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Clemmonsdogpark Brain Twister: Flipping Out


Welcome back, brave Clemmonsdogparkers! Best of luck with this week’s diabolical GRE Brain Twister. 🙂

If a fair coin is tossed six times, what is the probability of getting exactly three heads in a row?

(A) 1/16
(B) 3/16
(C) 1/8
(C) 3/8
(D) 1/2

See you on Thursday for the answer and explanation. 🙂


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11 Responses to Clemmonsdogpark Brain Twister: Flipping Out

  1. Ravi February 25, 2017 at 12:16 pm #

    If you don’t want to particularly write all possibilities then I got another method- its (1/2)^4 +(1/2)^5 +(1/2)^5+(1/2)^4=3/16….that’s all.simply multiply probability of particular outcome and add the cases. For eg. HHHT(HT)-The terms outside the braces are fixed so they have probability (1/2)^4 and terms inside the braces can be permuted viz. They could be HT TH TT HH so they have probability of 1 similarly for rest three cases

  2. tulu February 20, 2016 at 3:16 am #

    Why can’t I use binomial distribution here? I am getting 5/16

  3. Srivatsa M Hegdes September 26, 2015 at 8:32 am #


    I have a fundamental doubt. When we choose an event like getting the same number on 2 die in a single throw, why isn’t each trial counted twice (i.e a 5,5 on both die counted twice) are we not excluding one trial by considering it as a single trial ?

  4. VB February 11, 2015 at 6:24 pm #

    I tried that problem several times but I came to the answer which was 5/16, which isn’t any of those choices. I used the binomial distribution to solve this problem. 6C3 times (1/2)^(3) times (1/2)^(3).

    (1/2)^3 is for the head that is being tossed three times.
    (1-1/2)^(6-3) for the probability of remainder that should not contain the head.

    I am lost where I go wrong.

    • devajyothi February 28, 2015 at 10:10 am #

      your solution is for getting atleast 3 heads so it includes 6c4 6C5 6c6..remove possibilities of 6c4,6c5,6c6..ans will be 3/16

      • tulu February 20, 2016 at 3:14 am #

        no he applied the formula for exactly 3 heads.

  5. David February 11, 2015 at 1:58 pm #

    But how do you do this problem without having to think out all the possible combinations?

  6. David February 11, 2015 at 1:55 pm #

    Its B, because there are 12 possible “three heads in a row”

  7. Oxana February 11, 2015 at 12:57 am #

    The answer is A. There are 2^6 = 64 possible combinations of heads and tails. Only four of them meet the requirement of exactly three heads in a row: HHHTTT, THHHTT, TTHHHT, TTTHHH. Thus the probability is 4/64 or 1/16.

    • Chris Lele
      Chris Lele February 11, 2015 at 11:29 am #


      There is a little twist to the problem. You have listed some of the possibilities that fit the requirement, but there are actually a few more ways of arranging all heads.

      • Oxana February 12, 2015 at 10:58 pm #

        I see. So there are 8 more combinations like HHHTHT and THHHTH. Altogether 12/64 or 3/16. B

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