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  1. #16
    Monster of the deep Kevin Bonham's Avatar
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    Hi Barry, I've just sent you two solutions reaching the same result by very different methods. I'll drop onlookers a subtle hint about the general solution. Dr Evil will be carrying a very heavy bag.

  2. #17
    Reader in Slood Dynamics Rincewind's Avatar
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    Two prospective solution received so far. Still time for other interested parties to try out their problem solving skills.

    Deadline 24:00 Wednesday, 14-Jan (GMT+11)
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  3. #18
    Reader in Slood Dynamics Rincewind's Avatar
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    Make that three solvers so far. Don't delay for a chance to get in "on the ground floor" for the fad that is sweeping the online BB community.

    That didn't come across to self-obsessed or needy, did it?
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  4. #19
    Reader in Slood Dynamics Rincewind's Avatar
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    OK, Only 30 hours to the deadline. I have solution from MS and KB. I also have a solution from DR but he has subsequently sent me an PM that he would like to revise it. So unless I hear back I will assume the previous submission is withdrawn - leaving me with two solvers.

    I have also had the opportunity to prepare my solution as clear and correct as I can make it. Hopefully, it is of some use. Due to the symbology required I've published this as a PDF file and will be linking to it from my solution post.

    I'm also thinking of quoting the solvers' PMs directly in the solution post. I trust this is not a breach of protocol as I don't believe the solvers would have assumed confidentially in their messages, and would assume their PM would be posted anyhow.

    If anyone doesn't want me to do this, please let me know. I know Matt is away and I'll have to deal with his ire (if any) on his return. However, Matt's submission is very interesting and I really want to include it the solution post.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  5. #20
    Monster of the deep Kevin Bonham's Avatar
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    Quote Originally Posted by Barry Cox
    I'm also thinking of quoting the solvers' PMs directly in the solution post. I trust this is not a breach of protocol as I don't believe the solvers would have assumed confidentially in their messages, and would assume their PM would be posted anyhow.
    I certainly assumed this - it's not explicit from your previous posts but strongly suggested. Anyway Matt doesn't believe a PM is confidential anyway (past experience) so his consent may safely be taken for granted.

    However, Matt's submission is very interesting and I really want to include it the solution post.
    That's ... disturbing.

    I am worried that he may have got it right. :shock:

  6. #21
    CC International Master Cat's Avatar
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    final solution

    I've just sent my solution BJC

  7. #22
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by Kevin Bonham
    I certainly assumed this - it's not explicit from your previous posts but strongly suggested. Anyway Matt doesn't believe a PM is confidential anyway (past experience) so his consent may safely be taken for granted.
    Itwas clear from your post that this was the case. It is less so from Matt's message.

    Quote Originally Posted by Kevin Bonham
    However, Matt's submission is very interesting and I really want to include it the solution post.
    That's ... disturbing.

    I am worried that he may have got it right. :shock:
    I won't say too much more prior to the deadline as I don't want to confer any information about the possible solution. But I stand by my original posting, Matt's solution in indeed noteworthy.

    22:30 to go!
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  8. #23
    Reader in Slood Dynamics Rincewind's Avatar
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    Re: final solution

    Quote Originally Posted by David_Richards
    I've just sent my solution BJC
    Yep, message received. Three solvers registered so far. But it is not too late for someone elseto have a crack at it.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  9. #24
    CC Grandmaster arosar's Avatar
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    This is too painful Bazza. I never paid this thread much attention - but now I'm gettin' all a bit intrigued!

    AR

  10. #25
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by arosar
    This is too painful Bazza. I never paid this thread much attention - but now I'm gettin' all a bit intrigued!
    Why not have a go? 7:30 to go - it's not too late.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  11. #26
    Reader in Slood Dynamics Rincewind's Avatar
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    OK. It is the hour of reckoning. For my worked solution look at the following link. If you have any questions let me know here or by PM, whatever is your preference.

    Worked Solution

    The final scores were

    Matt: 6/8
    Kevin: 7/8
    David: 6/8

    Interestingly each person came up with a new approach on the problem all of which were basically correct. However, Matt and David went astray, I think by applying too much intuition. Perhaps, that's unfair in David's case as it may have just been a clerical error.

    Also it was noted that it was the exclusion of the 20/0 distribution which caused the probability to not be 0.5 This is a very interesting result and probably answers Matt's original question way back when.

