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An Ear Full of Cider

16 September 1999

By Donald Catlin

    The writer Damon Runyon had a cast of characters in his works that were based on real life characters he knew in New York during the 20s.  One of them, Nathan Detroit, you probably recognize as one of the principal characters in the Broadway show Guys and Dolls, which is based on Runyon's work of the same name.  In the show, Detroit offers another character, Sky Masterson, a $1000 bet that a deli called Mindy's sold more strudel than cheesecake the previous day.  Masterson replies as follows:

    "Let me tell you a story.  When I was a young man about to go out into the world, my father says to me a very valuable thing.  'Son,' the old guy says, 'I'm sorry that I am not able to bankroll you a very large start.  But not having any potatoes to give you, I am going to give you some very valuable advice.  One of these days in your travels, you are going to come across a guy with a nice brand new deck of cards, and this guy is going to offer to bet you that he can make the Jack of Spades jump out of the deck and squirt cider in your ear.  But, son, do not take this bet, for if you do, as sure as you are standing there, you are going to end up with an ear full of cider.' "

The father, of course, was speaking about so-called proposition bets, also called sucker bets, and the fact that you should never accept them.  A proposition bet is something that sounds like a sure thing and yet, if you take the bet, you always lose.  Here is a typical proposition bet.  You are in a bar when some guy tears a small hole in a napkin and offers to bet you that he can push a glass of beer through the hole.  "You mean to tell me that you are going to push a full-sized glass of beer right through that small hole?"  "Yup," he says.  "You're on," you say.  The guy proceeds to stick his finger through the hole and push the glass of beer saying, "Here I am pushing the glass of beer right through this small hole."  You lose!

    Lots of  sucker bets, like the one above, depend upon wordings that have double meanings or are misleading.  There are others that are worded clearly and accurately but rely on the bettor's misunderstanding of mathematics.  For example, there is a whole class of sucker bets that depend upon confusing the notions of mean and median; we will look at some of those ideas in later articles.  The most fascinating proposition bets, however, are those that don't rely on any sort of misunderstanding; they legitimately fool our perception and intuition.

    Did I ever fall for such a bet?  I'll say!  Years ago in Gainesville, Florida I was having a beer in some joint along 13th Street, it had the name 'Gator' something or another, when the bartender offered me the following proposition.  He dumped out a full carton of 24 beer cans on the bar and offered to bet me $5 that he could put 25 cans of beer, 3 packages of cigarettes, and 4 boxes of matches in the carton, then close it up completely without bending the sides, top, or bottom.  Everyone knows that there is nobody on earth smarter than a college sophomore or a mathematics graduate student, the latter being me at the time.  So, I took the bet fully convinced that what the barkeep had proposed was physically impossible.  Stuuuupid!  I watched while the bartender did just what he had promised to do - no trickery, he really did it.  You don't believe me?  Okay, I'll bet you $10 that now I can do it.

    Of all such proposition bets that defy our intuition and perception, there is none more intriguing than the famous Birthday Problem.  Let me set up this classic bet for you.  Suppose that you are in a room with 30 people.  Someone offers you the following wager.  "I'll bet you even money that at least two people in this room have the same birthday."  Should you take this bet?  After all, there are 365 days in a year (February 29 will be considered as March 1st).  Let's have a look.

    There are two notions that we need to introduce in order to address the Birthday Problem.  The first is quite easy.  Given an event E, let me denote the event that E doesn't happen by the notation E'.  Thus, E' is the set of all outcomes in the sample space S that are not in E.  The event E' is called the complement of E.  Notice that the simple collection {E, E' } forms a partition of the sample space S (see my July article at this site) so we have:

P(E) + P(E' ) = P(E or E' ) = P(S) = 1 (1)

where P represents a probability measure on S.  It follows at once from (1) that

P(E) = 1 - P(E') (2)

In words, relation (2) simply says that the probability of an event E is just 1 minus the probability that the event E doesn't happen.

    The second notion we need is that of conditional probability.  We are going to be using this idea in many applications other than the Birthday Problem, so it behooves us to take a good look at it.  To do so, we first have to address another matter, namely the meaning of the conjunction 'and'.

