyeung Ok, you don't believe me...so I wrote a comp program to simulate the process. Here ya go: There are a total of 11 coins, HHHTHTHHTTH 7 of which are heads, and 4 of which are tails. We will take a random selection of 7 coins: HHTHHTT <-- My Pile HHTH <-- Remaining Pile Now we flip everything we took out TTHTTHH The pile we chose now has 3 heads The other pile has 3 heads. SUCCESS!
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