## The Lottery Problem – Solved

On Friday I asked you the following riddle. Assume there are 2 different types of lottery. Lotto A makes you choose 6 numbers out of a possible 75,and lotto B makes you choose 5 numbers out of a possible 60 and 1 ‘mega’ number out of a possible 40. Which lottery gives you the best odds? Original problem here.

I’m sure a lot of you said Lotto B. And this is how you probably came up with the answer:

• Lotto A: 75 numbers, choose 6, so in order to choose the first number we have 75 options, for the second number, we have 74 options (since we just used up one number), etc. So we get:
$$75 \cdot 74 \cdot 73 \cdot 72 \cdot 71 \cdot 70 = 144,978,876,000$$ possibilities
• Lotto B: Out of 60 numbers, choose 5, then out of 40 numbers, choose 1. So the same logic as in lotto A, we get:
$$60 \cdot 59 \cdot 58 \cdot 57 \cdot 56 \cdot 40 = 26,215,257,600$$ possibilities

Congratulations if you got this far! This is technically correct IF the lottery forced you to choose the numbers in a specific order. The lottery (luckily) doesn’t force you to get the order right though, you can choose the numbers in any order. So then what are the chances for each lotto? Let’s first look at an easier example.

Choose 2 numbers out of 1-10. The first number has 10 choices, the second number has 9 choices. So we get a total of 90 different options ($$10 \cdot 9 = 90$$). BUT again the order doesn’t matter. Pretend your numbers were 1 and 7. We’ll call this pair (1,7) for simplicity. Selecting number 1 and then selecting number 7 is the same thing as selecting number 7 and then selecting number 1! So for each pair […(1,7), (1,8), (1,9),(1,10), (2,2), (2,3)…] we have 2 different options. So we must divide 90 by 2 to get the real outcome of 45 different outcomes.

Now let’s take it one step further and pretend we picked a 3rd number. Let’s also pretend this number is the number 4. Earlier we saw that the outcome (1,7) could either show up as 1 then 7, or 7 then 1. Let’s suppose the first occurred (1 then 7), then by adding the number we get 3 different choices as to where to put it: 4 then 1 then 7, 1 then 4 then 7, 1 then 7 then 4. [Note that the same thing can be said if we had chosen 7 then 1]. So this time instead of having $$10 \cdot 9 \cdot 8 = 720$$ possibilities, we have $$\frac{720}{6} = 120$$ possibilities. [We get 6 by multiplying 2 with 3. 2 since we can have 1 then 7 and 3 since we can put the number 4 first, middle, or last].

Now we can see that if we added a 4th number we would have $$\frac{10 \cdot 9 \cdot 8 \cdot 7}{2 \cdot 3 \cdot 4} = \frac{5040}{24} = 210$$.

The lottery follows this same principle. So applying this logic, lets reexamine our lottery:

• Lotto A: Since 6 numbers were chosen we must take our original number and divide appropriately:
$$\frac{75 \cdot 74 \cdot 73 \cdot 72 \cdot 71 \cdot 70}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} = \frac{144,978,876,000}{720} =201,359,550$$ possibilities
• Lotto B: Since 5 numbers were chosen the first time, and only 1 the second time we must take our original number and divide appropriately:
$$\frac{60 \cdot 59 \cdot 58 \cdot 57 \cdot 56 \cdot 40}{2 \cdot 3 \cdot 4 \cdot 5} = \frac{26,215,257,600}{120} =218,460,480$$ possibilities

So you have to buy over 17 million more tickets for Lotto B to give you a guaranteed winning ticket! So if you chose Lotto A, you just won the lotto!

## The Lottery Problem

New month, new riddle! Pretend you just found a \$1 bill on the floor and your instincts are telling you to go and buy a lottery ticket. You head to the nearest liquor store and ask for a lotto. The store owner asks you which kind you would like. There is lotto A where you have to choose 6 numbers out of a possible 75, or there is lotto B where you have to choose 5 numbers out of a possible 60 and 1 ‘mega’ number out of a possible 40. You don’t want to waste your dollar, so the question is: which lottery gives you the best odds?

Come back soon to find out!

