Coin Sequence Game

Source: Plus Magazine Issue 55

Problem: Let's witness a game. The game involves flipping a coin, which you assume has equal probability of coming up heads or tails. The game is played by two players, A and B, who each select a sequence of three flips. For example, assume that Player A selected "heads-heads-heads" (HHH) and Player B has selected "tails-heads-heads" (THH). Then the coin is flipped repeatedly, resulting in a sequence like the following:


The player whose sequence showed up first (HHH for Player A or THH for Player B) is declared the winner.

A chooses first and you have to help B. Show that given what A has chosen, B can always choose a sequence such that B is in a better position than A.

For example: If A chooses HHH, B chooses THH to make sure that now it has 7/8 probability to win.

Update (27/07/10): Solution: Posted in comments by me!


  1. The possible sequences that A can select and the particular sequence that B should select to have high probabilty of winning are

    HHH - THH
    HHT - THH
    HTH - THT
    HTT - THT
    THH - HTH
    THT - HTH
    TTH - HTT
    TTT - HTT

    To genaralize if A selects a squence like a1a2a3 B should select (-a1)a1a2

  2. @Bhanu: HTH - THT pair seems incorrect. Can you plz. check?

  3. @Shantanu.. you are correct.. Thanx..

    P(getting THT before HTH) = 1/2 {Symmetrical terms :P}

    @Bhanu.. nice try.. but this would not always work as pointed out by Shantanu..

    If you are interested, you can read the complete article here:


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