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abz_tech Group Title

Game of head and 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, how should the prize money be split?

  • 2 years ago
  • 2 years ago

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  1. JakeV8 Group Title
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    This looks like a probability problem, but it isn't. If the players split the prize money based on how many wins each has, then it's really a ratio problem.|dw:1352844957072:dw|

    • 2 years ago
  2. abz_tech Group Title
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    so the answer is just 2:3?

    • 2 years ago
  3. JakeV8 Group Title
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    sort of. Your problem is not very clear on what format the answer should be in. The ratio of player 1 (with 6 wins) to player 2 (with 4 wins) is 3:2. If you had to provide a fraction of the winnings that each player would get, remember that the whole here is 6 + 4 = 10. So player 1 would get 6/10 of the winnings, and player 2 would get 4/10. (and you could reduce those fractions if needed)

    • 2 years ago
  4. abz_tech Group Title
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    hmm, what i first thought too.. but seems.. a bit too easy

    • 2 years ago
  5. abz_tech Group Title
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    well any way thx for the opinion

    • 2 years ago
  6. abz_tech Group Title
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    i think the part '8 games' should fit somehwere in the answer..not sure how though

    • 2 years ago
  7. JakeV8 Group Title
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    yes, that bugs me a bit too. I believe that the "first person to 8 wins" is just a distractor to try to make you make an error. I'll think about it another minute... not totally convinced yet.

    • 2 years ago
  8. JakeV8 Group Title
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    If the players split all the prize money, then I think the 6/10 and 4/10 split (or the 3:2 ratio), is correct. You could try to make the claim that since neither got to 8, then the players should split something less than the whole prize money, in some sort of ratio based on how close they were to winning the 8 games needed to actually win. But that is going way beyond what is written in the problem... not usually a good idea.

    • 2 years ago
  9. abz_tech Group Title
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    if you where to do that how would it go?

    • 2 years ago
  10. JakeV8 Group Title
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    I don't think that 2nd approach is correct, but you could try it like: 6 is 3/4 of 8 games. 4 is half of 8 games. So 3/4 + 1/2 of the prize money should be paid out... 3/4 to Player 1, and 1/2 to player 2. But that obviously makes no sense... 3/4 + 1/2 is more than the total prize money.

    • 2 years ago
  11. JakeV8 Group Title
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    I don't know... I can't think of any way to make that 2nd approach work short of just making an arbitrary choice that you would only award some portion of the total prize money for incomplete games, and then apply your 3:2 proportion to the amount you made up. Like if it was supposed to be a $100 prize, you could decide to award $50 for an incomplete game, then player 1 would get $30 and player 2 would get $20... but you are totally making up everything but the 3:2 proportion here... none of the rest is given in the problem.

    • 2 years ago
  12. abz_tech Group Title
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    alright thank you very much

    • 2 years ago
  13. JakeV8 Group Title
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    Glad to help... sorry the problem wasn't a little more clear. Good luck!

    • 2 years ago
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