## How to find a prime

Last week we talked about what a prime number is, but we didn’t talk about a good way of finding what number is prime and what is not [other than checking if any number less than p divides that number]. This area of finding out what number is prime has been a big problem for many mathematicians and still has a lot of unanswered questions. The world of mathematics is still looking for an easy way to calculate whether a number is prime or not with a simple formula.

There are a few ways of finding primes, but the simplest (and one of the oldest) ways to finding a prime is by using a process called ‘the sieve of Eratosthenes’. Take out a piece of paper and make 10 rows and 10 columns and number them in order from 1 to 100. Row 1 should have the numbers 1-10 in order, row 2 should have 11-20 in order, etc. Then, starting from 2, cross out all the numbers that 2 divides without crossing out 2 (e.g. 4, 6, 8, 44, 96). Then go to the next number and cross out all the number that it divides without crossing that number out. (The next number is three so cross out 9, 15, 21, etc.) Then go to the next number not crossed out and cross out all the numbers that it divides without crossing that number out. (5 is the next number so we cross out 25, 35, 55, etc.) We keep going until we find all the primes less than 100. Write down all the numbers in order, and you’ll notice you have the same list that I listed last week! Cool! And that is basically the whole concept behind the sieve of Eratosthenes. You write down all the numbers and you go prime by prime removing all the ones that are not prime by going number by number. So if you wanted to find all the numbers less than 1,000 you would draw a 100 x 100 table of numbers and cross out each one as it came. This can be very time consuming! That’s why mathematicians, didn’t stop there.

A second way to see if a number is prime is to use what mathematicians like to call Wilson’s Theorem. This theorem states that if p is prime then $$(p-1)! + 1 \equiv \emph{0 mod p}$$ [Remember modulo? We’re basically saying the remainder when you divide p from (p-1)! +1 is 0.] And the theorem goes vice versa, so that if a number that is not prime is placed into the equation the remainder/modulo will not be 0. Kinda handy for looking at medium size numbers!

Another way of finding primes is to find what are called Mersenne primes. These primes were talked about by a French monk Marin Mersenne from the 17th century. They are primes of the form $$2^{m} – 1 = M_{m}$$ where $$M_{m}$$ is the mth Mersenne number. It turns out that this is a good way to find other primes. If the mth Mersenne number is prime the m is also prime. The issue is that this is a little cumbersome and doesn’t work in reverse (If m is prime it does not mean $$2^{m} – 1$$ is prime too.)

There are more complex ways of finding primes, but those are some of the most interesting and easy ones. Have a suggestion? Add a comment!

## Prime Numbers

A prime number by definition is a number p>1 such that p has no positive integer divisors other than 1 and p.

Wait… What?

Confuzzling! Let’s try to figure out what that means by breaking it down into smaller parts. Let’s first look at the section ‘is a number p>1’. This part is actually saying 2 things in 1. It’s first stating: let the ‘prime number’ be represented by the symbol ‘p’. So ‘p’ is our variable. It also states that p must be greater than 1. So, so far, p can be 1.1, 2, 940,300, $$\pi$$, etc.

The next part says ‘such that p has no positive integer divisors other than 1 and p’. First let’s look at what a ‘positive integer’ is. An integer is any whole number. Basically a number that doesn’t have any decimal part, and no fractions when simplified. So 2, 55, -7, 3.0, $$\frac{27}{9}$$, etc. are all integers. A positive integer means basically that the number must be greater than 0. Next let’s see what ‘divisor’ means. A divisor is an integer that divides another number into an integer. So since 2 divides 4 into 2, then 2 is a divisor of 4. Since 3 divides 99 into 33, then 3 is a divisor of 99. So a ‘positive integer divisor of p’ is a number that divides p, is a whole number, and is greater than 0. So if we let p be 4, then 1, 2, and 4 are all positive integer divisors of p.

But, we are stating that p does NOT have any positive integer divisors except for p and 1. So 4 is not prime since 2 is also a positive integer divisor. 2 is a prime number since it is greater than 1, and it’s only positive integer divisors are 1 and 2. 9 is not prime even though it is greater than 1 since it’s positive integer divisors are 1, 3, and 9. 6.5 cannot be prime since it is not an integer.