    The other interesting result is that when distributing an even number of things between two possibilities (say 20) the 12-8 distribution is more likely than the 10-10 distribution. This would come as no surprise to Bridge players but is an interesting general rule. With an odd-number (say 19) then 10-9 is more likely than any other single possible distribution.

    Anyway, pretty good results all round. I'll now post each submission in separate emails with some short comments.

    Remember, judges decisions are final yada yada yada, except as I said above I am willing to enter into correspondence.

    If anyone is interested I can come up with some other maths problems, not necessarily related to probability. If you're interested let me know.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  12. #27
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by Matthew Sweeney
    Ummm.....
    The bag of twenty is not randomly distributed beacause any 20/20 sames are missing.

    One white is drawn
    The remaining 19 distribution can be composed as :
    1 to 19 black ( there must be at least one black )
    0 to 18 white ( the first drawn white may have been the only white )

    The imposible composition is; 0 black to 19 whiute.

    In the case of normal B/W distribution bags of 20 marbles, the chance of having all white is given by
    =2^n
    =2^20
    =1048576
    => 1/1048576 = 0.000000090 = P(all white) = P(all black)

    However, the chance of this particular event occuring has been eliminated from the set of all posible outcomes.

    Also, for normal B/W distribution bags of 20 marbles, every draw has a P=0.5 for both balck and white.

    Therefore, we can add the chance of the Impossible composition (0 blcak) to the normal P(black)=0.5

    Thus, for the second marble after the first white draw
    P(black) = 0.50000000 + 0.00000090 = 0.50000090
    P(white) = 0.50000000 - 0.00000090 = 0.49999910

    Gee, I hope this makes sence
    It sort of does make sense. In fact I was quite surprised as to the simplicity and insightfulness of this line of reasoning.

    Unfortunately, Matt forgot one thing (the fact that removing the 2 possible distributions changed the basis of all the probabilities

    from 2^-20 to 1/(2^20 - 2), but that it all.

    However, this is the problem of relying too much on intuition. It can lead you to an original solution but the lack of rigour can cause you to make mistakes too. Better to hve the inspiration and then the technique to prove the insight correct.

    Regarding the scoring.

    (a) 2 points

    (b) 4 points - the insightfulness was remarkable. Unfortunately, the final answer was slightly off and there was a calculation error in the decimal side.

    1/1048576 is approximately 0.00000095 so the answer should have been 0.49999905

    Total Score: 6/8
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  13. #28
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by Kevin Bonham
    Hi Barry,

    (a) Black
    (b) (2^18 )-1/(2^19)-1, ie 262143/524287, ie 0.49999990 to 8 dps.

    I did this both the long way and the short way, but I did the long way first, then noticed the short way afterwards.

    Short way: We know that the probability of each marble in the bag being each colour is 0.5. So to start with there are (2^20) equally likely distributions of the marbles in the order that we are going to pull them out of the bag, except that 2 of these distributions are banned. Ignoring the colour of the first marble, there are 2x(2^18-1) legal distributions in which the first two marbles are the same colour, out of 2^20 distributions that are possible (again treating redraws as not redrawn). Then cancel 2s, and multiply ((2^18 )-1)/(2^19) by (2^19)/((2^19)-1) to deal with redraws (for full details see second para of long solution below) and you have the answer.

    This method leads to the general solution. Note: What this tells you is that if the chance of each marble being black or white is known to be 0.5 anyway, then drawing marbles out of the bag tells you nothing new about the distribution of the remainder - it is only the banning of monochrome results that has any impact on that.

    Long way, because it's the first way I did it (a bit like von Neumann summing the series to solve the fly and train problem) :

    As with previous solution first marble being white tells us nothing about what colour (if any) is the majority. Further the colour of the first marble tells us nothing about which of the many possible distributions of Maj-Min (19-1 down to 10-10 as 20-0 is banned) we have.

    Out of all the possible distributions of marbles in the random draw there is a 2x1/(2^20) = 1/(2^19) chance that the draw will be redone and hence a ((2^19)-1)/(2^19) chance that it won't. If it is redone the redrawing has no effect on the probabilities of other outcomes relative to each other - so find these probabilities assuming the draw is not redone and simply multiply that by (2^19)/((2^19)-1) to get the correct probabilities of eventually getting each permitted outcome.

    Let z = number of marbles in the Maj group, then probability of there being z marbles (ignoring redraws) is 2x20Cz/(2^20) for z=11 to 19 (the 2 is because either colour could have z number of marbles) and simply 20C10/(2^20) for z=10. (As a side note, the bag is more likely to have an 11-9 or 12-8 split than a 10-10).

