coverOn Friday I posted this puzzle….

100 pennies are on a table.  10 are tails.  With your eyes closed, how can you separate the pennies into 2 groups such that each group has the same number of tails?
If you have not tried to solve it, have a go now.  For everyone else the answer is after the break.
Turn over any 10 pennies. These pennies form group 1.   The rest of the pennies form group 2.  Did you solve it?
I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle(UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.

55 comments

  1. It’s easy to revise the solution so that both groups have an equal number of tails _and_ an equal number of heads…

  2. I know I’m missing something dead obvious, but I’d be grateful if someone could explain this to me. I googled it on Friday and was given the same answer… but I can’t see how it’s the same number of tails in each group if (say) you turn over ten pennies and they are all heads. Doesn’t that mean that group 1 has 0 tails? And group 2 has ten tails (in this example). I can’t see how that can be the same number.

    Sorry if I’m being stupid. Would like to understand this (especially given that the answer I read on Friday was on a website for children!).

    1. Separate out the coins into one group of 10 and one group of 90. As it happens, there are [for example] 3 tails in the group of 10, and 7 tails in the group of 90 (but you DO NOT KNOW THIS). Turn over the group of 10. They then become 3 heads and 7 tails, instead of 3 tails and 7 heads. After you have turned over the group of 10, the number of tails in the group of 10 will match the number of tails in the group of 90 – REGARDLESS of whether it’s 3, or 7, or 8, or whatever.

    2. If you pick 10 coins, and none of them are tails, then there are 10 tails in the other group. Turn them over, and all 10 coins are tails, just like the other group.

      If you pick 10 coins, and 4 of them are tails, then there are 6 tails in the other group. Turn the coins over, and there will be 4 heads and 6 tails. Equal amount of tails in both group.

      Group A has T tails: 10 – T = H (Heads)
      Group B has: 10 – T tails
      Turn group A over, then T becomes H, and H becomes T.

      It always works.

    3. Anne,

      My attempt to explain. 🙂

      In the most extreme example, the 10 coins that you move to one side were by chance the 10 tails. When you turn these over you have zero tails in each group.
      Or, the 10 coins you move to one side initially have zero tails for example. When you turn these all over you now have 10 tails in each group.
      Any number in between 0 and 10 gives the same result.

      Tim

    4. Thank you, geodetective and Tim. Very clear explanations. Once John had given me the idea of separating ten pennies BEFORE you turned them over it was as clear as day. Up until then I was struggling, I must admit. Good puzzle.

    5. If you know some algebra then it could be understood this way:

      The pile of ten has x tails and y heads and the pile of 90 has (10-x) tails and (90-y) heads.

      Now if we flip the pile of 10 we get (10-x) tails and (10-y) heads

      I.e. (10-x) tails in each pile.

  3. I guessed within about 5 seconds that the trick would be to turn over some of the coins, but it then took me another 3 minutes to work out the details of exactly how many. But I separated out the two groups *before* turning over the groups of ten.

  4. Err, I was more laterally thinking in that ALL coins have 1 head and 1 tail, so just divide into 50:50. The above solution is much cleverer, but I had thought that both groups had to have the same number of coins as well as tails. On closer inspection it doesn’t say that. READ THE QUESTION!!!

  5. this is solution #1.

    solution #2: pick 50 pennies. since every single penny has both a head and a tail, each group of 50 “has the same number of tails.” the wording of the puzzle did not specify the number of tails facing up.

    solution #3: US pennies have a face on the obverse, and a building with columns on the “tails” side. if you touch the tails side, you can clearly feel the parallel lines of the columns and thus can easily tell whether a US penny is heads or tails. it is then a simple matter of separating the pennies by touch.

    1. ah pillars allowing you to feel the difference. The obverse(?) of UK pennies have various bits of heraldic devices which are confusing to the finger tip, I shall start a campaign to make easier to detect H/T, Queenie’s head on one side and her royal backside on the other.

    2. I tried it with UK pennies – you can feel the difference. so about 2 seconds to think of it, 1 minute to find a 1p and test.

  6. It doesn’t have to be 10:100, either. 3:10 works, for example, and smaller numbers make it easier to understand, I think. I assume that any number combination would work, in fact …

    1. It works so long as the number of pennies you separate out and flip is the same as the number of tails you begin with. Also, the number of tails must be an even number. Not sure if that is what you were getting at or not.

  7. I scooped all the pennies into one large group in my pocket, leaving two groups of coins on the table. Both sets contained no coins.

  8. There is something deeply unsatisfying about this solution. As I previously said I thought that the problem was not sufficiently defined to obtain a solution. I would qualify this statement by stating that any problem that relies on people assuming rules about the problem is poorly defined. I would argue that the process of turning the pennies over changes the state of those pennies and therefore the subsequent set of 100 pennies is a different set and not just a reordered set. Maybe this is due to my formal mathematical training at university or my autistic brain not liking the ‘looseness’ of the problem.

