I had written an elaborate introduction to probability theory, but my post got to be a little bit long and I wanted to shorten it. It people are interested, I can post it later. For this post, I had a couple of interesting examples and wanted to cut to the chase.
Conditional Probabilities
Often times we want to know what the probability of something happens is given that we already have some information. For instance, we may ask "What is the probability that a team scores at least 2 runs in an inning given that we know they score at least 1?"
The formal way to generally state this question is "What is the probability of A given B?". This is given by the formula:
P(A|B)=P(A and B)/P(B)
where P(A|B) is the probability of A given B. For instance, what is the probability that the sum of two dice is 6 given that the first one is a 5? In this case A is the event that the sum of two dice is 6 and B is the probability that the first die is a 5. P(B)=1/6. What is the probability of A and B? Well, we must throw a 5 on the first die and a 1 on the second, so P(A and B)=1/6*1/6=1/36. Thus, P(A|B)=1/6.
Counter Intuitive Applications of Conditional Probability
Conditional probabilities often lead to counterintuitive results. The most famous of these is the Monte Hall Problem. Another famous example involves medical testing for a rare disease.
A particular disease has an occurance of one person in a million. A company develops a test for this disease that is 99.9% accurate (here we will assume all incorrect results are false positives). Given that a person has tested positive for the disease, what is the probability that they actually have it?
Many people would look at this problem and think that there is a very high probability that the person has the disease. Not so! We need to look at this through conditional probabilities. Here, A is the event that the person has the disease and B is the event that the person tests positive for it. By assumption, the person will test positive for the disease if he has it and so the probability that he has it and he tests positive for it is 1/1,000,000.
The probability that a person tests positive for the disease is a little bit more difficult to calculate, so we will look at it this way. On average, out of a million people that take the test 1 person will actually have the disease and out of the other 999,999 that don't have it, 0.1%*999,999 ~= 1000 will test positive and so P(B)~=1,001/1,000,000.
Then, we can just plug our number into the formula to get the probability that this person has the disease is roughly 0.1%. This notion seems almost absurd given a test of such high accuracy, but the reason is that the disease is so exceedingly rare that the chances of a false positive far outweigh the chance that the person actually has the disease.
This is another relatively famous application of this formula, although the formula is actually hiddent.
Given a room full of n people, what is the probability that at least two of them share a birthday?
For this particular problem, you want to rearrange the above equation to read P(A and B) = P(A|B)*P(B). This formula can be generalized somewhat and I won't get into detail about it, rather, I will just explain the reasoning behind the problem.
In essesnce, it is much easier to look at the probability that no two people have the same birthday. Well, if n=1, then the probability that no two people have the same birthday is 1 (duh). If n=2, the probability is 1*(364/365) since the first person can have any birthday and the second person may have any birthday except the birthday of the first person. If n=3, then the probability that none of them have the same birthday is the probability that the the first two don't have the same birthday times the probability that the 3rd person doesn't have the same birthday as the first two. Thus, we get 1*(364/365)*(363/365).
We can generalize this formula to the following way: Find the probability that the first n-1 people don't share a birthday, then multiply it by the probability that the nth person doesn't share a birthday with any of the first n-1. This gives the formula:
So then the probability that two people share the same birthday is just 1-p. The counterintuitive part of this problem comes when we ask what the smallest n is for which this probability is at least 50%. The answer is only 23. So if we put 23 people in a room, there is greater than a 50% probability that at least 2 people share a birthday.
Of course, if you put me in a room with Mickey Mantle, Snoop Dogg, Viggo Mortenson and John Krasinski and asked "What is the probability that all 5 people in the room have the same birthday?", the answer is 1. 
Compounding Probabilities in Sports
I'm not sure if statisticians have a term for this phenominon, but the birthday problem is an example of what I refer to as compounding probabilities. Basically, anytime you a number of different things to happen in sequence to achieve a larger goal, you have to start multiplying probabilities together and the result is that the final probability is small, even if the probabilities of each individual event were large.
This is part of the reason behind my sharks defeating my piranhas a large percentage of the time in extra inning games. In general, a team will win the game most of the time if they can simply score a single run. The sharks have a 1/7 chance of hitting a home run each time they come to the plate, which seems pretty small. On the other hand, the piranhas got on base .361 of the time. Unfortunately, they need string hits together (and compound their probabilities) in order to score a single run. This leaves the piranhas scoreless in 75% of innings, whereas the sharks are scoreless in only 63% of all innings.
