Results 1 to 3 of 3

Math Help - urn problem

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    3

    urn problem

    each of two players A and B toss N fair coins. let Pi be the probability that i coins of N coins player A (equivalently player B) tossed show heads.

    a) what is the probability the two players have the same number of heads?

    b) if the game is modified so that player A tosses N+1 coins ( player B still tosses N ) show that the probability taht player A has more heads than player B is 0.5.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by farukcan View Post
    each of two players A and B toss N fair coins. let Pi be the probability that i coins of N coins player A (equivalently player B) tossed show heads.

    a) what is the probability the two players have the same number of heads?

    b) if the game is modified so that player A tosses N+1 coins ( player B still tosses N ) show that the probability taht player A has more heads than player B is 0.5.
    a) Suppose the players toss N fair coins. Let X be the number of heads tossed by A and Y be the number of heads tossed by B. Then both X and Y have Binomial(N, 1/2) distributions, so

    P(X = Y) = \sum_{i=0}^N P(X=i) \; P(Y=i) =\sum_{i=0}^N \left( \binom{N}{i} (1/2)^N \right) ^2
    = \sum_{i=0}^N \binom{N}{i}^2 (1/2)^{2N}
     =(1/2)^{2N} \sum_{i=0}^N \binom{N}{i}^2
     =\binom{2N}{N} (1/2)^{2N}

    where we have used a well-known identity, \sum_{i=0}^N \binom{N}{i}^2 = \binom{2N}{N}.

    b) First, let's continue a) by finding P(X > Y) and P(X \geq Y).
    By symmetry, P(X < Y) = P(Y < X). Then
    1 = P(X < Y) + P(X=Y) + P(X >Y) = P(X=Y) + 2 P(X>Y)
    so
    P(X>Y) = (1/2) [1 - P(X=Y)] = (1/2) \left(1 - \binom{2N}{N} (1/2)^{2N} \right) .....(1)
    and
    P(X \geq Y) = P(X>Y) + P(X=Y) = (1/2) \left( 1 - \binom{2N}{N} (1/2)^{2N} \right) + \binom{2N}{N} (1/2)^{2N}
    = 1/2 + \binom{2N}{N} (1/2)^{2N+1} ....(2)

    Now consider the case where A has N+1 coin tosses. As before, let X = the number of heads in the first N tosses, and let's say Z is the total number of heads in all N+1 tosses.

    We find P(Z > Y) by conditioning on A's last toss. If A's last toss is a tail, then  Z > Y if and only if X > Y. If A's last toss is a head, then Z > Y if and only if X \geq Y. So

    P(Z > Y) = (1/2) P(X > Y) + (1/2) P(X \geq Y).

    I think that if you substitute the expressions for P(X > Y) from (1) and P(X \geq Y) from (2) in the above equation and do a little algebra, you will find that P(Z > Y) reduces to 1/2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1

    Simplification

    Quote Originally Posted by awkward View Post
    a) Suppose the players toss N fair coins. Let X be the number of heads tossed by A and Y be the number of heads tossed by B. Then both X and Y have Binomial(N, 1/2) distributions, so

    P(X = Y) = \sum_{i=0}^N P(X=i) \; P(Y=i) =\sum_{i=0}^N \left( \binom{N}{i} (1/2)^N \right) ^2
    = \sum_{i=0}^N \binom{N}{i}^2 (1/2)^{2N}
     =(1/2)^{2N} \sum_{i=0}^N \binom{N}{i}^2
     =\binom{2N}{N} (1/2)^{2N}

    where we have used a well-known identity, \sum_{i=0}^N \binom{N}{i}^2 = \binom{2N}{N}.

    b) First, let's continue a) by finding P(X > Y) and P(X \geq Y).
    By symmetry, P(X < Y) = P(Y < X). Then
    1 = P(X < Y) + P(X=Y) + P(X >Y) = P(X=Y) + 2 P(X>Y)
    so
    P(X>Y) = (1/2) [1 - P(X=Y)] = (1/2) \left(1 - \binom{2N}{N} (1/2)^{2N} \right) .....(1)
    and
    P(X \geq Y) = P(X>Y) + P(X=Y) = (1/2) \left( 1 - \binom{2N}{N} (1/2)^{2N} \right) + \binom{2N}{N} (1/2)^{2N}
    = 1/2 + \binom{2N}{N} (1/2)^{2N+1} ....(2)

    Now consider the case where A has N+1 coin tosses. As before, let X = the number of heads in the first N tosses, and let's say Z is the total number of heads in all N+1 tosses.

    We find P(Z > Y) by conditioning on A's last toss. If A's last toss is a tail, then  Z > Y if and only if X > Y. If A's last toss is a head, then Z > Y if and only if X \geq Y. So

    P(Z > Y) = (1/2) P(X > Y) + (1/2) P(X \geq Y).

    I think that if you substitute the expressions for P(X > Y) from (1) and P(X \geq Y) from (2) in the above equation and do a little algebra, you will find that P(Z > Y) reduces to 1/2.
    I now see (and should have seen earlier) that the answer in part b) can be obtained more simply without using the result of a) at all.

    As before, we know
    1 = P(X < Y) + P(X=Y) + P(X >Y) = P(X=Y) + 2 P(X>Y)
    so
    P(X>Y) + (1/2) \; P(X=Y) = 1/2.

    And
    P(Z > Y) = (1/2) \;  P(X > Y) + (1/2) \; P(X \geq Y)
    = (1/2) \; P(X >Y) + (1/2) \; [P(X>Y) + P(X=Y)]
    = P(X > Y) + (1/2) \; P(X=Y)
    = 1/2
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum