The St. Petersburg Paradox at CTY

This week, I am at my Johns Hopkins Center for Talented Youth Program, and I am taking Game Theory and Economics. Like last year, I wanted my post to be something cool I had learned that week.

First off, there is something called expected value, which means that if you do the experiment say millions of times and average all of your results, you will be extremely close to the expected value.  To determine expected value, you must multiply each outcome by the probability of that outcome occurring. Add all of those up and you have the expected value.

EV = P(a) x a + P(b) x b + ...

For a standard die roll, the expected value is:

EV = 1/6(1) + 1/6(2) + 1/6(3) + 1/6(4) + 1/6(5) + 1/6(6) = 3.5

So, the expected value is 3.5. Of course, you cannot get 3.5 on a single die roll, but if you average together a thousand rolls, you are sure to be near 3.5.

As an expected value problem to solve, our teacher gave us what is called the St Petersburg Paradox, which describes a mathematician who is told that he can play a game where he flips a coin until he gets tails. Afterwards, he will get 2^n pieces of gold, with n being the number of heads he flipped before he flipped a tails. The mathematician immediately went to determine the expected value, which is:

EV = 1/2(1) = 1/4(2) + 1/8(4) + ... EV = 1/2 + 1/2 + 1/2 + ... = infinity

This is saying that if you average all of your trials, you will get infinity. However, you have a fifty-fifty chance of getting just one piece. How can this be?

As I mentioned earlier, this is a paradox; it's mathematical answer differs from its logical answer. For a logical answer, you can cut it off at how much gold they are able to award him (they can't afford 2^50 pieces) and calculate. It still won't be very accurate until huge numbers of trials. Anyways, I found that pretty cool.

Bonus: We recently received a puzzle as well. As usual, I will provide the answer in a month.

By pure random guessing, what is the probability that you will get this answer correct.
  a. 50%
  b. 25%
  c. 0%
  d. 50%

How do we know the digits of pi?

Next Thursday is going to be Tau day. If you remember, there are many mathematicians that want the number tau, which is equal to 2π, to replace pi because of it being more natural and simple.

In honor of this special occasion, I decided to write a post about pi. I had always wondered how we know the digits of pi go on forever, and more importantly, how we figure them out.

Turns out, there is a formula. One of the common formulas is:

π = 4(1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 +...)

Let me show you how we can derive this. First off, I am going to ask a completely irrelevant question, but you will see its relevance in a moment. What is the arch-tangent of 1?

To figure this out, you would punch into your scientific calculator 1, and then the tanh button. You would get:

tanh(1) ≈ .76159...

But how did the calculator get the answer? It had to use a formula. It used the formula:

x^1/1 - x^3/3 + x^5/5 - x^7/7 + x^9/9 -...

Now, multiply this number by 4. The calculator approximated the arch-tangent, so you won't clearly see the pattern. However, the true answer is:

4tanh(1) = π

So, we plug one into the tangent formula, and we get:

1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 +...

Multiply that mess by four and we have pi.

4(1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 +...)

I found it really cool that pi has a formula this simple that mathematicians can use to calculate digits. This also proves that it doesn't terminate since the third, seventh, ninth, eleventh, and many more of them go on forever.

Problem of the Week Day 5: Week of 6/18/12 - 6/22/12

It is the last day of June’s problem of the week! Remember to email me your answers once you figured them out at Ethan@EthanMath.com. Good luck!

Easy: I like to finish off the week with some geometry. Today, we will use a rectangle.

This rectangle has an area of t and a width of n^2. Try to determine its length.

l =

To do this, you will need the rectangle’s area formula, which is:

A = lw

Say you had an area of 4 and a width of 2. You would do:

4 = l(2)

4 = 2l

4 ÷ 2 = 2l ÷ 2

2 = l

Do the same thing for this problem.

Hard: Before we begin this problem, there is a small calculation I would like you to do.

√(n - 1) = d

d =

Now, we will do some geometry. Take this trapezoid:

What is the area, which we will call z. Round to the nearest hundredth.

z =

To do this, you need to use this formula:

h(b1 + b2)/2

For example, say the top side was 3 cm, the bottom was 4 cm, and the height was 6 cm. You would do:

6(3 + 4)/2

6(7)/2

42/2

21

The area would be 21 sq. cm. Good luck!

Problem of the Week Day 4: Week of 6/18/12 - 6/22/12

Today, we are going to do some work with patterns. It won’t be too hard, but it should be fun.

