Tag Archives: math tricks

Math Tricks Prove You Have Famous Ancestors

Everyone’s been fascinated during recent years about delving into their ancestry and family history.

I did a little work on ancestry.com and it wasn’t too hard to trace my ancestry back to some people who fought in the United States Civil War, and then to some who fought in the Revolutionary War, which was pretty cool, since it qualified me to join the Sons of the American Revolution.

But then, I kept going and found that I am, among other things, descended from King Edward III of England and his son, Edward the Black Prince of Wales, and that Geoffrey Chaucer is my 16th great-uncle.

At first, this seemed amazing, and I thought, “Wow! What are the chances of such a thing?”

Actually, using some basic statistics and probability, the chances are very likely that you are related to someone famous from the Middle Ages (about 400 A.D. to 1500 A.D.) or earlier, in whatever region of the world your ancestry comes from. 

As an example, let’s assume you were born in 1980, and that on average, a new generation comes along every 25 years. Then, each of your two ancestors would be descended from two more parents, and so forth. So, a century before, in 1880, that you would have 2 x 2 x 2 x 2 ancestors, or 2 to the 4th power, which would be 16 great-great-grandparents.

Add another hundred years to trace back to 1780, and the number of spots to fill in your family tree would equal 2^8, or 256 potential ancestors.

In 1680, the number of spots to fill would increase to 2^12, or about 4,000 potential ancestors.

In 1580, 2^16 (about 64,000 ancestors)

In 1480, 2^20 (about a million ancestors)

In 1380, 2^24 (about 16 million ancestors)

In 1280, 2^28 (about 250 million ancestors)

Just for fun, go another 50 years to 1230 A.D., making the exponent a nice round 2^30, which is over a billion ancestors.

But in 1230, there were only about 400 million people living on Earth.  What does that mean?

It means that there are LOTS of people over-counted in that family tree, because the lines actually come back together over and over. Not just because second, third, and fourth cousins marry each other — if you go far enough in time, everyone is technically a cousin to everyone else.

So, this means that you are you are descended from a lot of different people by a lot of different lines of inheritance — lots of VERY DISTANT cousins marrying each other — no matter what your ethnic background is.  There are a billion slots that have to be filled in on the ancestral tree, and only 400 million people to put in them, so a lot of people have to be filling multiple slots on your ancestral tree.

Does this mean you are descended from every person who was alive in 1230 A.D.?  No — depending on your ethnic background, your ancestry could be focused on groups in the Americas, or south Africa, or southeast Asia, China, Japan, and so forth, since those groups did not largely mix with other groups worldwide until later in history; and plus, there were many people alive in 1230 A.D. who don’t have living descendants today.

But if you know some of your ancestors were from Europe or around the Mediterranean, then the odds are that anyone who was alive in 1230 A.D., lived in Europe or around the Mediterranean, and has living descendants, is likely one of your ancestors, including all of the Holy Roman Emperors, and the peasants they ruled; William the Conqueror and many of his knights; Lady Godiva; Pope Innocent III; and so on.  Basically, everyone you read about in the history books; the conquerors and the conquered, the famous and the unknown — if they had descendants, there’s a pretty good chance they’re one of your ancestors.

Pretty cool, huh?

Math tricks like this are fun to play with. If you enjoy learning things like this, you might enjoy taking the GMAT or the GRE. Contact me anytime to discuss options for test prep and tutoring on the SAT, ACT, LSAT, GMAT, GRE, and MCAT!

LSAT Logic Games – The “Einstein Puzzle”

An LSAT student of mine alerted me to the Zebra Puzzle, which he found and sent to me. It’s often called “Einstein’s Puzzle”, and it’s said that Einstein wrote it as a child and “claimed that only 2% of the population would be able to solve it.” It was originally published in Life International Magazine on December 17, 1962, and the contents make it unlikely that Einstein was the actual author, but it’s fun anyway! When my student sent it to me, I actually pulled over on the side of the highway and worked it out for 30 minutes at a gas station because it’s such a cool puzzle.  I made a couple of edits, inside brackets, for clarity.  The solution to the puzzle can be found here, but try not to look immediately!

  1. There are five houses [in a row].
  2. The Englishman lives in the red house.
  3. The Spaniard owns the dog.
  4. Coffee is drunk in the green house.
  5. The Ukrainian drinks tea.
  6. The green house is immediately to the right of the ivory house [as viewed from the street].
  7. The Old Gold smoker owns snails.
  8. Kools are smoked in the yellow house.
  9. Milk is drunk in the middle house.
  10. The Norwegian lives in the first house.
  11. The man who smokes Chesterfields lives in the house next to the man with the fox.
  12. Kools are smoked in [a] house next to the house where the horse is kept. [The original puzzle says “the” house, but “a” house is more clear.]
  13. The Lucky Strike smoker drinks orange juice.
  14. The Japanese smokes Parliaments.
  15. The Norwegian lives next to the blue house.