    1. The problem states, ” how can you separate the pennies into 2 groups such that each group has the same number of tails?” It does not state that the tails must be the ones that we started with. No rules are violated.

    2. Yes I did in fact read that. It is what is not stated that I have an issue with. It is an ill-defined problem. No rules are violated because the problem was ill-defined i.e. rules that should have been violated were not stated. It’s fine if you are happy with an ill-defined problem (game) like this but that’s not how real world problems are solved using science and for me that unsatisfying. It seems like I’m alone in thinking this ?

  9. the penny has just dropped, so to speak! I was all ready to write a moaning comment, but its like one of those magic eye images! even with the solutions posted I struggled with this one !

  10. Richard’s answer was not detailed enough to fully explain the reasoning, so even though I had started working in the right direction, the correct solution still evaded me. Thanks to those who laid it out clearly.

  11. First i had a feeling that it will be a very clever trick. But then i was sure thats not possible. I was fixed the two groups are inhalf, whats mean they have the same number of coins.

  12. Pick up each coin then place them together standing on their side to form two long cylinders of 50 coins each. Both groups will then have the same number of tails (i.e. zero)

  13. Call me old fashioned, but the puzzle didn’t say that the tails had to facing up or down. Forget putting the coins on end. Each coin has one head and one tail. The problem is “can you separate the pennies into 2 groups such that each group has the same number of tails?” Well, YES. Simply divide the pennies into any two groups, and they will have equal numbers of heads and tails. With all those people who had lightening solutions to this problem, I thought this MUST be the answer.

    You know how Richard’s problems so often tilt on an open interpretation of the question.

    1. I like your ‘solution’ as I didn’t think of it that way. We all think we know what the definition of a head and a tail is but that can be different for some people. Agreed that a coin has both a head and a tail but I think most were assuming (rightly or wrongly) that in this case a head meant a coin on a horizontal surface with the head of the coin facing up and it’s tail facing down with a tail defined in the same fashion.

    2. I thought of your answer too (after I got RW’s answer) but how does it incorporate the (admittedly badly worded) sentence “10 are tails” into its logic?
      Also , if you “Simply divide the pennies into any two groups….” they will not have the same number of tails (in your logic) unless they are of equal size. Would two groups one of 23 and one of 77, say, have “the same number of tails”?

  14. I changed the hundred pennies for a pound note, and framed it as they are very rare now. This was easier than finding the answer.

    1. Yeah Edgar, or you could open your eyes a little bit (so that other people thought they were still closed) and cheat.

    2. That’s a bit harsh. As many have been saying. No one said you couldn’t do it that way and no one said that there was only one solution. As the problem was ill-defined there are indeed many solutions one of which being this one.

    3. Steve,
      As I say above in my comment at 2.00am on your (questioned by me) solution of 1.12pm the problem is “admittedly badly worded”. I do not think it is ill defined however.
      It seemed obvious to me what Richard wanted us to do.

    4. If something has to be “obvious” then there are by definition things that are unsaid or unwritten with regard to the problem, That makes it ill-defined or if you like not defined sufficiently. When something is well-defined there is no need for assumption or to say that something is “obvious”. What is “obvious” to one person is not obvious to another. Some people come up with the similar ‘argument’ saying it’s ‘common sense’. So called common sense is far from common and what they are really saying is “if I can see it then everyone else should”. There’s something patronizing or arrogant in the use of “obvious” or “common sense”.

    5. Thanks Steve,
      Would it be “patronising” or “arrogant” (or indeed both) to point out that you have not yet responded to my post of 2.00 am on 12/6/13 querying your 1.12 pm post of 11/6/13?

    6. Hi Dress Rehearsal. Sorry I didn’t see your response before now. You said:

      “I thought of your answer too (after I got RW’s answer) but how does it incorporate the (admittedly badly worded) sentence “10 are tails” into its logic?
      Also , if you “Simply divide the pennies into any two groups….” they will not have the same number of tails (in your logic) unless they are of equal size. Would two groups one of 23 and one of 77, say, have “the same number of tails”?”

      I never gave an answer ! You are quoting Richard not me. I was just trying to confirm Richard’s logic as I think I understood it. That is a coin has a head and a tail. No matter what it’s orientation it still has a head and a tail. So if we have 100 coins on a table, there are 100 heads AND 100 tails. That is an equal number of heads and tails. I may have misunderstood Richard’s post but that was what I understood he was saying. Now I later went on to define what I think most people were assuming to be the definition of a head and a tail with respect to a coin on a horizontal surface with a head or a tail facing up. I apologize if I haven’t made myself clear or indeed misunderstood anyone’s comments.

  15. Hi CJS. Could you explain what you mean by non-sequitur with regard to my or others comments ? I would be sincerely interested to know where my argument is in error. I’m not interested in being right, just knowing what is right.

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