Puzzle time - penalty kicks
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wrote on 10 Jan 2021, 01:38 last edited by
At the start of the last year, soccer star Diego Primadona’s lifetime penalty kick percentage was less than 80%. By the end of the year it was over 80%.
Must there have been a time when it was exactly 80%?
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wrote on 10 Jan 2021, 02:44 last edited by
I think yes, but haven’t really thought about the math.
Just ran some numbers in my head.
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wrote on 10 Jan 2021, 04:14 last edited by
@jon-nyc said in Puzzle time - penalty kicks:
penalty kick percentage
How is "penalty kick percentage" defined?
The # of penalty kicks that result in a goal divided by # of penalty kicks attempted, is that right? -
@jon-nyc said in Puzzle time - penalty kicks:
penalty kick percentage
How is "penalty kick percentage" defined?
The # of penalty kicks that result in a goal divided by # of penalty kicks attempted, is that right? -
wrote on 10 Jan 2021, 05:28 last edited by
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Proof by counter example:
Start of year the footballer’s penalty kick percentage was 79/99*100%, so just a bit under 80%.
In that year, the footballer attempted only one penalty kick and that penalty kick scored a goal, so at the end of the year his pernalty kick percentage became 80/99*100%, a bit over 80%.
The footballer went from just a bit below 80% to just about over 80% without ever dwelling at exactly 80%. So there need not be a time when it was at exactly 80%.
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Proof by counter example:
Start of year the footballer’s penalty kick percentage was 79/99*100%, so just a bit under 80%.
In that year, the footballer attempted only one penalty kick and that penalty kick scored a goal, so at the end of the year his pernalty kick percentage became 80/99*100%, a bit over 80%.
The footballer went from just a bit below 80% to just about over 80% without ever dwelling at exactly 80%. So there need not be a time when it was at exactly 80%.
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wrote on 10 Jan 2021, 15:56 last edited by
Hm, this is trickier than it looks.
On first sight, it seems to be obvious that the percentage is a discontinuous function and can hence jump over .8.
But now I think it may actually be impossible. Let me think a little whether and why that's actually tre.
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wrote on 10 Jan 2021, 16:06 last edited by Klaus 1 Oct 2021, 17:21
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Among the pairs of numbers between 0 and 5000, there is no pair (x,y) such
that x/y < 0.8 and (x+1)/(y+1) > 0.8.
An engineer would stop here and consider the case closed. No soccer player ever scored more than 5000 penalty kicks. QED
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wrote on 10 Jan 2021, 18:55 last edited by
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There is something special about 90%, 80% and 50% ... if the quotient is 79% or 81%, or 49% or 51%, or 60% or 70% or 40%, etc., there are many pairs of integers that satisfy (x/y) < quotient and (x+1)/(y+2) > quotient. :man-shrugging:
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wrote on 10 Jan 2021, 20:35 last edited by jon-nyc 1 Oct 2021, 20:43
Lets do some arithmetic. We're looking for an 'a' and 'b' such that:
a/b<4/5
(a+1)/(b+1)>4/5The first equation simplifies to:
5a<4bThe second simplifies to:
5a+1>4b.So 5a+1 > 4b > 5a.
But that can't be true for positive integers a and b. So the answer is you can't skip 80%.
Ax notices some other numbers you can't skip. He noticed that 90%, 50% have the same property.
Can you generalize the property?
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Lets do some arithmetic. We're looking for an 'a' and 'b' such that:
a/b<4/5
(a+1)/(b+1)>4/5The first equation simplifies to:
5a<4bThe second simplifies to:
5a+1>4b.So 5a+1 > 4b > 5a.
But that can't be true for positive integers a and b. So the answer is you can't skip 80%.
Ax notices some other numbers you can't skip. He noticed that 90%, 50% have the same property.
Can you generalize the property?
wrote on 11 Jan 2021, 02:19 last edited by@jon-nyc said in Puzzle time - penalty kicks:
Can you generalize the property?
Yes, if you divide two integers (with the numerator smaller than the denominator) and you continually increase both by 1, you will eventually have the answer equal 0.5, 0.8, and 0.9!!
5555
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wrote on 11 Jan 2021, 03:04 last edited by
Yes you will, but what other percentages are unavoidable if you start below them and end above them?
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wrote on 11 Jan 2021, 03:09 last edited by
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Let's see ...
From:
a/b < A/B AND (a+1)/(b+1) > A/BWe get to Ba + (B-A) > Ab > B*a
If we are talking only natural numbers, the condition boils down to (B-A) = 1.
So fractions like 1/2 (50%), 2/3 (66.666...%), 3/4 (75%), 4/5 (80%), ... 9/10 (90%) ... all fractions that can be written as X/(X+1) are "special" that way.
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wrote on 11 Jan 2021, 03:10 last edited by
Ax got it!
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wrote on 11 Jan 2021, 03:12 last edited by
When I first saw this problem last week I thought it was stupid and the answer was obviously “of course you can skip 80”. I didn’t even sit down to play with it for a few days.
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wrote on 11 Jan 2021, 08:54 last edited by Klaus 1 Nov 2021, 12:03
@axtremus said in Puzzle time - penalty kicks:
FWWI, @Klaus might be happy to know that this puzzle got me to install Haskell.
There are infinitely many good reasons to install Haskell, but if all you want is list comprehensions, then you can do the same thing in many other languages, such as Python.
In Haskell you can of course write the program in a cooler and more general way using monads.
import Control.Monad [1..500] >>= \x -> [1..500] >>= \y -> guard (x/y < 0.8 && (x+1)/(y+1) > 0.8) >> return (x,y) -
wrote on 11 Jan 2021, 11:35 last edited by
I’m a little sad nobody commented on “Diego Primadona”.
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wrote on 11 Jan 2021, 14:21 last edited by
I have read @Klaus mentioning Haskell a few times in the past, and has been meaning to check it out. This puzzle is just the thing that finally got me to do. These days my primary computer programming language seems to be “go” (or “golang”), mostly because I found it quite convenient for dealing with web APIs.