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abz_tech
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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

abz_tech Group TitleBest ResponseYou've already chosen the best response.0
so the answer is just 2:3?
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

abz_tech Group TitleBest ResponseYou've already chosen the best response.0
hmm, what i first thought too.. but seems.. a bit too easy
 one year ago

abz_tech Group TitleBest ResponseYou've already chosen the best response.0
well any way thx for the opinion
 one year ago

abz_tech Group TitleBest ResponseYou've already chosen the best response.0
i think the part '8 games' should fit somehwere in the answer..not sure how though
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

abz_tech Group TitleBest ResponseYou've already chosen the best response.0
if you where to do that how would it go?
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

abz_tech Group TitleBest ResponseYou've already chosen the best response.0
alright thank you very much
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
Glad to help... sorry the problem wasn't a little more clear. Good luck!
 one year ago
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