Okay, so I’ve been trynna solve this puzzle since yesterday afternoon, and till now I’ve gotten Zero ideas..It just seems impossible, but I know for sure there has to be a way to solve it….!

A man comes back from a business trip with 100 coins to share with his two children. He places the coins on a table with 60 of the coins heads up, and the rest tails up. Then he turns out the light so that it is completely dark. He tells his son that he can do anything he likes with the coins on the table (flip them, rotate them, move them, etc.), but then he must split them into two groups. Then he tells his daughter that she may decide which of the two groups is hers, and which is her brother’s. They will then turn the light back on, and each child may only keep the coins in his or her group that are heads up. (Dad gets to keep all the tails up coins.)

When the light is off, the children cannot see the orientation of the coins, and it is impossible to distinguish the orientation by feel alone. The little boy is determined not to let his sister “win” by ending up with more coins than him, so he wants to split up the coins into groups that will *guarantee* that, no matter which group his sister picks, they will both end up with the same number of coins. What should the little boy do????




21 thoughts on “Puzzle

  1. Roggio says:

    I’m assuming you’re from the logic class. The kid would need to move 40 into a separate group and flip the remaining 60. Like that, he assures that there will be the same amount of face-up coins in each pile.

      • Alexander says:

        *Possibly a Spoiler Alert*

        Group them into 2 stacks and place them standing on the coins’ edges. That way none of them gets any.
        I just tried to google if I was right but get to say my answer instead, haven’t found it.

    • Roggio is right.

      Put 60 coins in one group. Let’s say that it has N coins in it that are heads-up. That means that he second group has (60 – N) coins out of 60 heads-up.

      Flip the coins in that second group (the group of 60)… now it has (60 – (60 -N)) coins that are face-up. In other words, it has N face-up coins, just like the other group.

      It is possible that (N = 0) if the children are unlucky. It also means that the most coins that the children might get is 40 (100 – 60), or 20 each.

    • {I posted a response earlier; don’t accept it– it was slightly mis-worded and wrong. This one is better, and is much less confusing. }

      Roggio is right.

      Split the coins into two groups.
      Group 1: 60 coins, lets say N of them are heads-up
      Group 2: 40 coins, (60 – N) of them are heads-up
      [This is because there started with 60 heads-up coins. N of them are in Group 1, so the remaining (60 -N) are in Group 2].

      Flip over all of the coins in Group 1.
      Group 1: 60 coins, (60 – N) are heads-up
      Group 2: 40 coins, (60 – N) of them are heads up
      [This is because Group 1 started with N heads- up coins and (60 – N) tails-up coins. When we flip all of the coins over, there are now (60 – N) heads-up coins. Group 2 is unchanged.]

      Both groups contain the same number of heads-up coins, (60 – N).

      It is possible that (N = 0) if the children are unlucky and all of the heads-up coins are originally split into the group of 60. It is also possible that the children both receive up to 30 coins, if both groups start with 30 heads-up coins.

  2. Radhakrishna says:

    Make two groups, 50 coins per group. Then flip each coin. Now by probability, each group should have 25 heads and 25 tails.So Dad gets 50, Brother gets 25 and sister gets 25.

      • I was Googeling some thoughts about this problem (instead of a solution). Radhakrishna is inspiring, but is he right? There are 100 coins on the table, mixed with 60 up, it is dark. Radhakrishna says the boy makes two groups of 50 coins, that could be with 50-10 up, 40-20 up, 35-25 up anything. Then flip them you get 10-50 up, 20-40 up, 25-35 up respectively. Sister may still choose and get more. I don’t get it.

  3. Yes I am, at Coursera 😉
    I am still thinking about the puzzle. Theoretically the boy may choose to have two groups of zero coins, but I don’t believe that is right.
    The best thing I could come up with is to turn over the coins at random. Then the chances are better for having a number < 60 I guess, but that would make it a statistics problem. What does the Bernoulli bell curve say about the proportion ratio?

  4. Hey hey, a real-life riddle with incomplete information! “He places the coins on a table with 60 of the coins heads up, and the rest tails up.” I’ve figured it out!
    What you got to know is: “The coins remain in two groups, one group head up, one group of coins head down, but the child (who may manipulate the coins) doesn’t know which group is which.”
    Father Michael let us sweat on that one. Now the riddle is solvable (and feels like having a nice punch, like all good riddlesl). Try again.

  5. Solution 1 (funny):
    Flip one group of coins, put the two groups together and divide them in two groups of 50 coins, flip over one group and you have two groups of 50 one heads up, the other heads down.

    Solution 2 (easy):
    Check how many coins each group has (here group A = 60 and B = 40), move half the difference over to the other side from the larger group, so (A+B)/2 = (60-40)/2 = 10 and A – 10 = B + 10 = 50).

    Hmm, seems there are two solutions, perhaps I am wrong here.
    My riddle for father Michael is:
    How can you change your coin riddle so it is solvable like Solution 1, but not by Solution 2 and it is clear that the coins aren’t mixed so no solution is possible (but A = 0 and B= 0).

  6. Gerard, both of your solutions are incorrect as they violate the fact that there is prior knowledge of the position of the coin. I think the problem has a pattern similar to the birthday paradox. The trick is to equate the number of heads and tails in one group with another. In such an event E, there could be 1 head, 49 tails in one instance, 2 heads and 48 tails in another instance etc. in one group. The trick now is to choose 50 coins so that E occurs.

  7. srujan says:

    Split the coins into two piles. One with 40 and other 60.
    Now flip the pile of 60 coins. Both will have same number of heads.
    h , 60-h, flip 60-h, you will get h

    • Kirem says:

      If there are two distinguishable groups, the boy should turn over the smaller group. This would mean that all the coins would be heads up an dad would get them all. The requirement is not that the boy gets the most, he must get equal what his sister gets.

  8. Lori says:

    100 coins 60 heads 40 tails slight off so all indistinguishable.
    Therefore, 60% are heads, 40% are tails.
    Because we don’t know any better we’ll assume on average they are equally distributed.

    The boy wants the same amount of coins as his sister, it doesn’t matter how many go back to dad.

    The boy splits the coins into 2 groups.
    1 group has 60 coins – (in the group 60%(36) are heads and 40%(24) are tails).
    2 group has 40 coins – (in this group 60%(24) are heads and 40%(16) are tails).

    He knows his sister will choose the big group. So he flips them.
    Now group 1 has 60%(24) heads and 40%(36)tails.

    Both group 1 and 2 have 24 heads, that the kids get to keep.

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