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Probability and ratio

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Game of Head & Tails between 2 people. The first one to 8 is the winner. If the game is interrupted: when one person has 6 wins and the other has 4 wins, how should the prize money be split?


    2. The attempt at a solution

    The obvious answer (at first) i got is 3:2
    But it just seems too easy.. i think the part 8 wins should fit in somewhere but i have no idea how =/.
     
  2. jcsd
  3. Nov 15, 2012 #2

    tiny-tim

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    welcome to pf!

    hi h0llow! welcome to pf! :smile:
    if the game was interrupted just long enough to place bets on it,

    what should the odds be? :wink:
     
  4. Nov 15, 2012 #3
    assuming it was 100$... person with 6 wins would have 75% chance of winning? and the other would have 50% chance..but if add 75+50 it adds up more than 100 =/
     
  5. Nov 15, 2012 #4

    tiny-tim

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    you mean, because 6 is 75% of 8, and 4 is 50% of 8 ?

    no!

    work out the probability that (starting from 0-0) there are 2 heads before there are 4 tails :smile:
     
  6. Nov 15, 2012 #5
    huh 0_o? can u rephrase that please =/?
     
  7. Nov 15, 2012 #6

    tiny-tim

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    if you keep tossing a coin,

    work out the probability that (starting from 0-0)

    the 2nd head comes before the 4th tail​

    (or, if you prefer the opposite, that the 4th tail comes when there are only 0 heads or 1 head)
     
  8. Nov 15, 2012 #7
    considering there's a 50% chance..during each toss.. 2nd head has 25% chance? and 4th tail has 6.25% chance?
     
  9. Nov 15, 2012 #8

    Ray Vickson

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    You need to account for the fact that there might not be a second head at all. Just "guessing" does not work in problems of this type; you need to write things down in detail.

    Also: the answer will depend quite a bit on whether the two-heads player tosses first or second.

    RGV
     
  10. Nov 15, 2012 #9

    tiny-tim

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    sorry, but that's no way to solve this :redface:

    try these

    what is the probability that a coin tossed 4 times will come out all tails?

    what is the probability that a coin tossed 5 times will come out 1 head and 4 tails, and that the head isn't the last one?​
     
  11. Nov 15, 2012 #10
    (I) 0.5 X 0.5 X 0.5 X 0.5 = 6.25% (1/2 chance it is tail)
    (ii) 3.125%?? (im not sure how exactly to do it)
     
  12. Nov 15, 2012 #11

    tiny-tim

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    yes :smile:
    write out all the possible combinations
     
  13. Nov 15, 2012 #12
    0.5 x 0.5 x 0.5 x 0.5 x 0.5??
     
  14. Nov 15, 2012 #13

    Ray Vickson

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    Don't keep guessing; that is no way to learn. As I said before: write things out in detail. By this, I mean: start enumerating some of the possibilities, until you see a pattern starting to emerge.

    The outcome strings can be lists of H and T. If the two-head player (A) starts, then passes the coin to player B, etc., a string such as HHTHT means A gets H on the first toss, then B gets H on the second toss (that is, on B's first toss), the A gets T on the third toss (that is, on A's second toss), etc. Player A wins on his second toss if the outcomes are HHH or HTH. Player A wins on his third toss if the outcomes are HHTHH, HHTTH, HTTHH, HTTTH, THHHH, THHTH, TTHHH, TTHTH, etc. You need to figure out all the probabilities and add them up.

    Then, you need to repeat this type of analysis if A goes second.

    RGV
     
  15. Nov 15, 2012 #14
    alright thank you
     
  16. Nov 15, 2012 #15
    so i should write out all the 120 possibilities? (5!)
     
  17. Nov 15, 2012 #16
    nvm.. i got it...can u tell me if i am right though?

    TT THHT HHHH
    HTT HTHT HTHHH
    HHTT THHHT THHHH
    HHHTT THTH HHTHH
    THT HHTHT HHHTH

    = 15 possibilities

    chance player A (T) wins = 10/15 = 66.67%
    chance player B (H) wins = 5/15 = 33.33%

    am i right?
     
  18. Nov 15, 2012 #17
    nvm.. i got it...can u tell me if i am right though?

    TT THHT HHHH
    HTT HTHT HTHHH
    HHTT THHHT THHHH
    HHHTT THTH HHTHH
    THT HHTHT HHHTH

    = 15 possibilities

    chance player A (T) wins = 10/15 = 66.67%
    chance player B (H) wins = 5/15 = 33.33%

    am i right?
     
  19. Nov 15, 2012 #18

    tiny-tim

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    hi h0llow! :smile:
    that looks right :smile:
    no, because not all of them have equal probability

    for example, the first one, TT actually is 8 times as likely as THHHH,

    because it includes TTHHH TTHHT TTHTH TTHTT THTHH THTHT THTTH and THTTT :wink:

    (which are all 5 long, so are all equally likely)

    ok now try to think how to work out the likelihood of all the others! :smile:

    EDIT: do the Ts look bolder than the Hs to you, or is it just my weird eyesight? :redface:
     
  20. Nov 15, 2012 #19
    TT = 25% likely
    TTHHH,etc.. = 3.125% likely (since there are 5(0.5^5))......X 8
    HTT = 12.5%(0.5^3) X 2
    HHTT = 6.25% X 4

    everything adds up to 100% 0_o..what now..

    the t's look fine :eek:
     
    Last edited: Nov 15, 2012
  21. Nov 15, 2012 #20
    is the answer by any chance 81.25% and 18.75% splits in prize?
     
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