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Michael Jordan is shooting free throws. He misses his first one. At the end of the day, he has made $99$% of his free throws.

At some point during the day, did he necessarily have a $50$% success rate?

My Attempt

Since he started at $0$% and made it to $99$%, he clearly had to pass $50$% at some point. Let's look at the shot right before.

If he's currently at an even number of shots, his proportion is $\frac{n-1}{2n}$.

  • If he misses his next shot, we don't care because he's more than one shot from potentially getting to $50$%.
  • If he makes his next shot, his proportion is $\frac{n}{2n+1}$. This brings us to the odd case.

If he's currently at an odd number of shots, his proportion is $\frac{n}{2n+1}$.

  • If he misses his next shot, we don't care because he's more than one shot from potentially getting to $50$%.
  • If he makes his next shot, his proportion is $\frac{n+1}{2n+2} = \frac{1}{2}$.

Therefore, he must have been equal to a proportion of $\frac{1}{2}$ at some point.

I don't think my solution is correct. It seems odd that I'm able to say he's one shot away at all times, otherwise we just don't care. I need help fixing my argument, because I'm fairly sure my answer is right, just my proof is wrong.

user01101001
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    Your argument looks good to me. In general, you can't "jump over" any proportion of the form $\frac k{k+1}$ Thus, Michael must have hit $\frac 23, ; \frac 34,\cdots$. – lulu Apr 12 '16 at 20:08
  • If $\frac{k}{n}<\frac{1}{2}<\frac{k+1}{n+1}$, then $2k<n$ and $2k+2>n+1$ so $n-1<2k<n$. Contradiction. – almagest Apr 12 '16 at 20:10
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    More likely Micheal Williams than Michael Jordan, but admittedly, not many people know about him. :-) – Brian Tung Apr 12 '16 at 20:23

3 Answers3

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Seems good to me; you can then forget about "missing the shot", as you say, because we are assuming that he makes that shot (and you are sure of the existence of such a shot)

just a couple of points:

First, why if the number of shot is even then the numerator must be $n-1$. The reason is that if it isn't, then that shot cannot be "the last one below 50%", but I'd say it more clearly.

Another point is that in the even case, the right conclusion is not "this brings us to the odd case" which doesn't really say anything, but you can say that the proportion is still below 50% contradicting the fact that that shot was the last one below 50%; hence the last shot below 50% has to be taken after an odd number of shots, and in that case it brings you to exactly 50%

Ant
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I think your proof makes sense, but I would try to simplify a bit.

We know that at some point Micheal passed the 50% mark by making a shot. Let $n$ be the number of missed shots he had before making the shot that got him over 50%. How many shots had he made at that point? Obviously, he had made $n$ shots. (If it were less than n, the next made shot wouldn't get him over 50%, if it were more than n, then he's already over 50%.) Since at that point he had both made and missed $n$ shots, his percentage was exactly 50%.

browngreen
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Well, I think this come down to is can the following have two integer solutions: $\frac {k}{n} < 1/2 < \frac{k+1}{n+1}$. ($k$ being the number of made shots and $n$ the number of total shots immediately before the rate was 50%.$

i.e. $k(n+1) = k(n+1) < \frac{n(n+1)}2 < n(k+1)$ so $k< n/2$ while $k + l > (n+1)/2$.

So $n/2 - 1/2 < k < n/2$ Or $n - 1< 2k < n$ which is impossible by archemedian principal.

Which I guess, is more or less, the same argument...

fleablood
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