Does this binomial probability pyramid have a name?

[Sariaht]

a charge can be either negative or positive, computers only understand once and zeroes and for every good thing there is a bad thing.

Imagine you have a coin with one 1-side and one 0-side. If you do not flip there is a 100% chance that you won't get anything. If you flip once there is a 50% chance that you will get a 1 and 50% chance that you will get a 0.

If you flip twice there is a 25% chance that you will get two 0, 25% chance that you get two 1 and 50% chance that you will get one 0 and one 1.

This is the beginning of a possibility pyramid, it looks like this:

------------------------------------1----------------------------0 flips
-------------------------------1/2-----1/2-----------------------1 flip
---------------------------1/4-----2/4-----1/4-------------------2 flips
-----------------------1/8-----3/8-----3/8-----1/8---------------3 flips
------------------1/16----4/16----6/16----4/16----1/16----------4 flips
--------------1/32----5/32---10/32----10/32---5/32----1/32------5 flips

What is awesome with this pyramid is that there is a pattern in in it. Does this possibility pyramid have a name? It's wonderful how it makes hard math so simple.

 

[shmoe]

This is Pascal's triangle, only you've multiplied the nth row by 2^(-n). (the top row is n=0, the zeroth row)

It gives the binomial coefficients, which are also the number of ways of choosing a heads/tails sequence. e.g. the row 1 4 6 4 1 tells us $$(x+y)^4=1x^4y^0+4x^3y^1+6x^2y^2+4x^1y^3+1y^4$$.

 

[Sariaht]

thanx, might binomial coefficients have something to do with possibility in coin flipping?

 

[shmoe]

Yes.

Let's look at where the coefficient of $$x^3y^1$$ in $$(x+y)^4=(x+y)(x+y)(x+y)(x+y)$$ comes from. When you expand out these brackets, you have a choice of picking x or y from each $$(x+y)$$ term. This corresponds to a sequence of x's and y's of length 4, you can think of this as a coin with "x" on one side and "y" on the other. The information encoded in the coefficient of $$x^3y^1$$ is the number of length 4 sequences with 3 x's and 1 y, namely xxxy, xxyx, xyxx, and yxxx. 4 of them, which is the coefficient of $$x^3y^1$$ when we expand.

Same idea for all the other coefficients.

 

[Sariaht]

Perfekt, thank you!