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Thread: Solve for X where sum of x's with consecutive exponents equals 1

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    Solve for X where sum of x's with consecutive exponents equals 1

    Hi. My first post and am hoping to get help. I have a problem where I have to distribute some prize money, halving the prize money each time until every participant receives their share and no money is left. For example, first place gets half the prize money. Second place gets half of what's left, etc. Last place gets remainder which was double the money given to the next to last place participant. The number of participants may vary for each event.

    Would this be the formula?
    .5(x5+x4+x3+x2+x) = 1
    assuming there are 5 participants?

    And how do you solve for 'X'?

    Thanks.
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Quote Originally Posted by sumdumgai View Post
    Hi. My first post and am hoping to get help. I have a problem where I have to distribute some prize money, halving the prize money each time until every participant receives their share and no money is left. For example, first place gets half the prize money. Second place gets half of what's left, etc. Last place gets remainder which was double the money given to the next to last place participant. The number of participants may vary for each event..
    I would do this: Start with $\mathit{A}=$ the total amount of the prize money for the particular.
    Then let $\mathcal{P}(k)=2^{-k}\cdot\mathit{A}$ where $k$ is the place the participant finishes.
    But if there were $L>1$ players the last player gets $\mathcal{P}(L)=2^{-(L-1)}\cdot\mathit{A}$ The last place gets the same as the next to last.
    Example: the third place finisher gets $\mathcal{P}(3)=2^{-3}\cdot\mathit{A}$ Here if the were four players then:
    $\mathcal{P}(4)=\mathcal{P}(3)$ thus If $\mathit{A}=\$64$ then $\mathcal{P}(1)\$32,~\mathcal{P}(2)=\$16,~\mathcal {P}(3)=\$8~\&~\mathcal{P}(4)=\$8$.
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Thanks. I had used your suggested formulas in Excel, but didn't give the last two players the exact same amount. That left a remaining amount of money equal to the difference between those two players. I'm still looking for a way to divide all of the prize money using the 'halving' technique. There has to be a way of 'halving' multiplied by some constant such that you're left with a zero balance. For example, by trial and error, I found that multiplying the exponent by the constant 1.032258065 gives the exact payout amount to 6 decimal places. How do I get that constant algebraically?
    Last edited by sumdumgai; Nov 27th 2018 at 12:02 PM.
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Or, how do I compute the negative exponent that I need to multiply by 2 to get the exact halving amounts?

    Thanks.
    Last edited by sumdumgai; Nov 27th 2018 at 12:10 PM.
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Quote Originally Posted by sumdumgai View Post
    Thanks. I had used your suggested formulas in Excel, but didn't give the last two players the exact same amount. That left a remaining amount of money equal to the difference between those two players. I'm still looking for a way to divide all of the prize money using the 'halving' technique.
    That is because you have to program it to do that.
    If $N$ is the number of players and $A$ is the total amount of the prize of the round then:
    t:= A,~ k:= 1
    while k<N-1
    $P(k)=\frac{1}{2}(t),~t:=P(k) ,~k=k+1$
    loop
    $P(N)=P(N-1)$
    end.
    Thanks from sumdumgai
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Isn't that same answer as you gave earlier, but programmatically; i.e., giving last two players the same? Thanks.
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    So are you saying that the prize money could look like (say):
    32, 16, 8, 4, 8 if 68 was the total ?
    ie, the last and third last will be the same, "Last place gets remainder which was double the money given to the next to last place participant."

    so that (when n=5):

    A = $\displaystyle x + \frac{x}{2}+\frac{x}{4} +\frac{x}{8}+2*\frac{x}{8} $
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Thanks, but no. If the total payout amount was $235.00 and there are N (5) contestants, the payouts would be 1st thru 5th, $121,29, $60.65, $30.32, $15.16, and $7.58, which all add up to $235.00. In this case, for N = 1 (first place) to 5 (fifth place), I multiplied 2 to the negative (N times some constant) power to get the payout for the player in Nth place. The constant was found by trial and error to be 1.032258065.

