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Difficult Math Problem #108 - Probability

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #108 - Probability

Post Mon Mar 12, 2007 12:02 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.

    1- Find the probability that the person who tosses first will win the game?

    2- What are the odds against A's losing if she goes first?



    oa coming when people respond with explanations. from diff math doc

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    gabriel Legendary Member
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    Post Mon Mar 12, 2007 1:31 pm
    800guy wrote:
    A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.

    1- Find the probability that the person who tosses first will win the game?

    2- What are the odds against A's losing if she goes first?



    oa coming when people respond with explanations. from diff math doc
    ok this is a tuff q....

    im not sure abt the solution... but here it is.... there will be more than one case in which the one who throws first will win

    1st case... he wins in the first try itself
    2nd case.... he throws a tail, then the next guy throws a tail and then the first guy throws a head ..... so it wuld be something like TTH

    3rd case... TTTTH
    4th case.... TTTTTTH
    5th case....TTTTTTTTH

    so the probability shuld be 1/2+(1/2)^3+(1/2)^5+(1/2)^7 +(1/2)^9..... again i am really not sure abt the solution...


    now i did not exactly understand what the 2nd case means... i am assuming it wants us to find the probability that the guy who tosses second ( in this case B ) wins...

    if that is what it means then...

    1st case ... TH
    2nd case... TTTH
    3rd case...TTTTTH
    4th case .... TTTTTTTH
    5th case.....TTTTTTTTTH

    therefore the probability shuld be ...(1/2)^2+(1/2)^4+(1/2)^6+(1/2)^8+(1/2)^10.... this was indeed a very good q....

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Mar 14, 2007 8:41 am
    here's the oa from the doc:

    look at the conditions; it says that the first person who tosses a head wins.

    Let's say A tosses first.

    what is the probability that he wins

    H + TTH + TTTTH + TTTTTTH + TTTTTTTTH

    i.e. either the first toss is head,
    or the first time A tosses the coin he gets a tail and B also gets a tail , n in the second throw A gets a head.....

    This continues for a max till 5 throws, because the game is for 5 throws only.
    So, 1. 1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9

    2. (1/2)^2 + (1/2)^4 + (1/2)^6 + (1/2)^8 + (1/2)^10

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Mar 14, 2007 8:41 am
    agreed, this is really tough

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