A basketball riddle

Anything goes, but keep it seemly...

A basketball riddle

Postby udosuk » Mon Aug 07, 2006 9:16 am

Hud will like this one...:D

Riddler wrote:Shaquillina O'Neal, daughter of a famous NBA star, joins the WNBA in the 2018 season... Unlike her dad, she is a very good free throw shooter, and after the first 5 games her free throw percentage stands at .900 (90%)...

To even improve the percentage, her dad decides to give her some "valuable advices" on the shooting movements... After that, her percentage drops dramatically, and stands at .650 (65%) after the first 25 games...

Determined to take matters into her own hand, Shaquillina refuses to listen to her dad anymore, and reverts back to her original movements... At the end of the regular season, her free throw percentage stands at .850 (85%)...

Here are 2 questions:

1. Must there be a particular instance from the 6th game to the 25th game that her free throw percentage stands at exactly .800 (80%)?

2. Must there be such an instance from the 26th game to the last game of the regular season?
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby MCC » Mon Aug 07, 2006 2:31 pm

As a wild guess, since I know nothing about basket ball, but going by the phrasing of the questions:D

udosuk wrote:1. Must there be a particular instance from the 6th game to the 25th game that her free throw percentage stands at exactly .800 (80%)?

No.


udosuk wrote:2. Must there be such an instance from the 26th game to the last game of the regular season?

Yes.

I'll leave the proof to others.


MCC
MCC
 
Posts: 1275
Joined: 08 June 2005

Re: A basketball riddle

Postby Cec » Mon Aug 07, 2006 2:33 pm

Riddler wrote:"...and after the first 5 games her free throw percentage stands at .900 (90%)...
"...her dad decides to give her some "valuable advices" on the shooting movements... After that, her percentage drops dramatically, and stands at .650 (65%) after the first 25 games..."

Firstly, my apologies for this joke udosuk but if Shaquillina is annoyed with her dad and gets caught swinging from the basketball net does she get "suspended" automatically.:)

Now for some serious thinking. In trying to work this out does the "first 25 games" include the "the first five games"?

Cec
Cec
 
Posts: 1039
Joined: 16 June 2005

Re: A basketball riddle

Postby udosuk » Mon Aug 07, 2006 3:27 pm

Cec wrote:Now for some serious thinking. In trying to work this out does the "first 25 games" include the "the first five games"?

Yes...

Sorry about the confusion... By "free throw percentage" I mean the total number of free throws made divided by the total number of free throws taken, cumulative from game 1 to the current time... So the .900 percentage is from game 1 to game 5, .650 is from game 1 to game 25, and .850 is from game 1 to the last game of the regular season, i.e. the whole regular season (where the highest percentage achiever would be awarded the "Free Throwing Queen" award:) )...

I don't know how much does the phrasing of my questions give away the answer but I'll refrain from commenting on MCC's answers until someone else gives a more elaborated attempt...
udosuk
 
Posts: 2698
Joined: 17 July 2005

Re: A basketball riddle

Postby Cec » Tue Aug 08, 2006 10:53 am

udosuk wrote:Q1. Must there be a particular instance from the 6th game to the 25th game that her free throw percentage stands at exactly .800 (80%)?

For Q1. For the first five games assume Shaquillina "nets" 45 of her 50 throws which would be a 90% success rate. For the next five games assume Shaquillina "nets" 35 of 50 which means for the first ten games she would have "netted" 80 of her 100 throws which gives her an 80% success rate. As this is a particular instance from the 6th to the 25th game meaning it occured at the end of game 10 then is this a correct answer for Q1?

Haven't studied Q2 properly yet.

Cec
Cec
 
Posts: 1039
Joined: 16 June 2005

Postby udosuk » Tue Aug 08, 2006 4:12 pm

Cec, you've shown that it is possible Shaquillina's FT% stands at .800 at one instance from the 6th to the 25th game. That's half the step you need to do for Q1. So congratulations.:)

What now you need to do is to show that if that is a must happening case or not, i.e. if Shaquillina must have such an instance from the 6th to the 25th game.

Basically, for both questions there are 3 scenarios:
(a) Impossible: Shaquillina cannot have any instance where her FT% stands at .800...
(b) Possible but not always: Shaquillina could have such an instance, but it's also possible she does not have one...
(c) Always/Must: No matter what Shaquilla must have an instance that her FT% stands at .800... (This is the hardest to prove...:!: )

So, you've successfully eliminated scenario (a) for Q1. Now you need to eliminate one of (b) or (c) for Q1 to get the correct answer.

