Percentages, discrete or continuous?

Question

I have this question I can't figure out. A basketball player starts a game. During the first period of the game, the success rate of his shots is less than $80$%. At the end of the game, his success rate is over $80$%. Is it certain, that at some point during the game, his success rate was exactly $80$% ? My intuition say not necessarily, because the number of shots is discrete and not continuous, however, I tried several examples, and in all I got $80$%, so the answer could be yes. In this case, I would love to get an explanation to why this is happening. Thank you.

Answer 1

Because 80% is a fraction of the form $\frac{N-1}{N}$, it is necessarily true that 80% is at some point achieved.

You can check what happens modulo $N$.

After $k$ sequences of $N$ throws, we must lag behind at least 1.

Even if the next $N$ throws are a score, we still end up at 80%.

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I'm not sure if it is entirely clear, so I'll try to explain a bit differently:

Once you lag at least 1 score, you need at least $N$ successive scores to catch up by 1.

One of these throws is number $k N$.

Either we caught up here at exactly (not crossing) 80%, or we still lag 1.

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So basically, it is the same argument as tossing a coin, where the first is heads, and you need to end up with more than 50% tails. You will always have 50% tails at some point.

Answer 2

In math: $$\frac{k}{n} < \frac{4}{5} < \frac{k+1}{n+1}$$ means $$ 5k < 4n $$ and $$ 5k + 5 > 4n + 4$$ so $$ 5k < 4n $$ and $$ 5k + 1 > 4n$$ which for integer numbers is a contradiction.

Answer 3

You're correct that a continuity argument is not enough by itself; as a silly example, if 80% were replaced by an irrational percentage, it's easy to see you're never at an irrational percentage. Even among rational percentages, you can find examples where you start below 60% and end up above and are never exactly at 60%. So if the result is true there's something special about 80% that makes it true.

In any case, the result is true. You can try visualizing it as follows: after $n$ shots, plot the point $(s_n, f_n)$, where $s_n$ is the number of successful shots and $f_n$ is the number of failed shots. This point only ever moves one unit to the right or one unit up. You're given that it starts below the line $y = 4x$ and ends above it, and want to know whether that means at some point it is exactly on the line. If you plot out what's happening on a sheet of graph paper, the answer is clearly yes: because the slope of the line is a positive integer greater than $1$, you can never "jump over" the line, since you can only go one unit to the right or up. More formally, no line segment of the form $(x, y) \to (x+1, y)$ or $(x, y) \to (x, y+1)$ passes between the region below the line and the region above the line.

This argument, made rigorous, shows more generally that $\frac{4}{5}$ can be replaced by $\frac{N}{N+1}$ for any positive integer $N$. These are the only ratios for which a result like this holds.