Qua locus

Keeping both eyes on the long game.

A whole lot of coins and two puzzles they pose…


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I love puzzles, here are a couple of I’ve come across recently. One needs some statistical knowledge to find a solution, the other is all logic. Please post your solutions in the comments, and your favourite puzzles too!

  • Bonus points if you solve this one in your head… It helps to know some statistics! I’m going to toss a coin 100 times, and every time I toss more than 60 heads I’ll pay you £1.
    • Intuitively, how much are you willing to pay to play my game?
    • After calculating, how much are you willing to pay to play my game?
  • And now a logical puzzle: You are blindfolded. On the table in front of you are a number of ordinary coins. You are told that exactly 26 coins show heads. Your challenge (while blindfolded): Divide the coins into two sets so that the same number of coins in each set show heads. There is no way for you to tell, by sight or touch, which coins are showing heads or tails.
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Written by CGF

May 13, 02012 at 18:50

Posted in Puzzles

Tagged with ,

8 Responses

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  1. a1) 10p
    a2) 2p

    b) I will mix all the coins(change positions on the table) as much as I can and then split them into two sets. If
    I do a good enough job or mixing, this should give me two sets with 13 coins showing heads.

    Akshat Rathi

    May 13, 02012 at 22:30

  2. I’m not playing! I would never win.
    I would toss all the coins blindfolded and split the lot in two – chances are I would get a 50/50 split in both piles.
    Seems logical, but I fear I have missed something.
    Mike

    Mike Fleming

    May 14, 02012 at 09:35

  3. Akshat, re: the first problem – nice! Were you as surprised as me by the statistical result? How did you find it?

    Akshat, Mike, re: the second problem… that was my first approach too, but you can do better and always split them exactly! A key clue is that the total number of coins on the table doesn’t matter. What’s more, if there were exactly say 52 coins (possible, since the total is left unspecified) then just randomly splitting them has only a low chance of splitting them evenly…

    cfogelberg

    May 14, 02012 at 11:46

    • re: first problem – It was surprising indeed. I used the pascal’s triangle.

      re: second problem – what’s the solution?

      Akshat Rathi

      May 15, 02012 at 08:07

      • re: first problem – interesting, tell me more! I approximated the binomial with a normal distribution, then I did some standard deviation calculations

        re: second problem – qwandor nailed it in his post

        Christo

        May 15, 02012 at 12:13

  4. Take 26 coins, flip them upside down, and call them set A. Call the rest set B.

    qwandor

    May 14, 02012 at 19:42

    • Nice! And concise 🙂

      Christo

      May 14, 02012 at 20:00

  5. For the record, here is a full solution to the coin tossing game. As Akshat hints at, there are others…

    At it’s core, this puzzle is binomial distribution problem, but binomials are hard to do for large numbers (especially on paper or in your head!). Fortuitously, the normal distribution approximates a binomial for large n. With that in mind…

    * A coin toss is always heads or tails (1 or 0), therefore std(1 coin toss) = 0.5
    * Variance = standard deviation^2, therefore var(1 coin toss) = std^2 = 0.25
    * 100 coin tosses are 100 independent samples and the total variance of n independent samples with individual variance v is n*v
    * Therefore var(100 coin tosses) = 25 and std(100 coin tosses) = 5
    * Therefore 60 heads is 2 standard deviations from the mean
    * ~95% of normal distribution samples are <=2 standard deviations from the mean
    * Therefore ~2.5% of 100 coin samples will have more than 60 heads
    * Solution: I should pay less than 3c (i.e. 2c or 2.5c) to break even

    Christo Fogelberg

    June 2, 02012 at 14:22


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