Tanya Khovanova's Math Blog

Suppose you meet a friend who you know for sure has two children, and he says: “I have a son born on Tuesday.” What is the probability that the second child of this man is also a son?

People argue about this problem a lot. Although I’ve discussed similar problems in the past, this particular problem has several interesting twists. See if you can identify them.

First, let us agree on some basic assumptions:

1. Sons and daughters are equally probable. This is not exactly true, but it is a reasonable approximation.
2. For our purposes, twins do not exist. Not only is the proportion of twins in the population small, but because they are born on the same day, twins might complicate the calculation.
3. All days of the week are equally probable birthdays. While this can’t actually be true — for example, assisted labors are unlikely to be scheduled for weekends — it is a reasonable approximation.

Now let us consider the first scenario. A father of two children is picked at random. He is instructed to choose a child by flipping a coin. Then he has to provide information about the chosen child in the following format: “I have a son/daughter born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?

The probability that a father of two daughters will make such a statement is zero. The probability that a father of differently-gendered children will produce such a statement is 1/14. Indeed, with a probability of 1/2 the son is chosen over the daughter and with a probability of 1/7 Tuesday is the birthday.

The probability that a father of two sons will make this statement is 1/7. Among the fathers with two children, there are twice as many who have a son and a daughter than fathers who have two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 1/2 for the second child to also be a son.

Now let us consider the second scenario. A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son. If he has one son, he is instructed to provide information on his son’s day of birth. If he has two sons, he has to choose one at random. His statement will be, “I have a son born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun.” If his statement is, “I have a son born on Tuesday,” what is the probability that the second child is also a son?

The probability that a father of differently-gendered children will produce such a statement is 1/7. If he has two sons, the probability will likewise be 1/7. Among the fathers with two children, twice as many have a son and a daughter as have two sons. Plugging these numbers into the formula for calculating the conditional probability gives us a probability of 1/3 for the second child to also be a son.

Now let us consider the third scenario. A father of two children is picked at random. If he doesn’t have a son who is born on Tuesday, he is sent home and another is picked at random until one who has a son that was born on Tuesday is found. He is instructed to tell you, “I have a son born on Tuesday.” What is the probability that the second child is also a son?

The probability that a father of two daughters will have a son born on Tuesday is zero. The probability that a father of differently-gendered children will have a son who is born on Tuesday is 1/7. The probability that a father of two sons will have a son born on Tuesday is 13/49. Among the fathers with two children, twice as many have a son and a daughter than two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 13/27 for the second child to also be a son.

Now let’s go back to the original problem. Suppose you meet your friend who you know has two children and he tells you, “I have a son born on Tuesday.” What is the probability that the second child is also a son?

What puzzles me is that I’ve never run into a similar problem about daughters or mothers. I’ve discussed this math problem about these probabilities with many people many times. But I keep stumbling upon men who passionately defend their wrong solution. When I dig into why their solution is wrong, it appears that they implicitly assume that if a man has a daughter and a son, he won’t bother talking about his daughter’s birthday at all.

I’ve seen this so often that I wonder if this is a mathematical mistake or a reflection of their bias.

How to solve the original problem? The problem is under-defined. The solution depends on the reason the father only mentions one child, or the Tuesday.

The funny part of this story is that I, Tanya Khovanova, have two children. And the following statement is true: “I have a son born on Tuesday.” What is the probability that my second child is a son?

How many different months’ lengths are possible?

For “simplicity” let’s stick to the Gregorian calendar.

Lennart Green is an amazing magician who performs card tricks. He is so good that the judges at the competition of the International Federation of Magic Societies didn’t believe his tricks. They assumed that he used the help of stooges, and unfairly disqualified him. At the next competition he used a judge to assist him and won first place in the cards category.

I saw his performance twice. Both times he brought a woman to the stage, but each time it was a different woman. It was clear that he talked to each woman before the performance, presumably asking her permission. Furthermore, during his performances, both of the women looked slightly bored, implying that it might be not their first time. My first impression was that the women were a part of the act. I was fooled just as the judges had been fooled.

What can I say? Lennart Green isn’t skilled at picking up the right women. Watch his performance at TED, and remember that he proved that the assistants were clueless.

When the Abel Prize was announced in 2001, I got very excited and started wondering who would be the first person to get it. I asked my friends and colleagues who they thought was the greatest mathematician alive. I got the same answer from every person I asked: Alexander Grothendieck. Well, Alexander Grothendieck is not the easiest kind of person to give a prize to, since he rejected the mathematical community and lives in seclusion.

Years later I told this story to my friend Ingrid Daubechies. She pointed out to me that my spontaneous poll was extremely biased. Indeed, I was asking only Russian mathematicians living abroad who belonged to “Gelfand’s school.” Even so, the unanimity of those responses continues to amaze me.