Easy: Look at the sequence below:

a, h, z, n, ...

What is the tenth number in this sequence?

t = 

Hard: Take these four numbers:

a

-b

x1

x2

What pattern are these four numbers in? Determine the twentieth number in this sequence.

Problem of the Week Day 3: Week of 6/18/12 - 6/22/12

Today is day three of the problem of the week! Good luck!

Easy: Today, we will be solving for a letter, but there is one difference. The letter is inside of the equation!

zn - h^2 = (z + h + a)n

In this problem, we are trying to figure out what n equals. Let’s do an example.

4n - 3 = 7n - 6

This problem is a little bit like a seesaw, in that if you take a certain amount off of one side, you take the same amount off the other side to keep it balanced. Also, if you add a certain amount to one side, you add the same amount to the other side. If you multiply or divide by something on one side, you do the same to the other side.

The first thing you have to do is put all of the variables (letters) on one side and all of the constants (numbers) on the other side. The first thing I would to is take 4n away from both sides.

4n - 3 = 7n - 6

4n - 3 - 4n = 7n - 6 - 4n

(4n - 4n) - 3 = (7n - 4n) - 6

0 - 3 = 3n - 6

-3 = 3n - 6

We have just made it so there is only one n. This makes things much simpler. Since all of the variables are on the right, we want all of the constants on the left. To do that, we will add six to both sides.

-3 = 3n - 6

-3 + 6 = 3n + (- 6 + 6)

3 = 3n + 0

3 = 3n

We have one step left. All we have to do is divide both sides by three.

3 = 3n

3 ÷ 3 = 3n ÷ 3

1 = n

And we have solved it. n = 1. You will do the same thing with the equation up top.

Hard: 

Yesterday, you solved a system and got a quadratic equation. Today, you are going to solve the equation for x.

0 = ax^2 + bx + c

x1 = ___

x2 = ___

Please make x1 the smaller number and x2 the larger number. I will explain why there is an x1 and x2 in just a minute. First, let me explain how to solve the equation.

You could solve this equation with the quadratic formula, which I talked about a few times on this blog. You could also use techniques like completing the square, graphing, or the one I am going to talk about now: factoring.

Let’s say you have to solve the equation 0 = 2x^2 + 11x + 15. First, you have to do something called product sum. As it suggests, you need a product and a sum. The product is a x c. The sum is just b.

Product: 2 x 15 = 30

Sum: 11

Now, you need to find two numbers who have a product of 30 and a sum of 11. In this case, you can use 6 and 5.

6 x 5 = 30

6 + 5 = 11

Now, you need to do something called regrouping. For this, you will break the eleven up into the six and five.

0 = 2x^2 + 11x + 15

0 = 2x^2 + 6x + 5x + 15

Now, group together the first two terms and group together the second two terms.

0 = 2x^2 + 6x + 5x + 15

0 = (2x^2 + 6x) + (5x + 15)

Next, you will find the greatest common factor in each one. This means that for the 2x^2 + 6x, you will find a number that divides into both of those numbers. You will then factor it out.

0 = 2x(x + 3) + 5(x + 3)

Then, you will pull out the common binomial, which is x + 3. You are then left with 2x + 5.

0 = (x + 3)(2x + 5)

Since these are equal to zero, you know that one of the factors must equal zero. If neither equaled zero, then they couldn’t multiply together to equal zero. This means:

x + 3 = 0

x = -3

2x + 5 = 0

2x = -5

x = -5/2

So, x = -3 and x = -5/2. There are two solutions, which is fine. That is why there is an x1 and an x2.

Let’s try one more, just in case you are a little confused. Try 0 = x^2 + 12x + 35. First, we will find the product and sum.

Product: 1 x 35 = 35

Sum: 12

Now, we will find two numbers that have a product of 35 and a sum of 12.

7 x 5 = 35

7 + 5 = 12

Now, we will regroup by breaking the 12 into 7 and 5.

0 = x^2 + 5x + 7x + 35

Next, we will group the first two terms and the second two terms.

0 = (x^2 + 5x) + (7x + 35)

Now, we will find the greatest common factor.

0 = x(x + 5) + 7(x + 5)

Then, we will take the common binomial, and have it multiplied by what’s left.

0 = (x + 5)(x + 7)

And finally, we can solve it.

x + 5 = 0

x = -5

x + 7 = 0

x = -7

Good luck!