Similarly, you have to eliminate 2 scenarios for Q2 (or prove that exactly one of them must be true)...

Carry on!:)
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby ronk » Wed Aug 09, 2006 1:22 am

udosuk wrote:Basically, for both questions there are 3 scenarios:
(a) Impossible: Shaquillina cannot have any instance where her FT% stands at .800...
(b) Possible but not always: Shaquillina could have such an instance, but it's also possible she does not have one...
(c) .............

The answer is (b). The following sequence transitions from 90.0% to 65.0% hitting 80.0% exactly along the way.
Code: Select all

 9 of 10 = 0.9000
 9 of 11 = 0.8182
10 of 12 = 0.8333
10 of 13 = 0.7692
11 of 14 = 0.7857
12 of 15 = 0.8000
12 of 16 = 0.7500
12 of 17 = 0.7059
13 of 18 = 0.7222
13 of 19 = 0.6842
13 of 20 = 0.6500


The following sequence makes the same transition "stepping over" the 80% number.
Code: Select all

 9 of 10 = 0.9000
 9 of 11 = 0.8182
 9 of 12 = 0.7500
10 of 13 = 0.7692
10 of 14 = 0.7143
11 of 15 = 0.7333
11 of 16 = 0.6875
12 of 17 = 0.7059
12 of 18 = 0.6667
13 of 19 = 0.6842
13 of 20 = 0.6500


To keep the examples simple, Shaquillina didn't get to shoot many free throws.:D
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby tarek » Wed Aug 09, 2006 3:06 am

Possible Scenario:
Code: Select all
 9 of 10 = 0.9000

12 of 15 = 0.8000
16 of 20 = 0.8000
20 of 25 = 0.8000
24 of 30 = 0.8000

26 of 40 = 0.6500

56 of 70 = 0.8000
60 of 75 = 0.8000

85 of 100= 0.8500


Now as you can see from the possibilities of .8 in the 1st 100 throws
Code: Select all
 4 of 5  = 0.8000
 8 of 10 = 0.8000
12 of 15 = 0.8000
16 of 20 = 0.8000
20 of 25 = 0.8000
24 of 30 = 0.8000
28 of 35 = 0.8000
32 of 40 = 0.8000
36 of 45 = 0.8000
40 of 50 = 0.8000
44 of 55 = 0.8000
48 of 60 = 0.8000
52 of 65 = 0.8000
56 of 70 = 0.8000
60 of 75 = 0.8000
64 of 80 = 0.8000
68 of 85 = 0.8000
72 of 90 = 0.8000
76 of 95 = 0.8000
80 of 100= 0.8000


The differences are increments of 1, to me that means that you must hit .8000 exactly as you climb from anything below .8000 to anything above .8000

tarek
User avatar
tarek
 
Posts: 2607
Joined: 05 January 2006

Postby udosuk » Wed Aug 09, 2006 7:15 am

Great analyses ronk & tarek!:)

Okay, so it's settled that Q1's answer is (b) and Q2's answer is (b) or (c). Tarek's explanation that (c) must be correct is not clear enough to me...:( Perhaps someone could post a more elegant reasoning?

PS: In a whole season it's not unusual for players to shoot 500-600 FTs...
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby ronk » Wed Aug 09, 2006 12:03 pm

udosuk wrote:PS: In a whole season it's not unusual for players to shoot 500-600 FTs...

Rather than pick numbers, we can generalize. Let 'M' be free throws Made ... and 'A' be free throws Attempted.

For Q1, when a free throw is missed ...

M/A >= 4/5 >= M/(A+1)

... and one of the equalities will occur when M is evenly divisible by 4.


For Q2, when a free throw is made ...

M/A <= 4/5 <= (M+1)/(A+1)

... but so far I'm unable to draw a conclusion from this. MCC [edit: and tarek] evidently believe one of the equalities must occur.
Last edited by ronk on Wed Aug 09, 2006 9:30 am, edited 2 times in total.
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby udosuk » Wed Aug 09, 2006 12:36 pm

ronk, what should I say... You're a half step from the destination now...:)

PS: The one you refered to should be tarek... MCC hasn't shared his wisdom on this one yet...:D
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby ronk » Wed Aug 09, 2006 2:26 pm

[edit: Incorrect conclusion for Q2 deleted. This "half-step" is a biggie for me.]
Last edited by ronk on Wed Aug 09, 2006 11:36 am, edited 4 times in total.
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby tarek » Wed Aug 09, 2006 2:34 pm

I'm not that good at maths, I'll use some of ronk's terminology here, I'm sure a more elegant explanation exists...