Now several years have passed and it does not seem that Alexander Grothendieck will be awarded the Prize. Sadly, my advisor Israel Gelfand died without getting the Prize either. I am sure I am biased with respect to Gelfand. He was extremely famous in Soviet Russia, although less well-known outside, which may have affected the decision of the Abel’s committee.

I decided to assign some non-subjective numbers to the fame of Gelfand and Grothendieck. On Pi Day, March 14, 2010, I checked the number of Google hits for these two men. All the Google hits in the rest of this essay were obtained on the same day, using only the full names inside quotation marks.

• Alexander Grothendieck — 95,600
• Israel Gelfand — 47,900

Google hits do not give us a scientific measurement. If the name is very common, the results will be inflated because they will include hits on other people. On the other hand, if a person has different spellings of their name, the results may be diminished. Also, people who worked in countries with a different alphabet are at a big disadvantage. I tried the Google hits for the complete Russian spelling of Gelfand: “Израиль Моисеевич Гельфанд” and got an impressive 137,000.

Now I want to compare these numbers to the Abel Prize winners’ hits. Here we have another problem. As soon as a person gets a prize, s/he becomes more famous and the number of hits increases. It would be interesting to collect the hits before the prize winner is announced and then to compare that number to the results after the prize announcement and see how much it increases. For this endeavor, the researcher needs to know who the winner is in advance or to collect the data for all the likely candidates.

• Jean-Pierre Serre — 63,400
• Michael Atiyah — 34,200
• Isadore Singer — 44,300
• Peter Lax — 118,000
• Lennart Carleson — 47,500
• Srinivasa Varadhan — 15,800
• John Thompson — 1,610,000
• Jacques Tits — 90,900
• Mikhail Gromov — 61,900

John Thompson is way beyond everyone else’s range because he shares his name with a famous basketball coach. But my point is that Gelfand and Grothendieck could have been perfect additions to this list.

I have this fun book at home written by Clifford Pickover and titled Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning. It was published before the first Abel Prize was awarded. Chapter 38 of this book is called “A Ranking of the 10 Most Influential Mathematicians Alive Today.” The chapter is based on surveys and interviews with mathematicians.

The most puzzling thing about this list is that there is no overlap with the Abel Prize winners. Here is the list with the corresponding Google hits.

1. Andrew Wiles — 64,900
2. Donald Coxeter — 25,200
3. Roger Penrose — 214,000
4. Edward Witten — 45,700
5. William Thurston — 96,000
6. Stephen Smale — 151,000
7. Robert Langlands — 48,700
8. Michael Freedman — 46,200
9. John Conway — 203,000
10. Alexander Grothendieck — 95,600

Since there are other great mathematicians with a lot of hits, I started trying random names. In the list below, I didn’t include mathematicians who had someone else appear on the first results page of my search. For example, there exists a film director named Richard Stanley. So here are my relatively “clean” results.

• Martin Gardner — 292,000
• Ingrid Daubechies — 76,900
• Timothy Gowers — 90,500
• Persi Diaconis — 84,700
• Michael Sipser — 103,000
• James Harris Simons — 107,000
• Elliott Lieb — 86,100

If prizes were awarded by hits, even when the search is polluted by other people with the same name, then the first five to receive them would have been:

1. John Thompson — 1,610,000
2. Martin Gardner — 292,000
3. Roger Penrose — 214,000
4. John Conway — 203,000
5. Stephen Smale — 151,000

If we had included other languages, then Gelfand might have made the top five with his 48,000 English-language hits plus 137,000 Russian hits.

This may not be the most scientific way to select the greatest living mathematician. That’s why I’m asking you to tell me, in the comments section, who you would vote for.

I got this problem from my friend, a middle-school math teacher, Tatyana Finkelstein.

We have N coins that look identical, but we know that exactly one of them is fake. The genuine coins all weight the same. The fake coin is either lighter or heavier than a real coin. We also have a balance scale.
Unlike in classical math problems where you need to find the fake coin, in this problem your task is to figure out whether the fake coin is heavier or lighter than a real coin. Your challenge is that you are only permitted to use the scale twice. Find all numbers N for which this can be done.

I would like to add an extra twist to the problem above. It is conceivable that there might be several different strategies for finding the direction in which the weight of the fake coin deviates from the real coins. In this case it is better to choose a strategy that can redeem as many coins as possible — that is, to identify the maximum number of coins as real.

The number of coins you identify as real depends on the outcomes of your weighings. Then what is the precise definition of the best strategy?

Let us call a strategy k-redeem if after the weighings you are guaranteed to demonstrate that k coins are real, but you are not guaranteed to demonstrate that k+1 coins are real. Your task is to analyze two-weighing strategies and choose the most profitable one — the strategy that guarantees to redeem the largest possible number of coins, that is, a k-redeem strategy for the largest k.

Sara was born in Boston on February 29, 2008 at 11:00 am. Her parents were quite upset that their calendar-challenged daughter would only be able to celebrate her birthday once in four years. Luckily, science can help Sara’s parents. How? Sara can celebrate her birthday every year at the moment when the Earth passes the same point on its orbit around the Sun as when Sara was born.