So by going number by number we can get a huge list of prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 …
To name the first few. You can go and look at the long list of prime numbers provided by sloane which is an online encyclopedia of integer sequences (or integer patterns). Here is a link: [Sloane]

At about this point you may want to exclaim ‘Wait a second! What about the number 1?! It can only be divided by 1 and itself too, why is 1 not classified as a prime?’ Originally the number 1 was considered a prime number, but a lot of mathematical concepts that have to deal with primes usually turned out funky when dealing with the number 1. When 1 was considered a prime, the concepts worked perfectly sometimes, but failed at other times, but they all succeeded if the number 1 was not considered a prime or a composite. So instead of keeping it as a prime number, it was eventually relegated to be just a number, neither prime nor composite (explained in a few!) That is why if you pay close attention to our definition we say our prime p has to be greater than 1.

So we mentioned that the number 1 cannot be composite either, what’s composite?! A composite number is just an integer c > 1 such that c is not prime. With all the information I gave you above, I’m sure you can figure this one out =D

Questionably yours,
The Cali Garmo

## Monty Hall Problem – Solved

Before the holidays I posted a question to you all to see what you thought. The question was the famous Monty Hall problem where there are 3 doors and a contestant is told 1 has a fabulous prize while the other 2 have goats. The contestant selects a random door and the host reveals one of the doors that have a goat and gives you a chance to switch doors if you want or not. The question was: is it better to keep the original door, switch doors, or are the odds the same? Here’s the link if you want a refresher.

So most people will look at this problem and say the chances are the same whether you switch or keep the same door. The logic usually follows like this: There are 2 doors left. One has a present, the other a goat, so there is a 50% chance that either of them will have a goat. Thus each door has the same probability of having the prize.

Although this is sound logic, there is only one big mistake, the assumption that the 3rd door doesn’t matter. Let’s start from the beginning and work our way up in order to see what the answer is. Since we first choose a random door let’s pretend it’s door A. We now have 3 options: the present is behind door A, the present is behind door B, the present is behind door C.

1. If the present is behind door A: We chose door A, so the host then chooses either door B or C (both have goats so it doesn’t matter) and opens it. If we switch to either B or C (whichever the host does not choose) then we lose, if we stay on A then we win.
2. If the present is behind door B: We chose door A, so the host must select door C to reveal the goat. So if we switch to B we win, and if we stay on A then we lose.
3. If the present is behind door C: We chose door A, so the host must select door B to reveal the goat. So if we switch to C we win, and if we stay on A we lose.

Since those are the only doors available those are only ways of playing. Now let’s look at the winning ratio. If we stayed the same we would have won 1 game. If we switched then we would have won 2 games. So by switching doors you have increased your chances of winning to 66%! So remember to always switch when given a choice!

## Monty Hall Problem

Time for a pop quiz! Say you are on a game show. You are presented with 3 different doors and are told that 2 of the doors contain goats and the 3rd door contains 6 quadrillion dollars! You are told to choose which door has the money, so you choose door number 1. At this point the game show host turns to you and says, “What a choice! Let’s look at what was behind door number 2. It was a goat! I’ll give you another chance since you know that the second door was a goat. Do you want to stay with door number 1 or switch?” What is the better choice: to stay with door number 1, to switch to door number 3, or do both doors have the same probability of having the goat/money?

Key points:

• you choose a random door out of 3 doors (2 of which are ‘bad’ and 1 of which are ‘good’)
• the game show host reveals that one of the other 2 doors is one of the ‘bad’ doors
• you are asked whether you would like to keep the original door or switch
• Which is better to stay, to switch, or are both the same chance?

This is called the Monty Hall problem because there used to be a game show back in the day that was put on by a guy named Monty Hall. I’m sure it was a super fun show!!!

[Solution after the holidays!]

## Triangles – Fun Fact

Fun fact! In the real world, if you take a triangle and add up the angles, they can equal more than $$180^{\circ}$$!

In school we always learn that the angles of a triangle always add up to $$180^{\circ}$$, but that is not always the case, especially when you look at our planet! Don’t believe me? Let’s prove it!