This type of thinking is common accross many sports and is the general reason that many standard strategies are too conservative.
Problems
I will repeat the same format as last time. I will have an easier problem based on what I wrote about and a hard one which may or may not be related.
1. A given pitcher has an opponent OBP of .250. What is the probability that he throws a perfect game? (You may assume that the probability of each individual hitter reaching base is .250 and ignore other factors that might change this).
2. Al Jefferson starts off the season by missing his first free throw. By the end of the season his FT% is around 99%. Is there necessarily a time when his FT% was exactly 75%?
Rules
- I don’t care if you cheat, but if you do, don’t post the answer.
- Even if you solve the problem (or have seen it before) please don’t immediately post the answer.
- Posting your solution after some time has passed is fine.
- Ask questions and state ideas about the problem to generate discussion.
- Posting solutions/hints in white text is always a good idea.
- I will post the solution eventually (assuming I know how to solve it) if nobody else does.
In a sense, all of the above are really variations of the following two ideas. I want the problem to provoke discussion and let the solution(s) evolve naturally as a result of the discussion. I don’t want people to show up 15 minutes after the problem has been posted just to see the solution posted below.
SBG Note: GreekHouse has an SBG email account (GH, your password is your first name -- all low caps -- please change it and then send me an e-mail to confirm at sbg at stickandballguy dot com) at greekhouse at stickandballguy dot com. If you want to correspond with him via this account, go ahead. If you want to post your answer before he has revealed the answer, put this
<font color="white">
before your answer and
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after your answer. You are encouraged to cut and paste that little bit of HTML.
Note to anyone with comment modification privileges: if you see an answer posted and not in all white, please edit the comment to add the whitening HTML.
Here are my answers, but before that, I'll add this little bit of filler to prevent the answer from showing up in the recent comments on the main page.
Thanks for the post, GH.
1. There are 27 outs. Given that each batter has a .250 OBP, the probabilty of getting any one batter out is .750. Therefore, the probability of getting all of them out is (.750)^27. The probability is about .0042.
2. Hmmm. I'm going to assume you mean exactly 99%. If Jefferson shot 100 free throws, the probability of ending at 75% is 1, because after four free throws, he would be at 75%. What about 200 free throws? If his second miss was before his fourth throw, he'd have to be six of eight at one point. How about 300 free throws? Yep. He'd either have to be 3/4, 6/8, or 9/12. It seems that the answer is yes.
Now, why is that? Because at representation of 75% can be reduced to 3/4 or more importantly, a numerator and denominator that each consists of a whole number. I don't think you can exceed 75% from a position less than 75% without at some point landing right on 75%. I do not, however, get how this applies to the lesson at hand.
For #1, your answer is correct. It is also interesting to look at 1/p. This gives you the number of games per perfect game. Now think about how a .250 OOBP is about a WHIP of 1.00 and that a starter with that WHIP would probably be the greatest of all time.
For #2, I don't mean "exactly". I edited the post to reflect that. In fact, it doesn't matter what his FT% is at the end of the season, just that it's > 75%.
Now think about how a .250 OOBP is about a WHIP of 1.00 and that a starter with that WHIP would probably be the greatest of all time.
You mean, a pitcher who would be potentially deserving of a Cy Young Award?
Yeah, that was one ridiculous season. What I really meant is that if a pitcher could do that over an entire career, he would probably be the best pitcher ever. If Santana had his 2005 season repeatedly, he would probably be just that.
SBG, copying and pasting your white-out code did not work.
before your answer and
Yup, copying and pasting SBG's code is not encouraged by me.
It works now. The problem is that when the post gets edited, the quotes get curly. Ugh. I'll have to add a plugin to get around that.
you roared into the driveway of our southwestern ranch-style house
on a new kawasaki, all yellow and black
fresh out of the showroom.
our house faced west,
so the big orange sun positioned at your back,
lit up your magnificent silhouette.
how much better?
how much better can my life get?
900 cubic centimeters of raw whining power.
no outstanding warrants for my arrest.
whoa-whoa. whoa whoa.
the pirate’s life for me.
The second question is kind of interesting.
Suppose that it is possible for Jefferson to go from below 75% to above 75% without passing 75% exactly. Then the following conditions must hold:
P = m/n < 3/4
Q = (m+1)/(n+1) > 3/4
Where m is the number of made free throws after n total attempts. After a few lines of algebra, you can convince yourself that:
-1 < 4m - 3n < 0
Since m and n must be integers (unless we are playing some sort of crazy non-integer basketball), there is no way to satisfy this inequality. Thus, Jefferson must pass through exactly 75% at some point when he starts at 0% and ends somewhere above 75%.
