# Basketball throws: a rather intriguing puzzle

I had run into this problem long time back. I ran into it again last week. Thought will post it. See if you can solve it without using Google to find the answer. If not, then Google it – pretty interesting, huh? (I will post the answer later).

A basketball player keeps track of his throws for a full calendar year. He, of course, misses a few shots and succeeds with a few more. He missed the very first shot of the year. But he ended the year with 83% successful shots. You have to prove that there had to be a point where is success rate was exactly 75%.

Hint: this is not necessarily true for any number – e.g. 60%. But it is always true for 75%.

In fact, can you guess what are the other % (other than 75%) for which this is also true?

Somshekhar BaksionOne answer to the second part is 0%, but thatâ€™s trivial. The other, I think, is 50%, because there is no way of going from a 50% without landing on 50.

Rajib RoyonSomshekhar, indeed. In fact you will always have to go thru the fraction (m) / (m+1), m being an integer

Somshekhar BaksionRajib Roy yes that seems right.

Somshekhar BaksionRajib Roy so 75% is just a specific example of a more general rule.

Rajib RoyonSomshekhar, I was wondering what is special about 75%. Then as I looked at the calculations, it became clear that m/(m+1) will work (for any m). This morning, I was able to write down the proof for that too. So, I am sanguine now.

Somshekhar BaksionRajib Roy yes, and I think it also means you run into a larger number of these ratios as you inch your way up into the 90s.

Rajib RoyonOf course!

Harry GoodnightonRajib Roy – you can take up a new career as a number theorist!

Sri GaneshonI don’t know the solution to the problem, but I know that NBA India Games 2019 will be held on Friday, October 4, and Saturday, October 5 2019. Yes, NBA is coming to India. đź™‚

Dhananjay NeneonIf he is one down and then eventually has to go up, then if he gets the next shot in, his success ratio will be 1/2, one more and it will be 2/3, yet another it will be 3/4 (which is the 75% you are talking about), one more and it will be 4/5

Each of these numbers will always figure however one chooses to arrange the hits and misses so long as the number of successes as a proportion of total hits is large enough to reach the number.

This is essentially a series of n/n+1 (and no I did not google it đź™‚ )

Rajib RoyonDanny, I am trying to follow the logic but am not able to. The n/n+1 is correct. But what I do not get is when you say that if he gets in the next one, he will be 1/2. He might miss the first thousand shots, right?

Dhananjay NeneonLet’s say we order all the shots as misses first, hits later.

Let’s say there are a total of m misses.

So early on his score is 0/m at the point in time all misses are over.

Now with hits his ratios with each successive hit is

1/m+1, 2/m+2, 3/m+3

If m = 1 this is *1/2*, *2/3*, *3/4*

If m=2 this is 1/3, *2/4*, 3/5, *4/6*, 5/7, *6/8*

If m=3 this is 1/4, 2/5, *3/6*, 4/7, 5/8, *6/9*, 7/10, 8/11,*9/12*

He always goes through the specific fractions of n/n+1 every m’th hit

Not sure if that helped

Dhananjay NeneonNote: ordering shots by misses first is just a simplifying assumption. The universe of fractional hit ratios he will go through will probably remain the same even if you were to reorder it as 1 miss, all hits, then remainder of all misses (haven’t verified it yet)

Rajib RoyonDhananjay, that is an interesting approach. I will think about this after office today.

Rajib RoyonDanny,

See if this logic works. The sequential way of putting hits or misses first will NOT work.

Take 10 throws. Starts with a miss and then finishes at 70%. Will he go thru 40%? The real sequence was â€śM H H H M H H M H H â€ś

If I sequence them the way you were thinking – M M M H H H H H H H – then it would appear you would hit 40% (at the end of 5th shot). 0%, 0%, 0%, 25%, 40%, 50% â€¦.

But in real life the percentages were 0%, 50%, 67%, 75%, 60%, 67%, 71%, 62.5%, 67%, 70%

Makes sense?

Dhananjay NeneonFair enough. Agree. The total universe of hit ratios is not stable.

But as you pointed out in the OP, there are some fractions .. represented by n/n+1 which are always there

Chandra M PendyalaonMonday Morning .. I will take a quick and dirty shot at it and see if I can think more end of day. The think I quickly notice is 83 is prime, and so final number of shots have to be a multiple of 100, but I needs 1s.. so let me try 3/4, 4/5 and 2/3 (since no fractions). I guess they cannot get to 17 fails without 75%,80% and 66.666%. Now I stop at 4/5th and not 6/7 th because 84. Very Quick Dirty intuition and a ton of guessing .. Let me think for rigor later in the eve

Chandra M PendyalaonHere is a more rigorous answer..

