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Turns out we've been using binary in the kitchen for a long time:
What are the chances that a family of X children will have Y girls?
Today's Question of the Day:
Let's take a look at that triangle:
Today the triangle is generally called "Pascal's Triangle", after Blaise Pascal, in part because he studied it so thoroughly, and published a paper with many of its properties; he and Pierre de Fermat were working on the foundations of probability and statistics together, and found it singularly useful.
Pascal used "his" triangle to compute probabilities, in part because of his interest in gaming and gambling. It turns out that this "arithmetic triangle" (as he called it) is very useful for counting things.
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Notice the powers of 2....
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Let's talk about stacking. In a way, the triangle is about "piling on":
The first diagonal is just 1. 1, 1, 1,.... It's just a single point, sitting there.
The counting numbers are what you get when you keep stacking one more onto the line of numbers you've already got.
The triangular numbers are the numbers you get by stacking the counting numbers on top of each other (as line segments).
The tetrahedral numbers are the numbers you get by stacking the triangular numbers on top of each other, as triangles.
Imagine flipping coins for each ball: heads, go left; tails, go right. Let's draw the "decision tree", which is just a tree which shows the results of some sort of decision at each step.
Pascal's triangle simply doesn't care if you go right then left, or left then right: you end up in the same plinko spot. So Pascal says "2 ways to do that". "1 way to get HH", "1 way to get TT", but two ways to get a head and a tail (HT and TH). Pascal's triangle just does the accounting: \[ 1 \hspace{1in} 2 \hspace{1in} 1 \] The next row is the ways of getting three heads, two heads, one head, and no heads: \[ 1 \hspace{1in} 3 \hspace{1in} 3 \hspace{1in} 1 \]
And, by counting the leaves of the binary tree, we see that it's always a power of 2! So the sum of the rows of Pascal's triangle has to be a power of 2.... Now we understand the "power of 2" rule;
but also the rule about where adding the "two above" to get the value in the next row. That's the plinko rule!:)
Now, if each vertex represents a unique individual (we name them), then Pascal's triangle tells us all the different Facebooks possible, of each type. If everyone's friends with everyone else, then the graphs we can get will look like these, the complete graphs:
Now how many different Facebook configurations are there for, say
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The same process works if we want to decide questions like Pascal's about boys and girls, such as
You have four extra tickets to a concert; in how many ways could you choose a set of four friends from your seven favorite friends? (Notice that it's the same number of ways as the number of ways of excluding three friends: the table is symmetric!)
\[ (a+b)^n \]
So, for example,
\[ (a+b)^0=1 \] \[ (a+b)^1=1a+1b \] \[ (a+b)^2=1a^2+2ab+1b^2 \] \[ (a+b)^3=1a^3+3a^2b+3ab^2+1b^3 \] \[ (a+b)^4=1a^4+4a^3b+6a^2b^2+4ab^3+1b^4 \]
(see your assigned reading).
I hope that you noticed Pascal's triangle peeking out from the right hand sides above!
Now we can see why the 11 powers work: \[ 11^0=(10+1)^0 = 1 \] \[ 11^1=(10+1)^1= 1*10 + 1*1 \] \[ 11^2=(10+1)^2=1*10^2+2*10*1 + 1^2 \] \[ 11^3=(10+1)^3=1*10^3+3*10^2*1+3*10*1^2 + 1^3 \] etc.!
For example, suppose that the probability that you have a cold on a given winter day is \(a=1/100\); then the probability that you don't have a cold is \(b=99/100\).
Let's suppose that winter is 90 days long. What is the probability that you have a cold-free winter? We need to compute the 90th row in the table, which has some really big numbers!
\[ \left(\frac{1}{100}+\frac{99}{100}\right)^{90} \]