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  • Monique
    replied
    R249



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  • Monique
    replied
    R248



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  • Monique
    replied
    R250 solution ...



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  • davis_greatest
    replied
    Originally Posted by snookersfun
    ok, mine eventually was this here:
    701000120
    160011002
    026110010
    202510010
    011351000
    011035001
    100000433
    000103241
    010101404

    We have to use coloured balls adding up to 99 points, meaning that those can be arranged in a minimum of 9 lines X 11 points each. Thus the smallest possible square is one of 9x9=81 balls.
    As we have 43 value balls, we need 38 white 'fillers', therefore we are in for 201-38=163 bananas...
    ... and I don't have my 163-banana offering any more, but I think the arrangement was pretty similar to that one!

    If you keep the balls as a "squashed" rectangle, with all the rows and columns aligned, then the 163 bananas is the best possible - but as abextra showed very nicely, more bananas are possible with more creative spaced out arrangements!

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  • davis_greatest
    replied
    While those are coming up, a quick one for those who dislike a bit of probability.

    Round 254

    In the Premier Ape League, matches consist of 6 frames.

    It doesn’t matter who wins – all that matters is the score. This means that there are four possible final scores: 6-0, 5-1, 4-2 or 3-3.

    In each match, each player in a pair has a fixed chance of winning any given frame, and each frame is independent of all those that come before.

    (a) When Rocketsalad plays Dragonfire, they are evenly matched (50% chance each of winning each frame). What is the most likely final score and how likely is it?

    (b) (harder) When Nuggins plays Windywhirl, Nuggins’s superior skill makes a 4-2 final score (which could be to either ape) exactly 3 times as likely as a draw. What is the chance that they will share the first two frames?

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  • davis_greatest
    replied
    Originally Posted by abextra

    I still hope Barry likes this arrangement... and don't forget how much Gordon likes bananas!
    Yes, nice picture! Very pretty!

    We have had the maximum possible bids - well done! ....so I'll invite all the answers for rounds 248 to 253 that have been sent to me to be posted on the thread...

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  • Monique
    replied
    Oh yes ... 281

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  • snookersfun
    replied
    Originally Posted by abextra
    Hehehe... how did I know it's coming! Ok, I bid 281... (sent out the explanation, hope there are no mistakes).

    Welcome back, d_g!
    lol, yes that 280 was too easy..., so 281 as well!

    and welcome back

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  • abextra
    replied
    Originally Posted by davis_greatest
    ... 280 is a good bid, but (slightly) higher is possible!
    Hehehe... how did I know it's coming! Ok, I bid 281... (sent out the explanation, hope there are no mistakes ).

    Welcome back, d_g!

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  • davis_greatest
    replied
    Thanks for all the answers so far to rounds 248 to 253... I will get to them all, but unfortunately can't tonight! (Sorry! )

    I think there have been plenty of correct answers (mainly from abextra, Monique and snookersfun!) to rounds 248, 249, 250, 251 and 252. So, well done! (I shall list them all.)

    Also some high bids to round 253! It looks like the highest bid submitted for round 253 is 280 (unless I've missed one), found by snookersfun, abextra and Monique! 280 is a good bid, but (slightly) higher is possible!

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  • snookersfun
    replied
    Originally Posted by Monique
    Finally got that 280 for Round 253, and 278 an 277 also ... well anything but 279
    see? Join the club
    Well, guess you could do a 279 (in a bit of silly version of the 280)...

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  • Monique
    replied
    Finally got that 280 for Round 253, and 278 an 277 also ... well anything but 279

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  • snookersfun
    replied
    Originally Posted by abextra
    I still hope Barry likes this arrangement... and don't forget how much Gordon likes bananas!
    wow, very creative...and rather squareish Another nice pic!

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  • abextra
    replied
    Originally Posted by snookersfun
    ...and while I am here, I shall put Monique's beautiful artwork to R.242 up:
    Well done, Monique, it's really beautiful!

    Here's my pic for Round 247 , 11 rows and 11 columns, the value of coloured balls adds up to 9 in each of them... ... you only have to count the balls in horizontals and verticals, not in diagonals.

    4 . 0 . 0 . 2 . 0 . 3
    . 3 . 1 . 0 . 0 . 5

    1 . 4 . 1 . 1 . 0 . 2
    . 1 . 6 . 1 . 1 . 0
    0 . 1 . 5 . 1 . 2 . 0
    . 0 . 1 . 7 . 1 . 0
    0 . 2 . 1 . 4 . 1 . 1
    . 0 . 1 . 1 . 6 . 1
    1 . 2 . 1 . 1 . 4 . 0
    . 5 . 0 . 0 . 1 . 3
    3 . 0 . 1 . 0 . 2 . 3


    I still hope Barry likes this arrangement... and don't forget how much Gordon likes bananas!
    Attached Files

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  • PaulTheSoave
    replied
    Parlez francais, seulement en PM s'íl vous plait!

    Originally Posted by Monique
    You are very welcome Paul! Plus on est de fous, plus on rit!
    I have not mentioned date or time! Could be a long wait. hehehehe

    Originally Posted by snookersfun
    archived now, so we are waiting

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