If we want to generalize this a bit (because, really, what's so special about 75%?), we can take an arbitrary fraction p/q instead of 75%. This leads to the following condition:
(p -q) < m*q - n*p < 0
So that as long as q=p+1, and p/q < 99%, Jefferson must pass through that fraction exactly on his way to 99%. Of course, then we can start to see fractions that Al doesn't have to pass through. If p/q is 5/7, then we get:
-2 < 7*m - 5*n < 0
Which we can satisfy with m=2, n=3.
900 cc’s of raw whining power,
no outstanding warrants for my arrest.
hi diddle dee dee.
god damn!
the pirate’s life for me!
You are too fast ubelmann!!!
I feel like there should be an alternate solution to this problem, but I'm not sure what it is.
900 cc's of raw whining power, baby.
Of course, if you put me in a room with Mickey Mantle, Snoop Dogg, Viggo Mortenson and John Krasinski and asked “What is the probability that all 5 people in the room have the same birthday?â€, the answer is 1.
Stick, you gotta sign this guy.
Of course, those folks only share the same birthday, not the same birth date. Virginia Madsen and I in a room together, on the other hand...
GH, you are assuming that there is an equal percentage of birthdays for each day of the year, which is not true. Or does the probability still hold, knowing that?
I actually have the same birth date as John Krasinski. Obviously, I can't share the same birthday with anyone else on that list since they were all born on different years.
This problem makes all sorts of simplifying assumptions that are false. For instance, it assumes that nobody is born on Feb. 29th. Also, more children are born in the summer, so the distribution isn't uniformly distributed. The first problem reduces the probability that 2 people share the same birthday, while the latter increases that probability.
I'm typing off the top of my head here, but I seem to remember that some date in October (14th?) is the most common birthday. Nine months after New Year's? Superbowl?
I'm sharing a birthday with Benjamin Franklin, John Jacob Aster, Al Capone, James Earl Jones, Don Zimmer, Muhammad Ali, Andy Kaufman, Jim Carrey, and Kid Rock, to name a few.
You have a very famous bday SBG! It's also sort of crazy that Carrey and Kaufman have the same bday.
My niece shares hers with LeBron James and Tiger Woods, an interesting pair.
Me and Jesse ain't got time to bleed on the birthday cake. Rembrandt is in charge of decorations. Linda Ronstadt handles the singing.
In 1099, Jerusalem fell to the Crusaders after a siege of just over one month The streets of the city ran with blood as the Crusaders slaughtered 40,000 and set fire to mosques and synagogues. In 1937, Buchenwald concentration camp opened. In 1973, Ray Davies announced retirement from Kinks then attempted suicide.
So I got all that going for me. Which is nice.
Well, since you mention it, on my birthday:
Led Zeppelin released their debut album
Vikings lose to Redskins in NFC Championship Game when Darren Nelson drops touchdown pass.
Vikings lose to Falcons in NFC Championship Game after 15-1 season.
The first Gulf War started (and it ended on my brother's birthday).
The Great Brinks Robbery Occurs.
Eisenhower gives his farewell address, warning about the dangers of the military industrial complex.
Gary Gilmore is executed.
Also, Steve Earle was born.
and mine goes without saying
“I love the Twins, ... I bleed the Twins. That's all I know. If I go anywhere else, I'd be a foreigner.â€
“I smell blood. I'm like a pit bull. Once I taste blood I'm on you ... and I'm about to taste blood.â€
"You can't beat Comerica Park. I don't know what's going on in Detroit, but when the All-Star Game comes to town, it's going to happen. I can come to Arkansas and people are going to throw parties, and you're going to have fun wherever you go. It doesn't matter if it's Detroit or whatever."
In regards to the first problem, I think it's interesting to note that having indentical hitters maximizes the probability of a no-hitter, though I don't think I can quite show it rigorously.
If you have two hitters, though, with odds p and q of making an out, on a "team" that has a probability r of making on out, we have:
(p+q)/2 = r
p*q = probability of a no-hitter
We can extremize p*q by taking the derivative of p*(2r-p) with respect to p. Setting that equal to zero gives us r=p. The second derivative is -2, indicating that this extremum is a maximum.
There's probably some induction argument that you can use to show that on a team with probability r of making an out, the probability of them getting no-hit is maximized when they each individually have a probability r of making an out, but I can't see it right now and I'm too lazy to finish what I started.
