Math Challenge

83 Million?

I don’t think this is a random act, therefore "odds" don’t apply.

What is the count? If 3-2 the batter may be trying to foul off (the only way he knows how to that spot>)

How many different speeds and pitches does the pitcher command – is each pitch identical?

How many fans are sitting in center field or right field corner– where they cannot catch that foul ball?

Is the batter or pitcher on steriods?

Are you asking for the odds before the events occur or after the first event?

If things were random...

The odds of it happening before the events occurred would be 1/9140 x 1/9140 or 1/83,539,600

The odds of it happening again after the first occurrence are 1/9140

My Answer

Might the answer be that there 1 out of 83, 557,881 chances that the same fan would catch the ball in mid-air on successive plays?

REASONING: Since there are 9,140 attendees at the game that translates to 1 out of 9,140 chances that some fan would catch the ball in mid-air. To that sum must be added the number 1 which represents the possibility that no fan would catch the ball before it hit the ground. Thus on a single play, there would be 1 out of 9,141 chances that a particular fan would catch the ball in mid-air. To calculate the possibility that this same fan would repeat the feat on the next successive play, I would adopt the same odds, namely 1 out of 9,141 possibilities, but with the additional necessity that the 2 fractions be multiplied together to obtain one product. Thus [(1 / 9,141) x (1 / 9,141)] = 1 / 83,557,881.

I should also point out that my solution excludes the possibility that some player from either team would be involved in catching the ball. It's been many years since I tried to solve this kind of a problem in school, so it will be interesting to learn whether or not I succeeded!

Answers from the Actuarial Minded

Part 1 of 2Here's my stab at answering the question. Predicated on lots of things being held constant between hitter and pitcher. Given the number of things that could be altered between pitches (speed, placement, type, bat speed, etc.), the actual odds have to be astronomically high.

See my son's response below (Part 2). His approach would be consistent with factoring in the variables I chose to ignore (pitch selection, speed, placement, batter's pickup of the ball, bat speed, etc.) in order to produce a basic estimate. How did you come up with 22 million pitch count? Did that factor in Canseco offering David Wells HGH? Bottom line, it's a damn rare thing.

I think your basic answer was that the event should happen once in every 3.5 million times. If the basic frame for this whole event was that no one remembers it happening before, then we should use a poisson distribution to predict the waiting time between occurrences. Over the last 30 years (a time when most fans should remember), there have been about 22 million pitches. Your prediction would suggest that the event would have happened 6.24857 times by now. According to a poisson distribution, there is only a .0128% chance that your prediction is right. A better estimate could be anywhere from 1 in 7 million to 1 in 40 million.C. Ford, Actuary & N. Ford, Statistics Student, NC

Answers from the Actuarial Minded

Part 2 of 2:Conditions:

1. Assume the pitcher delivered, and the hitter swung in identical fashion for both foul balls. Pitcher: Type of pitch, speed, location, style of delivery Hitter: Anticipated pitch v. actual pitch, location (point of contact on bat), bat speed, desired contact placement of a hit ball (intentional foul)

1A. What about a hole in one? Is this analagous to hitting consecutive holes in one directly in to the cup (no ball roll)? Golfer: Pin placement (distance), club selection, wind speed, wind direction, contour of green, ball spin, ball placement on green Green: Pin placement (distance), contour of green, moisture content of green

2. Assume that a seated fan "owns" a 3 foot by 3 foot footprint and that no other fan or object (roof, net) can interfere with this space

3. Assume that all 200,000 square feet of the stadium is available to have a ball hit to any 3 foot by 3 foot subsection ===>200,000 / 9 = 22,222 "catchable spots" exist in the stadium...9,594 "catchable spots" exist in foul territory of 86,350 square feet

4. Assume hitter will make contact with the ball sufficient to produce a fly, ground, or foul ball 40% of the time; the other 60% produces a ball, swing & miss, foul tip

5. Assume for every ball "hit" that it is equally likely to be hit to any of the 2,222 catchable spots in the stadium 6. Assume 300 pitches per average game; combined with 4. and assuming a fly, ground or foul ball are equally likely, then average of 40 foul balls per game.

Odds:

Of hitting a foul ball to your catchable spot anytime = 40 / 9,594 = 0.41693% or 1 in 240 foul balls Of it happening twice anytime during same game = once prob squared = 0.00174% or 1 in 57,528 foul balls Of two consecutive pitches producing foul balls hit anywhere = (40% probability of contact * 1 in 3 chance foul) all squared = 1.77778% or 1 in 56 sets of consecutive pitches Of two consecutive pitches producing foul balls, one to you = (40% probability of contact * 1 in 3 chance foul) all squared * 2 in 240 foul balls hit to you = 0.01481% or 1 in 6,750 sets of consecutive pitches Of it happening from 2 consecutive pitches during same game = (40% probability of contact * 1 in 3 chance foul * 1 in 240 chance to you) all squared = 0.000030864% or 1 in 3,240,000 sets of consecutive pitches Golf hole in one per 1999 Golf Digest article, one in a round by an average golfer 1 in 12,000, two players from same foursone, same hole 1 in 15 million (http://golf.about.com/od/faqs/f/holeinoneodds.htm)

Just a wild guess

In keeping with your theme of the math challenge, I thought the #5 highlight from yesterday’s ESPN SportsCenter would interest you:

http://espn.go.com/video/clip?id=4250446

My stab at the answer would be the odds are 1 in 196,000,000…just a wild guess.Larry Long, NC

9,140 to 1

9,140 to 1Vicki Paulus

100%

100%John J. Prendergast, CPA

Too many variables

[Editor’s Note… not having the math skills of my son, I forwarded the question to him.  This was his reply.  I do not take credit or responsibility for it.]

