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Toss a coin, probability

  1. Oct 10, 2011 #1
    Hello everybody. I have a problem, which is following.

    AA tosses a coin 3 times and BB 2 times. AA will win, if he gets more heads than BB.
    What is the probability that AA wins? Total probability is probably needed in this.

    My solution:

    first, this formula,
    P(b) = P(a) P(b | a ) + P(a^c) P(b | a^c)

    and events are;

    a = happens, that there will be heads
    b = happens, that there will be tails
    a^c = a won't happen


    In best case for AA;

    a = 1-3 * a
    b = 0-2 * a

    a wins

    In best case for BB;

    b = 1-2 * a
    a = 0-1 * a

    now, a won't win.


    P(b) = P(a) P(b | a ) + P(a^c) P(b | a^c)

    P(2) = P(3) P(2 | 3 ) + P(0) P(2 | 0)

    P(2) = P(3) P(2 | 3 ) + P(0) P(2 | 0)


    This is as fas as I can go.
     
  2. jcsd
  3. Oct 10, 2011 #2
    In this case you're interested in the number of heads AA gets compared to the number of heads BB gets. Just remember that if BB gets 0 heads, AA needs to get at least 1 head, and if BB gets 1, AA will have to get 2 or more, and if BB gets 2 heads, AA will have to get all 3 heads.
     
  4. Oct 10, 2011 #3
    Hi. Yes, that's what I meant with;

    a = 1-3 * a
    b = 0-2 * a

    a wins

    In best case for BB;

    b = 1-2 * a
    a = 0-1 * a.


    But maybe you meant more than that.

    And I want to do this in this method; P(b) = P(a) P(b | a ) + P(a^c) P(b | a^c) or something like that.
    I believe that conditional probability is an issue here.

    and my new events are;

    a = happens, that AA gets heads
    b = happens, that BB gets heads
    a^c = a won't happen


    Well, I think these are important, but these may be wrong..

    P(a) = P(a) P(b | a ) + P(a^c) P(b | a^c)

    P(b) = P(b) P(a | b ) + P(b) P(a^c | b )

    P(a^c) = P(a^c) P(a | b ) + P(b) P(a^c | b ) ?

    Wait a minute, both AA and BB have a 0.5 probability to have heads!

    And after that we must do some multiplication, I guess.

    I can't solve this problem.
     
  5. Oct 10, 2011 #4

    mathman

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    Science Advisor
    Gold Member

    For AA, P(3)=1/8, P(2)=P(1)=3/8, p(0)=1/8
    For BB, P(2)=1/4, P(1)=1/2, P(0)=1/4

    For AA to win - needs more heads than BB (I assume BB wins in case of ties)

    P(AA wins-no. heads in bracket) = 1/8[3 heads]+(3/8)(3/4)[2 heads]+(3/8)(1/4)[1 head]=1/2
     
  6. Oct 11, 2011 #5
    So probability that AA wins is

    P(AA wins-no. heads in bracket) = 1/8[3 heads]+(3/8)(3/4)[2 heads]+(3/8)(1/4)[1 head]=1/2
    and solution is this. This wasn't easy for me. I want to say; thank you very much!
     
  7. Oct 11, 2011 #6
    Does it mean that AA will always have equal chance 1/2 of winning no matter how many additional throws he gets more than BB?
     
  8. Oct 12, 2011 #7

    mathman

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    Science Advisor
    Gold Member

    No. If AA has more than 3 tosses, while BB has 2, then AA will win more often.

    Each combination needs to be worked out.
     
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