    In my June article at this site, I looked at the word 'or' and commented on the annoying fact that this same word is used for two different logical ideas.  To rid ourselves of such ambiguity, we agreed that whenever we write 'or' we will mean the 'inclusive or' which is sometimes written 'and/or'.  Happily, the word 'and' does not suffer such ambiguity.  Given two events A and B, we define the event A and B to be the event that both A and B occur.  Thinking of events as subsets of a sample space, which is what we always do, this means that A and B is the set of all outcomes that are in both A and B, that is, are common to both.  In set builder notation, we would write

A and B = { x | x is an outcome in A and x is an outcome in B} (3)

A simple example would help here.  Suppose that we consider a Roulette game and we make two bets, one on the first dozen numbers and the second on the odd numbers.  Let A represent the event of hitting the first dozen (so A is the set of numbers 1 through 12) and let B represent the event of hitting an odd number (so B is the set of odd numbers 1 through 35).  The event A and B represents the event of winning both bets so that

A and B = {1, 3, 5, 7, 9, 11} (4)

the numbers common to both.  With this notion in mind, let us return to conditional probability.

    Given two events A and B, we write the conditional probability of A given B as P(A | B).  But what does this mean?  The idea is that we have been given the information that the event B has occurred, and we would like to calculate our chances that A will occur (or has occurred) given that we know B has occurred.  Often this probability is obvious.  For example, consider an ordinary deck of shuffled playing cards and let the operation be that of dealing the top two cards in order.  Let B be the event that the first card drawn is a queen and let A be the event that the second card drawn is a queen.  What is P(A | B)?  Clearly if B has occurred the remaining deck has 51 cards of which 3 are queens.  The probability of A occurring in this situation is, therefore, P(A | B) = 3/51.  In a similar fashion, P(A | B' ) = 4/51.

    The trick in deriving an expression for P(A | B) is to observe that if event B has occurred, and then event A occurs as well, we know that event A and B has occurred.  In terms of outcomes, if I know that my outcome, call it x, occurred in B, and that event A also occurs, then x has to be in the event A and B.  It seems reasonable, therefore, to suppose that P(A | B) is proportional to, that is, is a multiple of the number P(A and B).  Let's make this assumption and see where it leads us:

P(A | B) = k P(A and B) (5)

To evaluate k, let us ask: "What is P(B | B)?"  What is the probability that B occurs given that B has occurred?  If B has occurred, then clearly it is certain that B has occurred.  Thus, P(B | B) = 1.  Setting A equal to B in expression (5) and noting that B and B = B (what is the set of outcomes common to B and itself?) we have

P(B | B) = 1 = k P(B) (6)

Solving (6) for k = 1/P(B) and substituting the result into (5) we obtain

P(A | B) = P(A and B) / P(B) (7)

Let's check the formula (7) using our card example above.  It is clear that P(B) = 4/52.  How do we calculate P(A and B)?  We know how to do that from last month's Keno article.  This is just the probability that we select 2 queens from a deck of 52 cards.  There are C(4, 2) ways to select 2 of the four queens and there are C(52, 2) ways of selecting 2 cards from 52.  Hence

P(A and B) = C(4, 2) / C(52, 2) (8)
  = [(4 x 3) / (2 x 1)] / [(52 x 51) / (2 x 1)]  
  = [(4 x 3) / (2 x 1)] x [(2 x 1) / (52 x 51)]  
  = (4 x 3) / (52 x 51)  
  = (3 x 4) / (52 x 51)  

Dividing the number at the end of (8) by 4/52, we end up with P(A | B) = 3/51, which, as we already know, is the correct answer.  It appears, therefore, that formula (7) is correct.  Wait a minute, though.  Doesn't the calculation we did in (8) look more complicated than what we did in the previous paragraph?  I'll say.  It reminds me of counting insects by counting their legs and dividing by 6.  The fact is that although the formula in (7) is useful for proving general facts (theorems) about conditional probabilities, and we will exploit it in such endeavors later in these articles, it is not frequently used to actually calculate a conditional probability.  So why am I bothering you with it?  Here's why.  If we rewrite (7) as

P(A and B) = P(A | B) P(B) (9)

we have a formula that can be used to calculate P(A and B).  Look how easily we obtained both P(B) = 4/52 and P(A | B) = 3/51 in our card dealing experiment.  Using these results we could have easily obtained the number in (8) using (9):

P(A and B) = P(A | B) P(B) = 3/51 x 4/52
= (3 x 4) / (52 x 51)
(10)

See how easy that was?  We will be doing a lot more with conditional probabilities later in these articles, but for now (2) and (9) are all we need to solve the Birthday Problem.