If one resident drinks water, which is it?  If one resident owns a zebra, who is it?

The LSAT includes an Analytical Reasoning section (which we call “Logic Games”) that tests your ability to make deductions and solve problems like the above (but a good deal less complex).  If you like problems like this, you might enjoy LSAT prep and going to law school.

Contact me for information about test prep classes and tutoring, either locally in Austin or online via The Princeton Review’s LiveOnline classrooms!

The “Choosing Among Three Doors” Problem

“On a game show, a new car is hidden behind one of three doors. You choose Door 1. The host opens Door 3 — empty — and gives you the option to change your choice. Would you switch to Door 2 or stick with Door 1?”

This question is often called the “Monty Hall Problem”, after the show “Let’s Make a Deal”, and was discussed in the movie “21”.

The question can be answered using knowledge of basic math probabilities.

Probability is always described as a number between zero (no change of something happening) and one (100% chance of something happening), and is calculated using the formula P(outcome you desire) = \frac{desired}{possible}.

When you make your initial choice between the three doors, the chance is \frac{1}{3} that your choice has the prize behind it. Thus, the probability that you have chosen correctly is \frac{1}{3}, and the probability that you have chosen incorrectly is \frac{2}{3}. So, there is a \frac{2}{3} chance that the car is behind either Door 2 or Door 3.

When the host opens Door 3, revealing that it does not contain the price, he has given you additional information, but the original odds remain the same.  There is now a \frac{1}{3} chance that your original choice (Door 1) was correct (Door 1) and a \frac{2}{3} chance that one of the other doors (now, only Door 2) contains the prize, and you should switch to Door 2.

This requires assuming, of course, that the host always offers the chance to switch and that you can’t gain any information from his behavior.

If it’s difficult to see why this makes sense, imagine the same scenario with a million doors.  You pick Door 1, and the host opens 999,998 other doors, showing them to be empty, leaving only your door and one other door.  Would you switch?  There’s a \frac{1}{1,000,000} chance that you picked the correct door, and a \frac{999,999}{1,000,000} chance that the other remaining door holds the prize.

The GMAT Test includes many questions that test your ability to break down and solve problems involving probability.  If you like problems like this, you might enjoy taking the GMAT and earning an MBA.  Contact Bobby Hood Test Prep for information about classes and tutoring, either locally in Austin or online via The Princeton Review’s LiveOnline classrooms!

The “Shared Birthday” Problem

“If a room has 23 people in it, what is the approximate chance that at least one pair of people in the room share the same birthday?” 

This question is similar to GMAT math questions that test your knowledge of probability theory.  It requires too much calculation to actually be used as a GMAT question, but the technique for solving it is exactly the same.

Often, it’s easier to calculate the probability of a set of events NOT happening than it is to calculate the probability of that set of events actually occurring. 

We can use the fact that, for any given set of events, the probability that it WILL occur plus the probability that it will NOT occur equals 1:

  • P (birthday is shared) + P (nobody shares a birthday) = 1

So, since it’s easier to solve for the probability that nobody shares a birthday, we can solve the equation for the probability that there is at least one shared birthday:

  • P (birthday is shared) = 1 – P (nobody shares a birthday)

So, what is the chance that nobody shares a birthday?  

Start with the chances of two people not sharing a birthday.  For the first person chosen, we choose from 365 choices of birthdays available: \frac{365}{365}, which equals a chance of 1.  For the second person, we have 364 remaining birthdays to choose from out of 365:  \frac{364}{365}.  Multiply the two together and you get a chance of \frac{364}{365}, or about 99.7%, that those two people have different birthdays, and \frac{1}{365}, or 0.3%, as the chance that they share a birthday.

For each person added, we have one fewer birthday to choose from (363, 362, 361, etc.) out of the 365 days; multiply them all together to get the result:

  • \frac{365}{365}\frac{364}{365}\frac{363}{365}\frac{362}{365}...\frac{343}{365} =

Do all the painful math (a spreadsheet helps a lot, unless you are a math savant), and you get about 49.3% chance that a birthday is NOT shared; subtract from 100% and you have a 50.7%, or about 1 in 2, chance that a birthday is shared between at least two people in the room.