Assuming that Sara lives her entire life in Boston and that the daylight savings time is not moved earlier into February, your task is to calculate the schedule of Sara’s birthday celebrations for 100 years starting from her birth. To simplify your homework, you can approximate one year as 365 days and 6 hours.

Joseph DeVincentis heard my prayers and created an index for MIT mystery hunt puzzles. He created it not because I requested it, but rather because he was on the writing team this year and they needed it. Anyway, finally there is an index.

I have to warn you, though, that this index was created for people who have already solved the puzzles, so the index contains hints for many of the problems and, on rare occasions, solutions.

Now I will do the math index for this year, and I promise that I will avoid big hints.

• Fun with Numbers — I got a very warm feeling from this puzzle. Some questions were right from my Number Gossip page and I knew some answers by heart.
• Planar Complex — A puzzle with complex numbers.
• Seeking Syren Scotchy — An advanced non-rectangular Nurikabe variation with colors and knight’s moves.
• Side by Side by Side — A massive logic puzzle.
• Counter Culture — So as not to reveal much info, I will just say that some math is involved in solving this puzzle.
• Expeditions — A logic puzzle that requires extra knowledge.
• Lighten Up — A Light-Up variation.
• Directions — I shouldn’t say.
• Message From Above — A math puzzle at the second stage.
• The Prodigious Riddle of Juno — A logic puzzle with a lot of if-then clauses.
• Bulls and Cows — Not surprisingly, a variation on Bulls and Cows
• Building a Mystery — Create a polyhedron.
• Trance State — Logic gates.
• Sequencer — Mirrors.
• 20,000 Leagues Under the Sea — A Battleship puzzle at the second stage.

One fine day in January 2010, John H. Conway shared with me his recipe for success.

1. Work at several problems at a time. If you only work on one problem and get stuck, you might get depressed. It is nice to have an easier back-up problem. The back-up problem will work as an anti-depressant and will allow you to go back to your difficult problem in a better mood. John told me that for him the best approach is to juggle six problems at a time.

2. Pick your problems with specific goals in mind. The problems you work on shouldn’t be picked at random. They should balance each other. Here is the list of projects he suggests you have:

• Big problem. One problem should be both difficult and important. It should be your personal equivalent to the Riemann hypothesis. It is not wise to put all your time into such a problem. It most probably will make you depressed without making you successful. But it is nice to get back to your big problem from time to time. What if you do stumble on a productive idea? That may lead you to become famous without having sacrificed everything.
• Workable problem. You should have one problem where it’s clear what to do. It’s best if this problem requires a lot of tedious work. As soon as you get stuck on other problems, you can go back to this problem and move forward on the next steps. This will revive your sense of accomplishment. It is great to have a problem around that can be advanced when you do not feel creative or when you are tired.
• Book problem. Consider the book you are working on as one of your problems. If you’re always writing a book, you’ll write many of them. If you’re not in the mood to be writing prose, then work on math problems that will be in your book.
• Fun problem. Life is hardly worth living if you are not having fun. You should always have at least one problem that you do for fun.

3. Enjoy your life. Important problems should never interfere with having fun. When John Conway referred to having fun, I thought that he was only talking about mathematics. On second thought, I’m not so sure.

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Birthdays are beneficial for your health. A new breakthrough statistical study unequivocally proved that the more birthdays one has the longer one lives.

* * *

We know through Erdös that “a mathematician is a device for turning coffee into theorems”. It thus follows by duality that a comathematician is a device for turning cotheorems into ffee.

* * *

– What do you do when you see a beautiful girl?
– I download her.

* * *

Programmers wear red T-shirts to match the color of their eyes.

* * *

We invented the decimal system, because humans have ten fingers on their hands; and 32-bit computers, because humans have 32 teeth in their mouths.

* * *

A general shows off a new tank and boasts:
– You see a tank supplied with the most modern computer technology.
– What is the speed of its computer?
– The same as the speed of the tank, of course.

I am so used to wise-men puzzles about hats, that I was pleasantly surprised when Leonid Makar-Limanov gave me a wise-men puzzle that didn’t include them.

A sultan decides to check how wise his two wise men are. The sultan chooses a cell on a chessboard and shows it to the first wise man. In addition, each cell on the chessboard either contains a rock or is empty. The first wise man has to decide whether to remove one rock or to add one rock to an empty cell. Next, the second wise man must look at the board and guess which cell was chosen by the sultan. The two wise men are permitted to agree on the strategy beforehand. What strategy can they find to ensure that the second wise man will always guess the chosen cell?

If you are stuck, there are many approaches to try. You can attempt to solve the puzzle for a board of size 1 by 2, or for a board of size 1 by 3. Some might find it easier to solve a version in which you are allowed to have a multiple number of rocks on a cell, and the first wise man is permitted to add a rock to a cell that already contains rocks.