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Math Help - 3 players, first heads wins, 2 methods

  1. #1
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    3 players, first heads wins, 2 methods

    First the problem, then the 2 solutions I have. The question: How do I know which method is correct? ie, test for independence and how would I have known to do that?

    Players A, B, and C take turns flipping a fair coin. The first to flip heads (1) wins. They flip in the order presented; A, B, C.
    -Describe the sample space
    -What is P(A wins)?

    Sample space:
    S = {1, 01, 001, 0001, ... , 000000...} where the game ends if a heads (1) is flipped.

    Method 1:
    A has to flip heads on the 1 mod 3 flip to win the game. However, if A is flipping, this means nobody has won the game. Every time A flips, we disregard all prior flips. Thus, every 1 mod 3 flip, P(A wins) = 1/2.

    Method 2 (infinite bernouli, I think):
    We sum the infinite converging series 1/2 + 1/8 + 1/64 + ... + 1/[2^(3k+1)] where k = 0,1,2,... representing each round/iteration, 3k+1 is A's kth flip. Thus, P(A wins) = 4/7. This math can be done for B and C as well, with all of their probabilities adding to 1.
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  2. #2
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    Re: 3 players, first heads wins, 2 methods

    Quote Originally Posted by tcRom View Post
    Players A, B, and C take turns flipping a fair coin. The first to flip heads (1) wins. They flip in the order presented; A, B, C.
    -Describe the sample space
    -What is P(A wins)?
    Sample space:
    S = {1, 01, 001, 0001, ... , 000000...} where the game ends if a heads (1) is flipped.
    No that is not the sample space for A winning.
    It is S=\{1,~0001,~0000001,~0000000001,\cdots\}
    Player A can win on the first, fourth, seventh, tenth, etc flip.
    But those are player A's first, second, third, fourth flip.
    So that is (.5)+(.5)^4+(.5)^7+(.5)^{10}+\cdots
    Can you find \sum\limits_{k = 1}^\infty  {\left( {0.5} \right)^{3k - 2} }
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  3. #3
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    Re: 3 players, first heads wins, 2 methods

    Plato, thank you for your time:

    I agree the sample space you've written is the outcome for A winning, but not for the game (which is what I've described). I believe I'm to describe the game, but I'm not certain.

    That point aside, I'm more concerned with P(A wins). I get 4/7 from your summation, the same as my second method, but I used  \sum\limits_{k = 0}^\infty  {\frac{1}{2^{3k + 1} } .

    Why is the first method incorrect? Treating each flip for A as if the game had reset and it was A's first flip again.
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    Re: 3 players, first heads wins, 2 methods

    Quote Originally Posted by tcRom View Post
    Plato, thank you for your time:
    I agree the sample space you've written is the outcome for A winning, but not for the game (which is what I've described). I believe I'm to describe the game, but I'm not certain.
    That point aside, I'm more concerned with P(A wins). I get 4/7 from your summation, the same as my second method, but I used  \sum\limits_{k = 0}^\infty  {\frac{1}{2^{3k + 1} } .
    Why is the first method incorrect? Treating each flip for A as if the game had reset and it was A's first flip again.
    Well what the h did you describe as the game?
    AB&C flip a coin in that order. The first to get a head wins.
    What is the probability that A wins?
    Is that not what you posted?
    If not, then what is the question?
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  5. #5
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    Re: 3 players, first heads wins, 2 methods

    The book doesn't make it clear if it's the whole set or just A winning that should be defined, so I'm not worried about that too much. Yes though, the game is A, B, and C flip a coin, in that order. The first to flip heads wins. So, the sample space for that would be S = {1, 01, 001, 0001, ... , 000000...}. For term 1, 2, 3, and 4 the winner is A, B, C, and A. The sample space I'm describing is all the possible outcomes of the game, not just where A wins.

    As for method 2, the summation, we're on the same page.

    Why method 1 would be wrong is what I don't understand.
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    Re: 3 players, first heads wins, 2 methods

    Quote Originally Posted by tcRom View Post
    Why method 1 would be wrong is what I don't understand.
    Because the probability changes on each flip. A has 0.5 probability of winning on the first turn but (0.5)^4 on A's second flip.
    Thanks from tcRom
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