The upshot being that it looks like is harder to throw a perfect game against a team of stars and scrubs than it is to throw a perfect game against a team where everyone has the same OBP, if both teams have the same team OBP.
"I can't say what I feel because all of a sudden I might get in trouble. Everybody wants that guy. I wish I was Torii Hunter's agent. That's my wish, just to be Torii Hunter's agent this year because he's in a great situation right now." -- Ozzie Guillen
I’m not sure if statisticians have a term for this phenominon, but the birthday problem is an example of what I refer to as compounding probabilities. Basically, anytime you a number of different things to happen in sequence to achieve a larger goal, you have to start multiplying probabilities together and the result is that the final probability is small, even if the probabilities of each individual event were large.
This is part of the reason behind my sharks defeating my piranhas a large percentage of the time in extra inning games. In general, a team will win the game most of the time if they can simply score a single run. The sharks have a 1/7 chance of hitting a home run each time they come to the plate, which seems pretty small. On the other hand, the piranhas got on base .361 of the time. Unfortunately, they need string hits together (and compound their probabilities) in order to score a single run. This leaves the piranhas scoreless in 75% of innings, whereas the sharks are scoreless in only 63% of all innings.
I believe you are correct and I think you can prove this using multivariable calculus.
This type of thinking is common accross many sports and is the general reason that many standard strategies are too conservative.
James Leer: Now, that is a big trunk. It holds a tuba, a suitcase, a dead dog, and a garment bag almost perfectly.
Grady Tripp: That's just what they used to say in the ads.
For problem #2, I think this is similar to SBGs approach, but a little more general.
take a fraction p/q that represents a percentage where q=p+1,
then starting from his 0-for-1 start, Jefferson must miss at least one of his next p shots or he will land squarely on a p/p+1 shooting percentage, i.e. p/q.
Assuming he misses one, which has him at p-1/q he must now miss at least one of his next q shots or he will land squarely on 2p/2q [(p-1+q)/(q+q)].
This continues on for as long as you have the patience. Thus, if Jefferson minds his p's and q's in order to avoid landing on a p/q percentage, he must always shoot below the p/q percentage. So he can't get to any percentage higher than p/q without shooting exactly p/q.
I don't think this works for p/q where q=p+n, if n>1, but I haven't proved that rigorously.
Grady Tripp: I'm beginning to think that the car wasn't exactly Jerry's to give.
Terry Crabtree: Ah, so who's car was it?
Grady Tripp: My guess? Vernon Hardapple.
Terry Crabtree: The hood jumper?
Grady Tripp: He said a few things that lead me to believe that the car was his.
Terry Crabtree: Such as?
Grady Tripp: "That's my car, motherfucker."
KOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOORN!!!
This seems like the general intuitive way to approach it, since it basically follows what you would try to do if you were trying to create a steak that somehow misses 75%.
If you wanted to formalize it a little bit more, you would let M be the total number of free throws that Jefferson misses, then for k=1,2,3,....,M, consider the most number of made free throws you can make and still have a FT% under 75%. Each of these will have a form similar to the one you described and so the next shot will be exactly 75%. Since he only misses M total, he must arrive at one of these situations eventually.
Excellent idea! I knew there was a second way to do this problem.
You are right also right about the second problem (I think). I will make this a bonus challenge problem. Change 75% to p/q in the example above where gcd(p,q)=1 and p<q -1. Show that there exists a sequence such that Jefferson does not ever have a free throw percentage of p/q (and of course, still ends up with a percentage higher than p/q).
KOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOORN!!!
MNF just had a a stat about the Colts pass defense. In terms of pass attempts, they allow the fewest of more than 20 yards, the fewest from 11-20 yards, and the most less than 10 yards.
In essence, they force the opposition to execute high probability plays many times and force them to work down the field. Although this can allow for many short gains, it is difficult to maintain a long drive and score against them. It's an excellent example of using compounding probabilities to your advantage.
West Coast Defense??
Revelation At Midnight
by Piet Hein
Infinity's taken
by everyone
as a figure-of-eight
written sideways on.
But all of a sudden
I now comprehend
that eight is infinity
standing on end.
Problem 1. Pr(perfect game) = prob of 27 consecutive outs. We are given the pitcher's Opponent OBP = 0.250, which implies the opponent out rate of 0.750. Pr(Out1)*Pr(Out2)*...*Pr(Out27) = 0.75^27*100 = 0.0423 pct.
