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A difficult and tricky problem

  1. Jul 31, 2011 #1
    1. The problem statement, all variables and given/known data
    You've 15 coins out of which 5 coins are turned with their HEAD facing upwards and the rest 10 have their TAILS facing upwards. Suppose you have been blindfolded. Now you've got to divide these 15 coins in two groups such that each group has the same number of coins have their HEADs facing up. But you cannot overturn any coin. How can you do it?? It may seem practically impossible but it can be really done!
    2. The attempt at a solution
    I thought it might be a problem of probability. I found the probability of HEAD which is 1/15 i.e. 1/3 which means one out of three. But actually I think it is impossible as to the fact that the number of HEADs is odd and we can't divide it into two equal groups! But in the question it says that it can be done! I suppose there must be some trick in it. Please help me.
    Last edited: Jul 31, 2011
  2. jcsd
  3. Jul 31, 2011 #2


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    Homework Helper

    Sorry to say but one difficulty is that if there is a trick in the question, it could easily be hidden in the less than perfect English used to supply the information here.

    Perhaps you divide the coins into two groups, of any size you like, but arrange the coins on edge, to form a couple of "cylinders". That way you have not overturned and coin, just stood all of them on edge??
    Last edited: Jul 31, 2011
  4. Jul 31, 2011 #3


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    Staff: Mentor

    I can't see why it is specified that you be blindfolded.

    15 coins? So maybe you cover over one of the heads with a tails-up coin?
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