Chandra M PendyalaonSince I did my thinking while driving without pen and paper, the binomial distribution aproach left me reaching for a paper, but this numerical equivalent felt conclusive, if you start with 0/1, and keep going up taking a snapshot of score board every 4 shots, the list of all possible outcomes [0-2]/4, [1-5]/8 etc.,will get to [11-20]/28 and then since no more than 17 missed per hunderd shots – the range gets trimmed both ends like [15-23]/32 ..etc., till [43-47]/60 and if you repeat it from other end ie 83/100 back [48-59]/60 proving it immpossible without 3/4 crossing .. this numerical approach does not clue me into other possibilities though

Chandra M Pendyalaonlooking forward to the right answer..

Rajib RoyonSolution

To solve this problem, we will use what is called â€śreductio ad absurdumâ€ť. Meaning we will assume that that indeed the player never reached 75% and then show that that assumption leads to impossible events. Therefore the assumption must be wrong.

If the player never hit 75% but started with 0% and reached 83%, it must be true that he crossed the 75% mark (without touching it). Letâ€™s say with the â€śnâ€ťth throw he was just below 75%. With the n+1st throw he went to above 75%. Therefore the n+1st throw had to be a successful one.

Letâ€™s say he had X successful baskets in that first n throw. So, X/n 3/4

From the first one,

X/n 4X < 3n (i)

From the second one,

3/4 3(n+1) 3n+3 3n < 4X+1 (ii)

Now look at (i) and (ii).

3n is an integer (since n is an integer)

4X is an integer (since X in an integer)

and 4X and 4x+1 are consecutive integers

So if you combine (i) and (ii), it say

an integer (3n) is greater than another integer (4X) but less than its next integer (4X+1)

which is an impossibility.

Ergoâ€¦

Instead of 75% (3/4), this can be done for all fractions s/(s+1)

Basically, you will have

(s+1)X < sn

and

s(n+1) < (s+1)(X+1)

which would imply that the integer sn lies between the integers (s+1)X and (s+1)X + 1

Hope you enjoyed it

Rajib RoyonSomshekhar, Thanks for the link. I just saw that. I was thinking they will extend that proof to show how it will work for all m/(m+1). BTW, have you ever read any of those books? I might get one just to see if the old brain can still do anything – getting mightily sluggish, I must say.

Somshekhar BaksionRajib Roy your brainâ€™s sluggish speed would be mine on, um, speed. So – while I can rarely say no to books – in this case the stuff you share on FB is enough to keep me adequately challenged.

Somshekhar BaksionBTW they have generalised it later in the narration.

Rajib RoyonSomshekhar, They did? I missed it completely.

rajibroy(Post author)onSolution

To solve this problem, we will use what is called â€śreductio ad absurdumâ€ť. Meaning we will assume that that indeed the player never reached 75% and then show that that assumption leads to impossible events. Therefore the assumption must be wrong.

If the player never hit 75% but started with 0% and reached 83%, it must be true that he crossed the 75% mark (without touching it). Letâ€™s say with the â€śnâ€ťth throw he was just below 75%. With the n+1st throw he went to above 75%. Therefore the n+1st throw had to be a successful one.

Letâ€™s say he had X successful baskets in that first n throw. So, X/n < 3/4 After the n+1st throw - where we know the last had to be a successful one - we have (X+1)/(n+1) > 3/4

From the first one,

X/n < 3/4 => 4X < 3n (i) From the second one, 3/4< (X+1)/(n+1) => 3(n+1) < 4(X+1) => 3n+3 < 4X+4 => 3n < 4X+1 (ii) Now look at (i) and (ii). 3n is an integer (since n is an integer) 4X is an integer (since X in an integer) and 4X and 4x+1 are consecutive integers So if you combine (i) and (ii), it say an integer (3n) is greater than another integer (4X) but less than its next integer (4X+1) which is an impossibility. Ergoâ€¦ Instead of 75% (3/4), this can be done for all fractions s/(s+1) Basically, you will have (s+1)X < sn and s(n+1) < (s+1)(X+1) which would imply that the integer sn lies between the integers (s+1)X and (s+1)X + 1 Hope you enjoyed it