Problem 2. Given (1) the first free throw is a miss and (2) tot. makes/(tot. makes+ tot. misses) ~= 0.99, must there be any point in the sequence of N = (tot. makes+ tot. misses) s.t. the running total of makes divided by the running number of attempts exactly equals 0.75?
This problem asks, what are odds of (not 3 makes out of 4) and (not 6 out of 8 ) and ... and (not exactly 75 pct of the largest multiple of 4 in N) conditional on conditions 1 and 2 above?
let's start with a numerical example with N=100. Let n= number of makes in N and m=number of misses. Here, n=99 and m=1 to meet the condition that the season pct ~=0.99. Since we are given that the one miss comes first, it follows that the next three attempts are all makes, so the FT pct after 4 attempts must be exactly 0.75.
now suppose 100
now suppose 200
and so forth with larger Ns and ms. It follows that there must come a point at which Al Jefferson's FT pct must exactly equal 75 pct.
Euclid
by Leo Moser
In the Greek mathematical Forum
Young Euclid was present to bore'em.
He spent most of his time
Drawing circles sublime
And crossing the pons asinorum.
This is not the cleanest proof ever, but it gets the basic idea. 9/10
Now, try the bonus problem listed above. I have solved it and it wasn't easy. I'll be eager to see if I've created a problem so hard that not even ubelmann can solve it!
As it turns out, that problem has a very simple solution as well. I made the problem way to complicated, although my solution generalizes to a more general case.
Did you get the freethrow problem from the 2005 Putnam exam? I distinctly remember sitting in a room for six hours trying to solve that problem and... uh... totally succeeding... Well, maybe it didn't go exactly like that, but close enough.
Yes, I basically took that problem and modified it somewhat.
For those who have never heard of it, the Putnam Exam is a very popular math competition for North American math students. It consist of two sets of 6 problems and you get 3 hours to work on each set. There are 10 points per problem for a max of 120 points. The winner usually scores around 100, but the median score is always 0 or 1. So if you didn't get this problem, don't feel too bad!
This problem was an "A1" problem from the test, meaning that it is the first problem from the first section. These problems are typically the easiest ones on the test.
\begin{brag}
In 2000, I scored a 21, which was high enough that the NSA sent me a letter offering me a summer internship. Unfortunately, I was going to study abroad in Budapest that summer and so I was only going to be in the US for about 2 months. I'm not sure which ones I got credit for. It was probably A1 and A2. I though I had gotten B1 and B2 as well, so I'm not sure exactly which ones I got most credit for and which ones I got little credit for.
In 2001, I managed a 12, but was disappointed because I thought I had about 4 problems right again.
\end{brag}
A 21 is quite impressive and probably better than I would have ever done. I wish I'd actually done the Putnam when I was an undergrad, though.
I bet you could have done pretty well. Both times I took it I looked at the sheet and said to myself "I don't know how to do any of these". Both times, I was able to make significant progress towards 4 problems. When you're sitting there for 3 hours at a time with nothing else to do, you can really start to figure things out!
You give the NSA this bit, GH?
LOL...I love that movie.
\begin...\end? You must be a LaTeX fan. Not that there's anything wrong with that, I use it for nearly everything typed these days.
Yep, being a math student I pretty much use it for everything as well!
At least two members of The Nation share the greatest day ever as a birthday.
I'm pretty sure I scored a 12 in 2004 on the Putnam, which only crushed my math advisor all the more when I chose to pursue chemistry in grad school.
for the supplemental problem, is it as simple as this?
For an irreducible fraction p/q where q>p+1, there exists a fraction (namely (p+1)/q) such that [p/q]< [(p+1)/q]<1. So if enough free throws are made in the first q shots the percentage ends up above the p/q line, and one just needs to make enough to get to the designated percentage (99% or whatever).
Since p/q is irreducible, there is no way to get that percentage in less than q shots, so, if you're above that percentage after q shots, you're home free.
Wasn't sure if the "white text rule" still applied.