    Things are not always as they seem.  There is a story about a Yankee who is driving through Mississippi.  He stops at a general store in a small rural town to buy a cold drink.  When he enters he sees several locals sitting around shooting the bull and one of them has a dog lying at his feet.  Trying to ingratiate himself with the local populace he addresses the man and asks, "Does your dog bite?"  "Nope," comes the reply, whereupon the man reaches down to pet the dog.  Immediately the dog snaps at him.  Startled, the Yankee says, "I thought you said your dog doesn't bite."  "Well," came the reply, "that there ain't my dog."

    The Birthday Problem wager seems like a good one.  After all, there are 365 days in a year and there are only 30 people in the room.  It sounds like there is less than one shot in ten that at least two people will have the same birthday.  But things are not always as they seem.

    Let E represent the event that at least two people among the 30 have a common birthday.  Then E' represents the event that among the 30 people, no two have the same birthday.  According to (2), therefore, we can calculate the probability of E happening by calculating the probability of E' happening and subtracting the result from 1.  In order to calculate the probability of E', I am going to solve a slightly more general problem.  Specifically, let Bk represent the event that among k people, no two have the same birthday.  Note that E' = B30.  I want to calculate P(Bk) for any k.  It will be convenient to think of the people lined up and entering the room one by one so that their birthdays can be determined in an orderly fashion.  I now define the event Ak+1 to be the event that when the k + 1st person enters the room and announces his or her birthday, it is different from the birthdays of the k people that had previously entered the room.  This said, I think it is easy to see that

Bk + 1 = Ak + 1 and Bk (11)

Let me say it in words and I think you'll get the picture.  If k people, having entered the room, all have different birthdays (event Bk ) and the k + 1st person enters the room and has a birthday different from each of the k people already in the room (event Ak + 1), then all k + 1 people now in the room have different birthdays (event Bk + 1).  Now from (9) and (11) it is easy to see that

P(Bk + 1) = P(Ak + 1 and Bk) = P(Ak + 1 | Bk) P(Bk) (12)

Relation (12) is known as a recursion relation.  Watch how nicely it works.  P(B1) = 1 since there are no other people in the room.  Now the next person enters:

P(B2)  =  P(A2 | B1) P(B1) (13)

What is P(A2 | B1)?  Well, there are 364 possible birthdays that are different from the first person, so P(A2 | B1) = 364/365.  Since P(B1) = 1, we have by (13)

P(B2)  =  364/365 (14)

Now from (12)

P(B3)  =  P(A3 | B2) P(B2) (15)

But P(B2) has already been calculated in (14).  What is P(A3 | B2)?  Easy!  If B2 has occurred, then there are two people in the room with different birthdays.  There are 363 birthdays left that differ from these first two, so P(A3 | B2) =363/365.  From (15), therefore, we have

P(B3)  =  363/365 x 364/365 (16)

Again from (12)

P(B4)  =  P(A4 | B3) P(B3) (17)

P(B3) was calcualted in (16) and I'm sure you see by now that  P(A4 | B3) = 362/365.  Thus

P(B4)  =  362/365 x 363/365 x 364/365 (18)

A bit of cogitating shows that, in general, we have

P(Bk) = 364/365 x 363/365 x 362/365 x ...
x (366 - k)/365
(19)

This is obviously a job for a computer or calculator.  Putting k = 30 in expression (19), our trusty computer comes up with

P(E) = 1 - P(E' ) = 1 - P(E30) = 1 - 0.22937 = 0.7063 (20)

Wow, there is a 70% chance that two people have the same birthday.  Using this figure it is easy to calculate that the proposer's edge in an even money bet with 30 people in the room is around 41.26% (0.2063 / 0.5).  This is an edge akin to the Quick Draw game from last month's article.  If you raise the number of people to 34, the proposer's edge jumps to 59.06%; in this case, the probability of at least two people having the same birthday is almost a 4 to 1 shot.  Incredibly, and you can check this yourself using a hand calculator and (19), the bet is slightly better than even with just 23 people in the room.  Hard to believe isn't it?

    So, here's a question for you.  Suppose you are in a state where the license plates end with two or more numbers so that any plate ends in one of the 100 numbers 00 to 99.  Would you accept an even money wager that by the time 15 cars pass at least two will both have license plates that end in the same two digit number 00 to 99?  Things are not always as they seem!  See you next month.

Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers