Puzzle time - penalty kicks

[jon-nyc]

@axtremus Yep

[Axtremus]

:::

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%.

:::

[jon-nyc]

@axtremus You forgot to increment the denominator. Your example ends at 80%, not above.

[Klaus]

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.

[Klaus]

:::

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
:::

[Axtremus]

:::

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.

:::

[jon-nyc]

Lets do some arithmetic. We're looking for an 'a' and 'b' such that:

a/b<4/5
(a+1)/(b+1)>4/5

The first equation simplifies to:
5a<4b

The 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?

[taiwan_girl]

@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

[jon-nyc]

Yes you will, but what other percentages are unavoidable if you start below them and end above them?

[Axtremus]

:::

Let's see ...

From:
a/b < A/B AND (a+1)/(b+1) > A/B

We 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.

:::

[jon-nyc]

Ax got it!

[jon-nyc]

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.

[Axtremus]

FWWI, @Klaus might be happy to know that this puzzle got me to install Haskell.

[Klaus]

@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)

[jon-nyc]

I’m a little sad nobody commented on “Diego Primadona”.

[Axtremus]

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.