Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A problem in probability

  1. Jun 6, 2004 #1
    Hi members,
    I have traveled this forum sometimes, But this is my first question. I hope to get your help so that I can prepare better for my GRE Math test.

    Following is my question.

    In a game two players take turns tossing a fair coin; the winner is the firt one to toss a head. The probability that the player who makes the first toss wins the game is:
    A)1/4
    B)1/3
    C)1/2
    D)2/3
    E)3/4

    Thanks in advance.
    LuuTruongHuy
     
  2. jcsd
  3. Jun 6, 2004 #2
    HI

    Here's my solution....

    H = head
    T = tails

    (AH denotes "A got a head")

    Suppose A starts first. Then the different possibilities are tabulated thus:

    AH (A gets a head, game stops)
    AT,BT,AH (A gets tails, B gets tails, A gets heads, game stops)
    AT,BT,AT,BT,AH (A gets tails, B gets tails, A gets tails, B gets tails, A gets heads, game stops)

    and so on....

    So the probability is given by the sum,

    [tex]\displaystyle{\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} + ...} [/tex]

    the kth term is

    [tex](\frac{1}{2})^p[/tex]

    where p = (2k+1) for k = 0, 1, 2, .... note that there are (2k+1) continued products in the kth term)

    The game goes on as long as A and B get tails and stops as soon as A gets a head, since A was the one who started the game first.

    This is an infinite sum, the value of which is given by

    [tex]SUM = \displaystyle{\frac{1/2}{1-(1/4)}} = \frac{2}{3}[/tex]

    I think 2/3 should be the answer, but I could be wrong (as usual) ;-)

    Someone please correct me if I'm wrong. If any part of the solution is wrong/not clear, please let me know. (I have assumed that you are familar with addition and multiplication in probability and also with geometric progressions, esp containing an infinite number of terms--the kinds that appear in such problems.)

    Cheers
    Vivek
     
    Last edited: Jun 6, 2004
  4. Jun 6, 2004 #3

    uart

    User Avatar
    Science Advisor

    The answer is 2/3

    The first player has a probability of 1/2 that both he will take a first toss AND that he will win on that toss.

    The second player only has a probablity of 1/4 that he will both take his first toss and win on that toss.

    The first player then has a probabilty of 1/8 that he will both require his second toss and win on that toss.

    Continuing on like this the first player has a probabilty of 1/2 + 1/8 + 1/32 + ... and the second player has a probability of 1/4 + 1/16 + 1/64 + ... of winning.
     
  5. Jun 6, 2004 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For those who like clever answers, you can skip the infinite series. :smile:

    Suppose the first player's first flip is a tails. Now, if you look at how the game proceeds, it is identical to the original game, except the first and second player are reversed.

    So if p is the probability that the first player in the game wins, then once the first player flips a tails, the second player has a probability p of winning. (and probability 0 of winning otherwise)

    Since there's a 1/2 chance the first player will flip tails, the second player has a probability p/2 of winning, and the first player probability p.

    Thus, p = 2/3.
     
  6. Jun 8, 2004 #5
    Thank All for very nice answers !!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A problem in probability
  1. Probability problem (Replies: 3)

Loading...