Proof:
First let’s define what a triangle is. A triangle is 3 ‘straight’ lines that intersect at 3 points. By straight we mean a line that doesn’t curve to the right or to the left. We also notice that since we are using Earth as our reference point, we are doing all calculations on a sphere.
Ok, so let’s start! Pretend you are standing somewhere on the equator on Earth. Also, pretend you have outrageously long legs that allow you to travel really really far really quickly. Now walk north until you hit the North Pole. That’s 1 straight line, 2 more to go. So make a $$90^{\circ}$$ turn right and walk south all the way back down to the equator. That’s 2 lines, 1 more to go. Now in order to walk to our starting point we need to make a right turn. In fact we need to make a $$90^{\circ}$$ turn. (Cause we are heading directly south and now need to go directly west) Once we get to our starting point we notice that this final angle also equals $$90^{\circ}$$. (Since we are looking west, but must turn north in order to ‘retrace’ the triangle) And that’s 3 straight lines. Adding up the angles we get: $$90^{\circ} + 90^{\circ} + 90^{\circ} = 270^{\circ}$$

Oh yeah! Take that geometry. I just kicked your butt!!! This type of geometry is called Spherical Geometry and a lot of mathematicians study it in order to better understand the universe.

And thats all from me until after the holidays. Armenian Christmas isn’t until January 6 so Ill be gone for the next 2-3 weeks and will return with some more math funness afterward. *lates*

## Modulo

Modulo! Mod you who? Modulo: A complicated way to say the remainder when you divide two numbers!

So what is modulo exactly and how does it work? The easiest way to break it down is to look at an equation: $$a \equiv \emph{b mod m}$$

What this means is that basically if you take a and divide m then you get a remainder of b. Let’s try a few examples to understand it better. Say $$a = 10$$ and $$m = 7$$. Then $$\frac{10}{7} = 1 \cdot 7 + 3$$ So $$b = 3$$. Therefore we have the modulo equation: $$10 \equiv \emph{3 mod 7}$$ We can do the same with negative numbers! Let $$a = -5$$ and let $$m = 7$$ again. So now we have: $$\frac{-5}{7} = -1 \cdot 7 + 2$$. So we have $$b = 2$$ and $$-5 \equiv \emph{2 mod 7}$$

Well, that was simple you may be thinking, but when would I use stuff like this in real life?! We actually use it DAILY without even knowing it! Want a quick example? Just look at a clock. What happens after you hit 12:59 (or 23:59 for some of you)? It goes straight to 1:00 (or 0:00). We use modulo to keep track of time! If its 3:00 right now and we go 30 hours in the future, then what time is it? Well we have $$a = 30 + 3$$ and $$m = 12$$ Then $$\frac{33}{12} = 2 \cdot 12 + 9$$ [or] $$\frac{33}{24} = 1 \cdot 24 + 9$$ and we have that the time is b which is 9!

What else can we do with modulo in the world of mathematics? Well it turns out that with such a simple concept we can actually compose a whole lot of theorems! Here are some theorems to keep you entertained. If you dare, try and prove them! (Note for those who try and attempt these, proofs require additional knowledge in division and how certain numbers divide other numbers.)

Theorem 1:
Let a, b, l and m be integers and let l and m be greater than 0. Now let $$a \equiv \emph{b mod m}$$. With this information we can show that $$a^{l} \equiv b^{l} \emph{ mod m}$$

Theorem 2 (By John Wilson):
Let p be prime. Then $$(p – 1)! \equiv \emph{-1 mod p}$$

Theorem 3 (By Fermat):
Let p be prime and a be a whole number that is not divisible by p. Then $$a^{p – 1} \equiv \emph{1 mod p}$$

These theorems go on and on and there are hundreds that span from the concept of modulo. In the real world we also use modulo when it comes to creating/breaking codes, ISBN numbers, computer programming languages, and a ton of other places you probably never thought of!

For a good introduction to these types of numbers look to the following book: Elementary Number Theory and its applications by Kenneth H. Rosen (AT&T, 2005)

## Infinity

Say you have an infinite number of camels walking through the desert and a little baby camel is born, how many camels are there now? Surely there are still an infinite number, and surely since we added 1 more camel this infinity is greater than the original infinity. But is it even possible for infinity to be greater than infinity? So confusing!