Why shouldn’t I work for the N.S.A…. that’s a tough one. But I’ll take a shot. Say I’m working at N.S.A. and somebody puts a code on my desk, something nobody else can break. So I take a shot at it and maybe I break it. And I’m real happy with myself, ’cause I did my job well. But maybe that code was the location of some rebel army in North Africa or the Middle East, and once they have that location, they bomb the village where the rebels were hidin’- fifteen hundred people that I never met, never had no problem with get killed. Now the politicians are sayin’, oh, “Send in the marines to secure the area†’cause they don’t give a sh!t. It won’t be their kid over there, gettin’ shot, just like it wasn’t them when their number got called, ’cause they were pullin’ a tour in the National Guard. It’ll be some guy from Southie over there, takin’ shrapnel in the @$$; he comes home to find that the plant he used to work at got exported to the country he just got back from, and the guy who put the shrapnel in his @$$ got his old job, ’cause he’ll work for fifteen cents a day and no bathroom breaks. Meanwhile he realizes the only reason he was over there in the first place was so that we could install a government that would sell us oil at a good price, and of course the oil companies use the little skirmish over there to scare up domestic oil prices- a cute little ancillary benefit for them, but it ain’t helping my buddy at two-fifty a gallon. They’re takin’ their sweet time bringin’ the oil back, o’ course, maybe they even took the liberty of hiring an alcoholic skipper who likes to drink martinis an’ f*****’ play slalom with the icebergs;
This is definitely a nice, simple solution to this problem. The other simple solution I came up with was for Jefferson to miss his first FT (as stated in the problem) and then make every free throw after that until he had made enough. Each of his percentages is of the form k/(k+1). Since gcd(k,k+1)=1 and the numerator is one less than the denominator, this cannot be equal to p/q. Also, it asymptotically approaches 1 so eventyally we will reach a percentage higher than p/q and will have jumped past it without hitting it.
The harder version of this problem comes if Jefferson takes some undetermined number of free throws at the beginning and starts with some FT% a/b < p/q. At the end of the season he has some FT percentage d/e > p/q. The general idea behind proving this version is similar to what you were saying and is a three step process:
1. Miss a bunch of FTs
2. Make a bunch of FTs
3. Miss a bunch of FTs
The idea is that if you miss the right number at the beginning, there is some point where he's taken s free throws and made r (for a ratio of r/s) then, after he makes his next one he will have a ratio of (r+1)/(s+1). The key is to show that such number exist and is not an easy task!
As you can see, neither of our solutions will be sufficient in this case because he might have taken more than q FTs at the beginning and you certainly can't rely on his percentage being of the form k/(k+1).
it ain’t too long ’til he hits one, spills the oil and kills all the sea life in the North Atlantic. So now my buddy’s outta work, he can’t afford to drive, so he’s got to walk to the f*****’ job interviews, which sucks ’cause the shrapnel in his @$$ is givin’ him chronic hemorrhoids, and meanwhile he’s starvin’ ’cause every time he tries to get a bite to eat the only blue plate special they’re servin’ is North Atlantic cod with Quaker State. So what did I think? I’m holdin’ out for somethin’ better. I figure f*** it, while I’m at it, why not just shoot my buddy, take his job, give it to his sworn enemy, hike up gas prices, bomb a villiage, club a baby seal, hit the hash pipe and join the National f***’ Guard? I could be elected President.
The harder version of this problem comes if Jefferson takes some undetermined number of free throws at the beginning and starts with some FT% a/b < p/q. At the end of the season he has some FT percentage d/e > p/q.
My solution follows those three steps you laid out but just to write it out:
Given a starting free throw percentage a/b, and an irreducible fraction p/q such that q>p+1 and a/b < p/q:
Short version:
One must first get to a (np-1)/nq shooting percentage, then make p+2 out of the next q shots to get to [(n+1)p+1]/[(n+1)q] > p/q.
n is just a whole number. Note that (np-1)/nq < p/q, so it is possible to reach that percentage without shooting exactly p/q. Also, since p/q is irreducible there is no possible way to get exactly a p/q percentage unless an exact multiple of q shots have been taken. So when the percentage goes from below to above the p/q line in the span of q shots, it can't possibly equal p/q within that span.
Longer version:
The first x shots must be missed, where x = n(q-p)+1-(b-a). n(q-p)+1 gives the total misses at the (np-1)/nq goal. Then the (b-a) number of misses to start with are subtracted.
Next, the shooter must make y = np-1-a shots, which will get them to a (np-1)/nq shooting %.
Then by making p+2 of the next q shots, the shooter gets to a [(n+1)p+1]/[(n+1)q] percentage which is greater than p/q.
Thus, it is possible to shoot free throws in such a way that the final shooting percentage d/e > p/q without ever shooting exactly p/q.
For example:
if we start out 17 for 29 but need to get above 77.78% (7/9) shooting;
miss the first shot (n=6, 6*(9-7)+1-(29-17) = 1),
then make the next 24 (6*7-1-17 = 24) to get to 41/54 (75.93%).
Then making 7+2 out of 9, we end up at 50/63 (79.37%) without hitting the magic percentage.