Flip a coin 2n times; chance of n heads

[Jamin2112]

Homework Statement

I don't my book to give me the exact wording of the question, but the essence is finding a formula for the probability of n heads after flipping a coin n times.

Homework Equations

"n choose k" = n! / k!(n-k)!

The Attempt at a Solution

Obviously you have a 1:2 chance of flipping n heads. But I'm not sure how to show this, and I think I'm supposed to use the above equation. The number of possible combinations will be 2²ⁿ; half of those will have n heads.

 

[Dick]

The odds of flipping exactly n heads in 2n flips isn't 1/2. If n is large it's extremely small. Now if you forget trying to prove it's 1/2, can you solve the problem? You have some of the right ingredients.

 

[Jamin2112]

Guys, guys, guys ...

I'll draw a picture out for y'all

After 2 flips, a 1/2 chance of getting 1 head

After 4 flips, a 6/16 chance of getting 2 heads

Not sure where those numbers come from, but ...

When there's an even number of flips --- for example, 3 --- it's impossible to have the same amount of heads as tails. Since you can have 2 heads or 2 tails, though, you can use the paths that give you 2 heads/tails and have a 1/2 chance of getting the same number of heads an tails when that path continues into the 4th flip.

(See what I'm saying? If you look at the 2nd T from the left on the 3rd flip line, you see that 2 H's preceded it, meaning you will have 2 H's and 2 T's for one of the paths that extends from that T to the next flip.)

 

[Dick]

I'm not sure what you are trying to say here. Your diagram method is correct. The odds of getting two heads in four flips is 6/16. The point of the problem is to compute that without drawing the whole diagram. Use the "n choose k" number and the 2^(2n) total possibilities observations to write a formula for the number of ways to get n heads in 2n flips.

 

[Jamin2112]

I'm not sure what you are trying to say here. Your diagram method is correct. The odds of getting two heads in four flips is 6/16. The point of the problem is to compute that without drawing the whole diagram. Use the "n choose k" number and the 2^(2n) total possibilities observations to write a formula for the number of ways to get n heads in 2n flips.

Well, yeah, there are 2²ⁿ possibilities to choose from and we want to choose the number that have half as many heads as total coins flipped. How do I find that number?

I understand how "n choose k" works, but don't see how it fits into this problem because somewhere, somehow, I have to sum up the number of heads.

 

[Hurkyl]

You're thinking of generating a piece of data by:

(2n)-many times:

Flip a coin.
Write down whether it was heads or tails.​

Count how many heads there are.
If there are n heads, keep this sequence. Otherwise discard it.​

You are obviously having trouble counting the set of all possible results of this procedure. Can you think of any other way to generate the same set of data?

 

[Jamin2112]

You're thinking of generating a piece of data by:

(2n)-many times:

Flip a coin.
Write down whether it was heads or tails.​

and counting how many sequences have n heads in them.

You are obviously having trouble analyzing this procedure. Can you think of any other way to generate the same data?

Well, yeah. I could use Stats, but this a Math Reasoning class.

This keeps track of how many heads:

+1 +0

The average is: (+1+0)/2 = 1/2

After 2n flips, the expected value is (1/2)*2n = n

The Standard Error is: [ ( (1-1/2)²+(0-1/2)² ) / 2 ]^1/2 * √(2n)

etcetera ...

 

[vela]

If you write down the 16 possible outcomes from four flips with 2 heads, you get

H H T T
H T H T
H T T H
T H H T
T H T H
T T H H

Can you see how \begin{pmatrix}4 \\ 2\end{pmatrix}=6 might apply here?

 

[Hurkyl]

Well, yeah. I could use Stats...

You haven't changed how to generate the data at all -- all you changed is what information you are extracting from the data.

 

[Jamin2112]

If you write down the 16 possible outcomes from four flips with 2 heads, you get

H H T T
H T H T
H T T H
T H H T
T H T H
T T H H

Can you see how \begin{pmatrix}4 \\ 2\end{pmatrix}=6 might apply here?

Oh ...

2=n, and that's the number you need to choose (since 2n=4).

But where did the 2^2*2 come into play for this problem?

 

[owlpride]

But where did the 2^2*2 come into play for this problem?

That's the 16 possible outcomes of 4 flips, out of which 6 = (4 choose 2) are favorable.

 

[Jamin2112]

That's the 16 possible outcomes of 4 flips, out of which 6 = (4 choose 2) are favorable.

So the probability P(n) would be

"2n choose n" / 2²ⁿ = [ (2n)! / n! (2n-n)! ] / 2²ⁿ

 

[Dick]

So the probability P(n) would be

"2n choose n" / 2²ⁿ = [ (2n)! / n! (2n-n)! ] / 2²ⁿ

That's it.