The biggest error most people make when talking about infinity is to think of infinity as a number. In reality, infinity is not a number, it is a concept. By definition infinity is a quantity without end. So if you take a number and you keep adding 1 to it, then you are going toward infinity, but you never actually hit infinity since infinity is not a number.

So since infinity is not a number, what happens when you ‘add 1’ to infinity. As we saw earlier if we keep adding one we still have an infinite amount. So by adding 1 we are still at infinity. Now looking at the number of camels, even though we added 1 camel to the group, we still have an infinite number of camels and so the ‘number of camels’ has not changed. It’s a weird concept to grasp, but this concept (created by Georg Cantor) helps us understand infinity.

So then we are naturally inclined to ask, can we ever make a certain infinity greater than another infinity? Weird concept to think about, but thanks to Cantor and a bunch of other super smart mathematicians, we now know the answer is yes!

This concept is created in Set Theory where we look at the number of elements in a set. Cardinality is just the number of elements in a set, and a set is just a collection of objects. So if you have a group of 10 people, then the cardinality of the group is 10. The cardinality of the number of months is 12, the cardinality of the number of weeks is 52, the cardinality of the set of birthdays is 366. Well how about the cardinality of a set like the natural numbers? The natural numbers are the numbers 1, 2, 3, 4… So what is the cardinality since it seems like there is no end to the number of items in the set.

This is where the concept of infinity comes into play. Since the number of items in the set of natural numbers never ends, there are an infinite number of them. We call this type of infinity, countably infinite. The reason we call it countable is because we as humans can sit there and count them. We may never be able to hit the end (since there is none), but we can count them. (We can first say 1, then say 2, then say 3, etc. And if you say 1 number per second for 70 years you’d hit 2,209,032,000 !)

Well are there more countably infinite sets? Yes, integers and rational numbers are also countable! We can count integers by counting them in this way: 0, 1, -1, 2, -2, 3, -3, etc. How about rational numbers?! There seems to be a crazy amount of rational numbers, and in between 0 and 1 there are an infinite number, so they must not be countably infinte, surely. But they are!

How did mathematicians find this out? They took the rational numbers (which are all integers and fractions) and put them into a grid like the following:

 $$\frac{1}{1}$$ $$\frac{2}{1}$$ $$\frac{3}{1}$$ … $$\frac{1}{2}$$ $$\frac{2}{2}$$ $$\frac{3}{2}$$ … $$\frac{1}{3}$$ $$\frac{2}{3}$$ $$\frac{3}{3}$$ … … … … …

Now in order to make it countable just go in diagonals. So our sequence would be: $$\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, etc.$$. Then in order to get all the negative numbers, just switch in between the two (positive, then the same number negative like we did for the integers.)

Now how about the set of all numbers (aka real numbers, which is the rational numbers with the irrational numbers such as $$\pi$$) It turns out that the real numbers are actually not countable. We can see this because there is no way to sit and count the real numbers. For example let us start with the number 1. What number do we choose next? 1.01 is wrong, because 1.001 is less and is also a real number. But we can’t choose that since 1.0001 is also a real number, so no matter what number we choose there is no logical next number to choose since we can keep finding a smaller number closer to 1.

Wait, now we have 2 different types of infinity! Countable and uncountable infinities. The cool thing is that if you look further into set theory you’ll find out that uncountable infinity has a higher cardinality than countable infinity! So there are the same number of whole numbers as rational numbers, but there are more real numbers than any other type of number. So cool!

If you want further information on Set Theory to be able to see different types of infinity I recommend A transition to advanced mathematics by Douglas Smith, Maurice Eggen, & Richard St. Andre (Books/Cole 2001) Have fun counting!

## And then there was light…

Do you have an unquenchable thirst for mathematics? From Euclid to Newton to Einstein, you want it all. You want to know the details about topologies and about the intriguing nature of prime numbers and Fibonacci numbers. Or maybe you are sitting in your algebra class confused as all else and can’t understand a word the teacher says. Well you are at the right place! The Cali Garmo was placed upon this world to help you understand math like you’ve never understood it before, learn concepts you’ve never learned before, and run free in a world you’ve never experienced before. So turn on your noggin, hunker down with pen and paper and get ready to experience the funnest subject on the planet in this weekly blog!