    (Note for uninitiates in case this is posted on the BB: nCr = number of possible combinations of r successes from a sample of n success-failure events = n!/r!(n-r)! )

    (Note for even more uninitiated for same reason - n! = 1x2x3x4....xn)

    For each value of z between 10 and 19, the probability of the first two marbles being the same is (z/20)x((z-1)/19) + ((20-z)/20)x((19-z)/19)

    (ie Maj, Maj + Min, Min, except that for z=10 both Maj and Min groups consist of 10 marbles).

    This probability simplifies to (2(z^2)-40z+380)/380, ie (((x-10)^2)+90)/190 which is, incidentally, 90, 91, 94, 99, 106 ... 171 all/190 for z=10 to 19 in order.

    So for each value of z we can now obtain both the probability of their being z marbles in the largest group, and the probability of the first two marbles having the same colour if there are z marbles in the largest group. Multiplying these together and cancelling out I get:

    z=10, 21879/(2^18 )
    z=11, 20111/(2^17)
    z=12, 31161/(2^18 )
    z=13, 5049/(2^16)
    z=14, 2703/(2^16)
    z=15, 1173/(2^16)
    z=16, 3213/(2^19)
    z=17, 417/(2^18 )
    z=18, 77/(2^18 )
    z=19, 9/(2^18 )

    Mutliplying out and adding up the total is 262143/(2^19) - but this is still assuming no redraw.

    To cater for redraws multiply by (2^19)/((2^19)-1), cancel (2^19)s to get

    262143/524287.

    Interestingly had redraws been allowed, the probability of a white marble on the second draw becomes 262144/524288 = 0.5, ie the first marble picked being white would have had no effect on the likely colour of the second marble. The stipulation that the marbles are of different colours changes this.

    Let me know how I went or how badly I explained myself.
    Almost full marks! I like the long way as this was basically my starting point too. The short version is good too. Unfortunately with this solution there was again an error with decimal side of the calculation (probably just a transcription error).

    262143/524287 is approximately 0.49999905 (8 dpl)

    I assume you got carried away with the 9's and included one too many. Anyway, just subtracting a point for that the score is...

    (a) 2 points
    (b) 5 points

    Total score: 7/8 - and our winner of round 3.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  14. #29
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by David_Richards
    ok, this time!

    1 white ball has gone leaving 19 balls in total.

    Using the formula C=n!/r!(n-r)!, where n is the total no. of balls in left in the bag (19), and r is the number of white balls, Dr Evil has produced bags in the following combinations (C)

    NO. WHITE BALLS IN BAG(a), NO. BALLS(b),NO. WHITE P=axb


    18/19, 19, 18
    17/19, 171, 153
    16/19, 969, 816
    15/19, 3876, 3060
    14/19, 11628, 8568
    13/19, 27132, 18564
    12/19, 50388, 31824
    11/19, 75582, 43758
    10/19, 92378, 48620
    9/19, 92378, 43758
    8/19, 75582, 31824
    7/19, 50388, 18564
    6/19, 27132, 8568
    5/19, 11628, 3060
    4/19, 3876, 816
    3/19, 969, 153
    2/19, 171, 18
    1/19, 19, 1

    --------------------- ---------------------

    Total 524286(all balls) 262143(white)

    But there is 1 bag remaining with 19 black balls so the total combination is
    524286+19=524305


    so the answer is

    a) black

    b) 262143/524305 = 0.49998188

    A valiant effort working from second principles but not quite right. Your table is almost complete I would have added the following line

    0/19, 1, 0

    Which would have accounted for the all black possibility and lead to the result 262143/524287.

    However, everything else (including your calculation of the answer to 8 decimal places) was accurate so I'm scoring your answer on a par with Matt's - on the right track but not quite right. Possibly, lack of rigour in some of the reasoning leading you astray - possibly just an oversight. Score...

    (a) 2 points
    (b) 4 points

    Result 6/8
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  15. #30
    Monster of the deep Kevin Bonham's Avatar
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    Quote Originally Posted by Barry Cox
    262143/524287 is approximately 0.49999905 (8 dpl)

    I assume you got carried away with the 9's and included one too many.


    I was using a calculator which couldn't hold all the decimal places, so I had to subtract the .499999 off to get the rest, and must have simply written one two many 9s down then not checked it. Dumb.

    Careless errors like that were always my weakness but at least I had the method(s) right. :?

    Feel free to post more, except I'm allergic to trig, geometry and calculus.

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