So, next time you’re at a party and you count 23 people, it’s a fun bet to make that there will be a shared birthday.  32 people brings the odds up to 75%, and 40 people puts it right about at 90%.

(Yes, I know we’re ignoring the poor folks who were born on February 29 and only have a quarter as many birthdays to celebrate as the rest of us, but they change the odds only a tiny amount.)

If you enjoy math problems like this, you would probably enjoy, and do well on, the GMAT test.  Contact Bobby Hood Test Prep and I’d be happy to discuss the various class and tutoring options for the GMAT, both locally in Austin and online through The Princeton Review’s LiveOnline classrooms.

The “Ages of Three Children” Puzzle

A census taker approaches a woman leaning on her gate and asks about her children. She says “I have three children and the product of their ages is 36. The sum of their ages is the number on this gate.” The census taker does some calculation and (correctly) claims not to have enough information. The woman enters her house, but before slamming the door tells the census taker, “I have to see to my eldest child, who is in bed with measles”. The census taker then departs, satisfied, knowing the answer.

This question tests your ability to factor a number.  Whenever a problem discusses “products” or “factors”, or “divisibility” or a “remainder of zero”, that problem likely involves breaking down a number into its factors.

The factors of 36 are 1 and 36; 2 and 18; 3 and 12; 4 and 9; and 6.  (1, 2, 3, 4, 6, 9, 12, 18, 36). So, each child’s age must be one of those factors, and they must multiply to 36 and add up to the mysterious “number on the gate”, which is not given. 

The possible ages of the children, then, are:

1, 1, and 36 (and the number on the gate would be 38)

1, 2, and 18 (and the number on the gate would be 21)

1, 3, and 12 (and the number on the gate would be 16)

1, 4, and 9 (and the number on the gate would be 14)

1, 6, and 6 (and the number on the gate would be 13)

2, 2, and 9 (and the number on the gate would be 13)

3, 3, and 4 (and the number on the gate would be 10)

Based on the above, the census taker must not have been able to figure out the ages using the number on the gate — so the number on the gate must be 13 (the only sum that could come from more than one sets of factors). So, the census taker knows that the children are either 6-year-old twins and a 1-year old, or else a 9-year-old and 2-year-old twins

Once the woman mentions an “eldest child”, the census taker knows that the correct answer is the 9-year-old and 2-year-old twins.

The GMAT Test includes many questions that test your ability to break down and solve problems involving probability.  If you like problems like this, you might enjoy taking the GMAT and earning an MBA.  Contact me for information about classes and tutoring, either locally in Austin or online via The Princeton Review’s LiveOnline classrooms!


The “Dog Running Between Two People” Problem

“Jim & Jan, 600 feet apart, walk toward each other. Jim walks at 100 ft/min, while Jan walks 50 ft/min. Their dog is standing next to Jim, and when Jim starts to walk toward Jan, the dog runs toward Jan at 300 ft/min, then turns around and returns to Jim, and so forth until all three meet.  Approximately how many feet does the dog run?”

The first time you encounter a question like this, it sounds like it involves calculus, or at least a lot of algebra. The GMAT loves including questions like this on its test.

One of the common question types on the GMAT is the “rate problem”: you are given rates for people moving or completing a task. When two people are both doing the same “job” — for instance, walking — you can simply add their rates together to find out their combined speed.

In this case, Jim and Jan are traveling toward each other at a combined rate of 100 ft/min + 50 ft/min = 150 ft/min.  You can use the Distance formula (D = rt) to calculate that the time they will talk is equal to the Distance of 600 feet divided by their combined rate of 150 ft/sec, or 4 minutes. They won’t meet exactly in the middle — Jim will travel 400 feet, while Jan will travel 200 feet — but we don’t need to calculate this to solve the problem.

Calculating the actual distance the dog traveled would be difficult using calculus, and would require using functions and limits. However, if you know that the dog will be traveling at a constant rate of 300 feet per second until they meet, it doesn’t matter in what pattern he runs; he will be running for four minutes exactly, and in those two minutes he will travel 1,200 feet. (Ignoring any time lost for the dog turning around to change direction.)

The moral to this story:  when on a test like the GMAT, don’t fall for the test’s attempt to get you to use difficult calculations.  Look for a simple way to solve the problem; virtually all the problems can be solved in under 3 minutes with one of several basic strategies.

If you enjoy math problems like this, you would probably enjoy, and do well on, the GMAT test.  Contact Bobby Hood Test Prep to discuss the various class and tutoring options for the GMAT, both locally in Austin and online through The Princeton Review’s LiveOnline classrooms.