    All of the prize money is distributed and every player gets something. The first place player gets twice as much as the second place player, who gets twice as much as the third place player, and so on.
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Last place gets remainder which was double the money given to the next to last place participant.
    What does that mean then?
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Sorry, I misspoke. Should have said 'Last place would get whatever is left after everyone else is paid, and that amount would be exactly half of the money given to the next-to-last-place player.'

    So, if A = total prize money and N = number of payouts, then the payout (P) for the nth place is:
    P(n)*A = .5**(n*x) where x is some constant that increases the exponent so that the sum of all payouts = A.

    Confused? I am. But there should be a way to calculate x so that:
    .5**(5*x) + .5**(4*x) + .5**(3*x) + .5**(2*x) + .5**(1+x) = 1
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Quote Originally Posted by sumdumgai View Post
    Sorry, I misspoke. Should have said 'Last place would get whatever is left after everyone else is paid, and that amount would be exactly half of the money given to the next-to-last-place player.'

    So, if A = total prize money and N = number of payouts, then the payout (P) for the nth place is:
    P(n)*A = .5**(n*x) where x is some constant that increases the exponent so that the sum of all payouts = A.

    Confused? I am. But there should be a way to calculate x so that:
    .5**(5*x) + .5**(4*x) + .5**(3*x) + .5**(2*x) + .5**(1+x) = 1
    Yes I am confused. What does ** mean?
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Actually, there should be a way to calculate x so that:
    (.5**5)*x + (.5**4)*x + (.5**3)*x + (.5**2)*x + (.5**1)*x = 1
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Quote Originally Posted by sumdumgai View Post
    Hi. My first post and am hoping to get help. I have a problem where I have to distribute some prize money, halving the prize money each time until every participant receives their share and no money is left. For example, first place gets half the prize money. Second place gets half of what's left, etc. Last place gets remainder which was double the money given to the next to last place participant. The number of participants may vary for each event. Would this be the formula? assuming there are 5 participants?.
    Quote Originally Posted by sumdumgai View Post
    Sorry, I misspoke. Should have said 'Last place would get whatever is left after everyone else is paid, and that amount would be exactly half of the money given to the next-to-last-place player.'
    Quote Originally Posted by sumdumgai View Post
    Actually, there should be a way to calculate x so that:
    (.5**5)*x + (.5**4)*x + (.5**3)*x + (.5**2)*x + (.5**1)*x = 1
    Frankly you sowed great confusion with your examples. I still don't think that we understand the setup.
    Look at this: $5++4+3+2+1=15$ So $\dfrac{5}{15}+\dfrac{4}{15}+\dfrac{3}{15}+\dfrac{ 2}{15}+\dfrac{1}{15}=1$
    In a race with five runners you could award the runner in the $k^{\text{th}}$ place $\dfrac{6-k}{15}$ of the total amount.

    IN GENERAL:
    In a race with $N$ runners you could award the runner in the $k^{\text{th}}$ place $\dfrac{N+1-k}{P}$ of the total amount where $P=\dfrac{N(N+1)}{2}$

    Look at this Calculation.
    Eight runners with a total of $\$362.88$.
    Last edited by Plato; Nov 27th 2018 at 04:29 PM.
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Quote Originally Posted by sumdumgai View Post
    Thanks, but no. If the total payout amount was 235.00 and there are N (5) contestants, the payouts would be 1st thru 5th, 121,29, 60.65, 30.32, 15.16, and 7.58, which all add up to 235.00.
    BUT your 1st post says this: "For example, first place gets half the prize money":
    121.29 is NOT half of 235 !

    Are you fooling around or something?
    So far, you gave 2 WRONG information...
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    Re: Solve for X where sum of x's with consecutive exponents equals 1

    Your silly problem works like this:
    p = prize money = 235
    a = 1st place share

    a + a/2 + a/4 + a/8 + a/16 = p = 235
    so:
    16a + 8a + 4a + 2a + a = 16*235
    31a = 3760
    a = 3760/31 = 121.29
    Last edited by DenisB; Nov 27th 2018 at 04:27 PM.
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