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Topic: Free Throws (Read 446 times) |
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ThudnBlunder
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Free Throws
« on: Dec 7th, 2004, 12:42am » |
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Shaq's coach has been keeping a record of the number S(N) of successful free throws he has made in his first N attempts of the season. Earlier in the season S(N) was less than 80% of N, but later in the season S(N) was more than 80% of N. Was there necessarily a time in between when S(N) was exactly 80% of N?
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| « Last Edit: Dec 7th, 2004, 12:45am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Icarus
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Re: Free Throws
« Reply #1 on: Dec 7th, 2004, 12:40pm » |
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Yes: Let S, N be the values immediately before reaching or passing the 80% mark. Then 4N - 1 <= 5S < 4N. Since 5S is an integer, it must be the only integer in that range. I.e. 5S = 4N-1. So S+1 = (1/5)(4N-1) + 1 = 1/5 (4N-1+5) = 4/5 (N+1).
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Foolish
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Re: Free Throws
« Reply #2 on: Dec 7th, 2004, 4:46pm » |
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Why does this work for 80% and not for all values?
35% for example 3/9 -> 4/10 skips over that value.
40% however, it seems to hold for. Is it the fact that .35 isn't made of prime/prime?
There must be something underlying that I am missing.
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ThudnBlunder
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Re: Free Throws
« Reply #3 on: Dec 7th, 2004, 9:22pm » |
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on Dec 7th, 2004, 4:46pm, Foolish wrote:| Why does this work for 80% and not for all values? |
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It works for all percentages of the form 1 - 1/k, where k is integer.
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| « Last Edit: Dec 7th, 2004, 9:26pm by ThudnBlunder » |
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Foolish
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Re: Free Throws
« Reply #4 on: Dec 7th, 2004, 9:35pm » |
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on Dec 7th, 2004, 9:22pm, THUDandBLUNDER wrote:
It works for all percentages of the form 1 - 1/k, where k is integer.
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Ah, that makes sense. Thanks :)
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Barukh
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Re: Free Throws
« Reply #5 on: Dec 8th, 2004, 2:08am » |
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on Dec 7th, 2004, 9:22pm, THUDandBLUNDER wrote:
It works for all percentages of the form 1 - 1/k, where k is integer. |
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... and I think it has something to do with continued fractions.
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