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    (Original post by Iron89)
    You just need to work out the expected value of the game:

    Expected value is the integral between 0 and B (where B is the bid) of this:

    (1.5V - B)*(1/500)dV

    With V being the actual value of the chest and the 1/500 coming from the assumption that the probability for the value is uniformly distributed between 0 and 500.

    From differentiating this you find the only way to break even would be a bid of 0, since there is no maximum turning point in the expected value.
    no, that would be wrong even if you were summing instead of integrating - you only get the extra V/2 if V<B...
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    For the pirate puzzle, it depends on whether you keep your bid when you win or not as to what your optimum bid should be.
    Call your bid B, and number of coins in the chest N

    I think if you never keep your bid you should bid 0
    And if you do keep it you should bid 500

    If you never keep your bid, whatever your bid is (B), N can take any value 0<N<B for you to win. The average value of N is then B/2...and even when w...e times that by 1.5, we get 3B/4... less than B.

    If you do keep your bid, you might as well bid 500, so you win every time.

    Makes sense?
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    (Original post by Chewwy)
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    (Original post by Lorchii)
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    (Original post by Altruistic1)
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    (Original post by Lorchii)
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    (Original post by Iron89)
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    (Original post by Swayum)
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    Hey guys. I've quoted those of you who tried to solve this, as I think I have solved it (see my post directly above).

    Just in case you're still interested
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    Just use a calculator, Wall Street full of unethical *******s anyway.
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    (Original post by M_E_X)
    For the pirate puzzle, it depends on whether you keep your bid when you win or not as to what your optimum bid should be.
    Call your bid B, and number of coins in the chest N

    I think if you never keep your bid you should bid 0
    And if you do keep it you should bid 500

    If you never keep your bid, whatever your bid is (B), N can take any value 0<N<B for you to win. The average value of N is then B/2...and even when w...e times that by 1.5, we get 3B/4... less than B.

    If you do keep your bid, you might as well bid 500, so you win every time.

    Makes sense?
    The answer is certainly 0, yes. I worked it with looking at your expected return, and getting -aB^2-cB(a and c rational positive numbers, can't remember what they were/it didn't much matter).
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    (Original post by Slumpy)
    The answer is certainly 0, yes. I worked it with looking at your expected return, and getting -aB^2-cB(a and c rational positive numbers, can't remember what they were/it didn't much matter).
    I think a better/more clear line of reasoning is...

    Whatever your bid B, the chests you can win have an average value of B/2, and so 1.5*(B/2) = 3B/4, < B... so on average you lose B/4, therefore you should bid 0.
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    (Original post by M_E_X)
    I think a better/more clear line of reasoning is...

    Whatever your bid B, the chests you can win have an average value of B/2, and so 1.5*(B/2) = 3B/4, < B... so on average you lose B/4, therefore you should bid 0.
    Except that's not actually right. If you bid B, then your expected gain is the sum from 0 to B of 3/2*p-B (where p is the variable we sum over), divided by the possible number of coins(I'll ignore this bit in future as it's just a coefficient of 1/500 or w/e.)
    Summin we get 3/2*1/2*B(B+1)-B(B+1)=-1/4*B(B+1), which is negative(but not B/4).
    Intuitively you could say that if you bid B, for 2/3 of values of the chest you'll make a loss, but the scale of the loss obviously depends.
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    (Original post by Slumpy)
    Except that's not actually right. If you bid B, then your expected gain is the sum from 0 to B of 3/2*p-B (where p is the variable we sum over), divided by the possible number of coins(I'll ignore this bit in future as it's just a coefficient of 1/500 or w/e.)
    Summin we get 3/2*1/2*B(B+1)-B(B+1)=-1/4*B(B+1), which is negative(but not B/4).
    Intuitively you could say that if you bid B, for 2/3 of values of the chest you'll make a loss, but the scale of the loss obviously depends.
    Say we bid 100 coins

    The values of chests we can win are 0, 1, 2, 3...99. The average of all these numbers is 50 (B/2)

    Is that not right? For any bid B, the values of chests we can win are 0 < N < B, and so N = B/2 on average.
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    (Original post by M_E_X)
    Say we bid 100 coins

    The values of chests we can win are 0, 1, 2, 3...99. The average of all these numbers is 50 (B/2)

    Is that not right? For any bid B, the values of chests we can win are 0 < N < B, and so N = B/2 on average.
    I think we also win if there's 100 coins, but that's by the by(In fact the average of 0,1,2,...99 is 49.5, as can clearly be seen by pairing the numbers up, and its only 50 if we have 100 as well).

    N=B/2 on average is correct, yes, or rather, the expected value is B/2. But the profit/loss we gain isn't linear, it's 3/2*p-100, so we can't just take the profit of the average(the loss at 0 is 100, but the profit at 100 is only 50, we would require them to be the same for your method to be valid).
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    (Original post by Slumpy)
    I think we also win if there's 100 coins, but that's by the by(In fact the average of 0,1,2,...99 is 49.5, as can clearly be seen by pairing the numbers up, and its only 50 if we have 100 as well).

    N=B/2 on average is correct, yes, or rather, the expected value is B/2. But the profit/loss we gain isn't linear, it's 3/2*p-100, so we can't just take the profit of the average(the loss at 0 is 100, but the profit at 100 is only 50, we would require them to be the same for your method to be valid).
    The value of our bid has to exceed the number of coins in the chest, it says in the question.