M = Free throws scored
A = Free throws attempted
FT percentage= M/A

A free throw = 1
M+1 => A+1 the reverse is not true....

.8000 percent can be expressed in its simplest form in basketball terms as M=4 & A=5 (you cant express that in NBA with smaller numbers)...

Note that A-M=1 (which is the equivelant of 1 free throw)
the next .8000 would be the 8/10 (M=8 & A=10) ...here A-M=2

Note that (A2-M2)-(A1-M1)=1

This goes on for any expression 0f .8000 that follows (the same rationale would decribe an ratio expressed by x/x+1, examples include 1/2=.5 2/3=.6667 4/5=.8 5/6=.8333 6/7=.8571 7/8=.875 8/9=.8889)

at .65 A0-M0 would be an integer
at .90 A1-M1 is also an integer

(from what I mentioned earlier if A1 is around M0 then this is the fastest approach to .9000)

Anyway as all integers are multiples of 1........ climbing from .65 to .8 should always produce an integer A-M that is why it will she will pass through 2/3=.6667 4/5=.8 5/6=.8333 6/7=.8571 7/8=.875 8/9=.8889 & that is why you see those ratios frequently in NBA


tarek
User avatar
tarek
 
Posts: 2607
Joined: 05 January 2006

Postby MCC » Wed Aug 09, 2006 3:23 pm

As I said in my post, I was going by "the phrasing of the questions", in fact, I originly had the answers the other way round, yes to 1 and no to 2.

I won't use probabilities as I will probably mess it up, but in general -

Let's say the question was about the average height of trees in a forest.

You know from a booklet you been given that the average height is 80ft.

If you took a random sample of 50 trees and measured the average, what is the probability you'll come up with 80ft as the average:?:

Compared that to a random sample of 1000 trees:?:

The more you have in a sample the more accurate you can be.

So my thinking (probably totally up the creek) was that the more throws you had, the closer to an actual figure (here 0.80%)you could get.


MCC
MCC
 
Posts: 1275
Joined: 08 June 2005

Postby udosuk » Wed Aug 09, 2006 4:11 pm

Before MCC confuse us more with his forestry analogy (an interesting topic nonetheless), which is not that relevant to this problem, perhaps I'd better demonstrate it myself (as I think most have essentially "grabbed" the main concept here)...

Q1:
ronk has already given 2 separate examples to show that Shaquillina could or could not has her FT% touching the .800 mark. So the answer is (b), neither impossible nor a must.

For it to happen, either M/A=4/5 or M/(A+1)=4/5, i.e. 5M=4A or 4A+4, which is possible only if A or A+1 is a multiple of 5 and M is a multiple of 4.

For it to not happen, M/A > 4/5 > M/(A+1), i.e. 4A < 5M < 4A+4, so there must be a multiple of 5 among 4A+1, 4A+2, 4A+3. For example, A=7, then 4A+2=30 and M could be 6, or A=38, then 4A+3=155 and M could be 31. (Check: 6/7 > 4/5 > 6/8, 31/38 > 4/5 > 31/39...):idea:

Q2:
The answer is (c), i.e. Shaquillina's FT% must at one stage be exactly .800. Suppose the contrary is true, i.e. she "jumped" from below .800 to above .800 without stepping on the line:) , then we must have:

M/A < 4/5 < (M+1)/(A+1)

Which leads to the following 2 inequalities:

5M < 4A so 4A-5M > 0
4(A+1) < 5(M+1) so 4A-5M < 5-4 so 4A-5M < 1

Since both A and M are integers, 4A-5M must be an integer. There is no integer existing between 0 and 1, so we reach a contradiction. Therefore Shaquillina's FT% could not have "jumped" from .800- to .800+ without hitting .800 at least once...

Note that there is 1 exception to the above reasoning... Suppose A=0 and M=0 (i.e. she hasn't shot a FT at all), from basketball convention her FT% is defined as 0% (although mathematically speaking 0/0 is not a valid fraction). If she makes her 1st ever FT, then her new FT% is (0+1)/(0+1)=1=100%, and she has "jumped" over the .800 line...

However, for this particular problem we know it is not the case (she was trying to "jump" from .650 to .850), so this reasoning stands and we can be sure of the answer...:idea:

Therefore both tarek and MCC has got the correct answers to both (though not with the elegant proof I had in mind)... I'm sure ronk and others could more or less guess it anyway...:)
udosuk
 
Posts: 2698
Joined: 17 July 2005

Next

Return to Coffee bar