That question has too many variables and not enough known variables, there are multiple answers that could be given, but without more information, the odds cannot be accurately calculated.  However, there are two limitations that help us guess possible statistics…. The number of fans, and the fact that it is on two consecutive pitches.

First, let’s estimate some variables that can be rounded due to statistical rarity

Assuming the pitcher never changes during the middle of a count, we can assume the odds of the same pitcher for two consecutive pitches, when we know the first pitch did not change batters, to be 1

Assuming the batter isn't subbed for mid-count, we can assume the odds of the batter swinging again after hitting a foul ball to be 1

Now, Working with the absolute known we have 1 fact:

What are the odds that a foul ball flew towards a specific fan in the stadium? 1/9140

In order for the specific fan to be determined, the first event had to have happened.   What are the odds of hitting a second foul ball to a specific fan when the first pitch was not hit foul? 0… so we must assume the odds of hitting a ball to a fan to be 1

Thus you could argue that the odds of this happening simply being 1/9140

Now, there are other variables in this scenario that are unknown, thus we cannot truly define the answer, but I will elaborate some of them for you.

Base probability a ball arriving at a specific fan 1/9140

Suggested Variable              EST Variable Odds       Cumulative OddsOdds of hitting a Ball at all?          ½                       1/18280Odds of a hit ball going foul?          ½                       1/36560Odds of that foul ball being caught?    ½                       1/73120

Now, if we consider the odds from the beginning of the scenario, and assume that we want the odds of two consecutive pitches, not just the odds of the same thing happening after the first one does, we need to square the odds (as we have already decided that the pitcher and batter won’t change) 

That results in odds of about 1 in 5,346,534,400

Now roll that back and assume this fan is an amazing catcher and would never drop a fly ball… the odds become 1 in 1,336,633,600, or about 4 times better.  Basically, every variable you introduce that is 50/50 increases these odds 4 fold.

After I calculated this, I found this article, which re-emphasizes the abundance of variables to calculate a solid answer.

http://blogs.wsj.com/numbersguy/a-pair-of-lucky-baseball-fans-337/Dick Spencer, CPA, WI

Astronomical

AstronomicalChris Wood, OH

Almost 100%

The probability is very, very high, almost 100%. In fact, some would say that since the two catches were made (they have already happened) and you did not predict that in advance, the probability is exactly 100%.Paul W. Guy, PhD, CA

?

one over 9140 squared?Terence Wong

Answer-9140

Answer-9140Barry S.

References

Here are some references for more reading material on the subject...http://www.fangraphs.com/blogs/index.php/odds-of-catching-a-foul-ball/http://www.fangraphs.com/blogs/index.php/foul-balls-again/http://blogs.wsj.com/numbersguy/a-pair-of-lucky-baseball-fans-337/

Michael Eikenberry, IA

Astronomical

...astronomicalFrank Page

Same man in the seats

In a game where 9,140 people attended, the batter hit a foul ball into the right-field corner, which was caught in the air by a fan. On the next pitch, the batter hit another foul ball--caught in the air by the same man! Same batter, same pitcher, same man in the seats catching the ball in mid-air...and on consecutive pitches! What are the odds of this happening? 50%Norris Groves

Lucky fans

Check out this article (http://blogs.wsj.com/numbersguy/a-pair-of-lucky-baseball-fans-337/). Carl Bialik, The Numbers Guy, in an article entitled "A Pair of Lucky Baseball Fans" The Wall Street Journal, May 13, 2008. He talks about the odds of fans sitting in adjacent seats catching consecutive foul balls, which probably equates to the odds of a single fan being able to catch consecutive balls (similar ball placement proximity). The result of his discussion is not definitive, but it sounds like it would be 1:1,000,000, based on a major league stadium. Of course, I am not sure what impact there would be to the odds by reducing the total number of potential fans, and using some type of factor for the lucky fan's catching ability.James Maurer

Can't happen

The answer is ……it can’t happenJerry Derstein, PA

1 in 763,551,944,000

1 in 763,551,944,000Rob Lee

Most Unlikely

Most unlikely. To 1.Randal R. Carson

Incalculable

The answer must be huge because the number of placement options that a pitcher can throw toward home plate, the position of the batter & direction in which a batter can swing his bat are almost incalculable. The distance in which a ball can go after being hit by a batter is within a range of 575 feet (Guinness Book of World Records, although in another publication I have it says Babe Ruth hit a ball 602 feet) from home plate. The number of spectators & players can’t be used to divide the area within this 575 feet because the spectators are restricted to the stands & the 9 defensive players have the field.