    I rounded the 49.5 up to 50 (you can't have half a gold coin, right? ).

    The expected value of N = B/2. And the expected value of our winnings is hence 3B/4. I think because we are just multiplying N by 1.5, and not squaring or anything, it is linear.


    Can you plot a spreadsheet or anything to show what you are talking about? Or write your result out properly, I don't quite get it at the moment. Is it true that your expected return is -B/4 ? That is the result that my method gives, do yours?
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    (Original post by M_E_X)
    The value of our bid has to exceed the number of coins in the chest, it says in the question.

    I rounded the 49.5 up to 50 (you can't have half a gold coin, right? ).

    The expected value of N = B/2. And the expected value of our winnings is hence 3B/4. I think because we are just multiplying N by 1.5, and not squaring or anything, it is linear.


    Can you plot a spreadsheet or anything to show what you are talking about? Or write your result out properly, I don't quite get it at the moment. Is it true that your expected return is -B/4 ? That is the result that my method gives, do yours?
    Ok, when I did this question if it was equal you also got it, but no matter. Also probably shouldn't round.

    (You don't get -B/4 either in fact, you get -B/2000, but since I'm ignoring the 1/500, I'll let you off:P)

    How about an example-bid 2.
    We win if there's 0 or 1 coins in the case(or 2 my way, but I digress).
    If 0 coins, overall gain=-2
    If 1 coin, overall gain=-0.5
    So we have (-2-0.5)/500=-5/1000. Your method we average -B/4 for 2/500 cases, which gives us -B/1000, which is not the same.
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    (Original post by Slumpy)
    Ok, when I did this question if it was equal you also got it, but no matter. Also probably shouldn't round.

    (You don't get -B/4 either in fact, you get -B/2000, but since I'm ignoring the 1/500, I'll let you off:P)

    How about an example-bid 2.
    We win if there's 0 or 1 coins in the case(or 2 my way, but I digress).
    If 0 coins, overall gain=-2
    If 1 coin, overall gain=-0.5
    So we have (-2-0.5)/500=-5/1000. Your method we average -B/4 for 2/500 cases, which gives us -B/1000, which is not the same.
    In that example, our expected return is -1.25, right? (our possible returns are -0.5 and -2, both equally likely, so can average to -1.25)

    And as B = 2, that is -5B/8

    Which is not -B/2000

    So I still don't quite get your method?


    e: interestingly, if we 'win' if B=N (as in your method), our expected return here is -B/4.
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    (Original post by M_E_X)
    In that example, our expected return is -1.25, right? (our possible returns are -0.5 and -2, both equally likely, so can average to -1.25)

    And as B = 2, that is -5B/8

    Which is not -B/2000

    So I still don't quite get your method?


    e: interestingly, if we 'win' if B=N (as in your method), our expected return here is -B/4.
    In fact, if we do add the case where we win with the same bid, the overall return is -B/4*(B+1)/501
    In the other case, it's -B/4*(B+3)/501
    You just sum the possible wins(3/2*p-B) for p=0 to B-1, and divide by the number of possible outcomes(501).

    Edit-Also, you keep forgetting that the average return when we 'win' the box is not the average return, there are still all the cases where it's worth more than we bid, in which case we gain and lose nothing.
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    (Original post by Slumpy)
    In fact, if we do add the case where we win with the same bid, the overall return is -B/4*(B+1)/501
    In the other case, it's -B/4*(B+3)/501
    You just sum the possible wins(3/2*p-B) for p=0 to B-1, and divide by the number of possible outcomes(501).

    Edit-Also, you keep forgetting that the average return when we 'win' the box is not the average return, there are still all the cases where it's worth more than we bid, in which case we gain and lose nothing.
    Well, when N>B, I don't know what happens. It does not state in the question and it is not clear. I think it is most likely that we just lose our bid (makes the game even less worth playing). If we just get our bid returned, which seems unlikely, then we should probably multiply our expected 'win' return by the chance of winning (=1/(500-B)?)
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    (Original post by M_E_X)
    Well, when N>B, I don't know what happens. It does not state in the question and it is not clear. I think it is most likely that we just lose our bid (makes the game even less worth playing). If we just get our bid returned, which seems unlikely, then we should probably multiply our expected 'win' return by the chance of winning (=1/(500-B)?)
    I'll go through the question as it was told to me:

    I have a chest, with 0-500 coins in it, uniformly distributed.
    You bid a number. If your number is equal to or more than the number of coins in the box, you buy the box for your bid. You also know someone else, who will pay you 1.5 per coin, once(if) you've bought the box. What is your optimum bid?

    The chance of winning is actually B+1/501.(Or, sorry, B/500 in the question as originally posted here).
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    The correct answer is 0 which is the answer I gave. This was only a round one question, it only get harder for there.
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    (Original post by paulak1985)
    The correct answer is 0 which is the answer I gave. This was only a round one question, it only get harder for there.
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    (Original post by paulak1985)
    The correct answer is 0 which is the answer I gave. This was only a round one question, it only get harder for there.
    like? everyone loves puzzles
    (well when a job isn't dependant on the right answer anyways)
 
 
 
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