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Little explorations with HP calculators (no Prime)
12-27-2018, 08:31 PM
Post: #321
RE: Little explorations with HP calculators (no Prime)
(12-27-2018 06:52 PM)pier4r Wrote:  David I am not sure I understood your test. I am not sure I decoded the data properly.

That's fair, because I'm not clear on exactly what it is I'm supposed to be looking for with the original problem. Smile I was simply experimenting to determine some of the possible outcomes for a series of randomized lists.

(12-27-2018 06:52 PM)pier4r Wrote:  For example:

1    60726
2    60743
3    60133
4    59444
5    58566
6    57093
7    55206

It means that out of 1 million attempts, 60726 of them had at least one list of 70 elements that matches the requirement? 60743 times there were two matching lists?

Almost. I would change your description slightly to make it "out of 1 million randomly-ordered lists of 140 balls, 60726 of them had exactly one sublist of 70 elements that matches the requirement, 60743 times there were exactly two matching sublists..."

In the original problem description, you write:
pier4r Wrote:Could they find a way to ensure that Anna and Berta pick both 50 blue balls and 20 white balls? If yes, is a special arrangement of the balls needed to achieve it?

I believe the answer to the first question is simply "yes", as has been pointed out by ijabbott. There will always be at least one contiguous sublist of 70 balls containing the proper mix of 50 blue and 20 white balls. I also believe the answer to the second question is "no", because it does not appear that any special arrangement matters.

I'm not mathematically skilled enough to offer a rigorous proof of those assertions, so I should probably just call them "opinions". I am , however, mathematically curious enough to experiment with the results of repeated trials to see what patterns emerge. Hence my testing to see the results of counting the number of matches in random lists.

It's been a long, long time since I completed my last statistics course, but it appears to me that the distribution of match counts fits the right half of a normal curve (ie. the peak is at/near the lowest value in this case). I'm sure that has some significance, but I'm still not clear on exactly what that tells us. The shape of the curve seems to be more than a coincidence, though.
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RE: Little explorations with HP calculators (no Prime) - DavidM - 12-27-2018 08:31 PM

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