There are no statistics of where foul balls go when they are hit, so a pattern could not be established there. This leaves best guess probabilities, which are certainly not reliable.

Now a pitcher could pitch to a batter on the outside of the plate, and for a right handed batter the odd are the ball will be hit  to the right field side, up the 1st base line.  But whether it’s hit fair or foul is mostly determined how far outside the plate the ball is pitched and at some point the batter won’t swing at the ball because his purpose is to hit the ball into the fair play portion of the field.

So this is the reason the answer is incalculable.  Maybe, it’s just me!Randy Gloyd, CA

1 in 83,539,600 (9140 squared)

1 in 83,539,600 (9140 squared)Fred P. Curry, CPA, NM

1 out of 9,140.

1 out of 9,140.Dean H. Hiltzik, NY

1 in 9,140

[P(batter hits a foul ball) * P(foul ball is caught by a fan) * 1/9140]^2.

Where P(batter hits a foul ball) is the probability that the batter hits a foul ball on the next pitch and P(foul ball is caught by a fan) is the probability that the foul ball is caught by a fan.

Richard G. Fleissner

1 in Infinity

1 in Infinity... In doing a probability analysis in this case, one must consider every possible pitch (e.g., speed, angle, style, location (e.g., across the plate, wild pitch)); every possible swing (e.g., speed, angle, ball impact) based on the pitch; every possible location on the bat that the ball could hit based on the pitch (if the ball is hit at all); every possible location the ball could land based on the pitch, swing, and location on the bat where the ball connects; every possible scenario that the fan would catch the ball (e.g., speed and angle of the ball, sitting/standing position); etc., etc....

Never in a million games...

I loved your story about the back-to-back foul ball grabs at a baseball game. I’ve been to a million games in my life and have seen a lot, but never THAT!! Don’t know about the odds – my head still hurts too much from tax season to think about that… Chris Harper,WA

1 in 9,140

The probability of the same person catching the second foul hit is 1 in 9,140 (or approx 0.00010941).Patric Edward, ME

1 in 9140

From the few facts you give it would be a 1 in 9140.  It’s like playing the lottery, no matter how many tickets you buy, it’s still the same probability of winning?  But it this case you have some many different variables that you have to use a multiple regression model to obtain a scientific model of the probability.  But if you ever saw Darrel Strawberry hit, he could hit the ball to the same spot every single time.  Of course he had personal problems.

You are missing some information from your question to give a better answer

How big is the stadium?What’s the weather conditions? Temperature, humidity, wind speed, time of day, etc.How are the fans in the stadium distributed?How far is the right field from the batter’s box?What kind of glove does the fan have?Is the fan sitting by himself?On average how far does this player hit fouls ball?What are his batting percentages against this pitcher?Where in the line up is he batting?Are there men on bases?What is the score?What’s the furthest this batter has hit any kind of ball?What is the pitchers pitching numbers?  What inning was this?When did the pitcher start pitching this particular inning?Is he a left or right hitting batter?

Just a few things I could think of.

Miguel Gil,CA

One in 9,140

One in 9,140Bruce Braithwaite

There's more to this one than meets the eye.

Any "correct" answer to this one requires up front assumptions and none of them are air tight in this scenario.  So I don't see this as a pure math problem, but a baseball problem where math is involved.

The underlying assumption is that the location of each fan has an equal probability of being the target of a batted ball for any given at bat.  Well, we all know that isn't reality.  I don't know what the true distribution is for batted balls in the stadium in question, but for smaller stadiums it would seem that every seat not screened off for safety has at least a probability greater than zero.  But some sections of seats have a far greater probability of being hit than others.  Certainly the first few rows down the third base line with a left handed batter has considerably higher odds of being hit than the outfield upper corners down the first base line.  These odds change by batter, pitcher, weather conditions, whether a switch hitter chooses to bat left or right, the game situation, etc.

For consecutive pitches, one would expect the weather to hold constant.  The game situation would likely hold constant, although if the first pitch was strike 2 versus strike 1 and it's late innings in a tied game, that may change compared to say, the 2nd inning of a 7-0 game.  In other words, it's possible then that the batter is swinging with a different intent on the subsequent pitch, but probably in most cases that would hold true.

I have ignored the fact that fans are mobile and indeed to get up and travel for the opportunity to catch balls, sometimes trampling little kids and grandmothers in the process.  And so then there is the aspect that regardless of what seat a foul ball or homerun hits, not all fans are equi-probable of catching the ball even once, let alone twice.

In the contest in question, with nothing to go on but what is given, about the only thing you can do is just make the easy assumption that every fan has an equal probability of being the target of any given batted ball.  In which case, the odds are 1/9140 X 1/9140 (or 1/(9140 squared).  So if you select a winner based on those submitting an answer of 1/83,539,600 then mathematically you are sound.Michael Eikenberry, IA

1.2 in 10,000,000

Robert W. Peddy, CPA