Expected Value with 2 coin flips performed multiple times

Question

I am trying to make sense of Expected Value. Assume coin flips.

Each trial is two coin flips: { HH, TT, HT, ... }

The probabilities are:

P(H) = 1/2, P(T) = 1/2

The trial is repeated N = 8 times and forms pairs.

I guess a possible outcome of 8 trials N:

N = { HH, HT, TH, TT, HH, HT, TH, TT }

I calculate the probabilities and assign to random variable X the number of heads (H) x:

x = 0: P(TT) = 1/4
x = 1: P(HT or TH) = 1/2
x = 2: P(HH) = 1/4

I next look at what each probability yields:

For x = 2:
P(HH)N = (1/4)(8)= 2 => { HH, HH }

For x = 1:
P(HT or TH)N = (1/2)(8)= 4 => { HT, TH, HT, TH }

For x = 0:
P(TT)N = (1/4)(8) = 2 => { TT, TT }

Next, I calculate the Expected Number of Heads He:

He = P(TT) x N x (0)+ P(HT or TH) x N x (1)+ P(HH) x N x (2)
= N[ P(HT or TH)(1)+ P(HH)(2) ]
= N[ (1/2)(1)+ (1/4)(2) ]
= N[ 1 ]
= 8[1]
= 8

I calculate the Expected Value E:

E = He/N = 8/8 = 1

That is 1 Head on average in 1 Pair (2 coin flips)

Another way of defining He (as per our text) is:

He = Np(x), where p(x) = 1/2

But that does not make much sense to me when working with trial pairs.
I then get:

He = p(x)N = (1/2)(8)= 4 faces

Instead I would like to get (with 8(2) = 16 faces)

He = p(x)N = (1/2)(16)= 8 faces

But then N is given two different meanings:

• Trials
• Faces

I cannot see where I am going wrong with this or how to change the way I think about it.

Answer 1

If a trial corresponds with flipping $2$ coins then - if there is only one trial - we can do it with probability space $(\Omega,\wp(\Omega),P)$ where $\Omega=\{HH,HT,TH,TT\}$ and $P$ is determined by $P(\{\omega\})=\frac14$ for every $\omega\in\Omega$.

Then we can define $X:\Omega\to\mathbb R$ (the number of heads that show up) by stating that:

• $X(HH)=2$, $X(HT)=X(TH)=1$ and $X(TT)=0$ resulting in:
• $P(X=2)=\frac14$, $P(X=1)=\frac12$ and $P(X=0)=\frac14$

This leads to: $$\mathbb EX=0P(X=0)+1P(X=1)+2P(X=2)=0\cdot\frac14+1\cdot\frac12+2\cdot\frac14=1$$

However, if there are $8$ trials like that then we need another probability space. We can do with $(\Omega',\wp(\Omega'),P')$ where $\Omega'=\Omega^n$ and $P'$ is determined by $\{(\omega_1,\dots,\omega_8)\}\mapsto\left(\frac14\right)^8$.

On that probability space we can define $X_i$ for $i=1,\dots,8$ as the number of heads that show up at the $i$-th trial. If $\pi_i:\Omega'\to\Omega$ is prescribed by $(\omega_1,\dots,\omega_8)\mapsto\omega_i$ then actually $X_i=X\circ\pi_i$.

Observe that the total number of heads that show up is then $Y:=X_1+\cdots+X_8$ and with linearity of expectation we find $\mathbb EY=\mathbb EX_1+\cdots+\mathbb EX_8=8$.

"Another way of defining He (as per our text) is: He = Np(x), where p(x) = 1/2"

That refers to $N$ trials at which there is a success or not (so $2$ possible outcomes).

We are not in that situation here, because there are $3$ outcomes here: $0$ heads, $1$ head or $2$ heads.

Actually we use a generalization of He = Np(x). In our case it is He = NE(x) where E(x) denotes the expectation at each trial. In the special case with 2 possible outcomes we have p(x)=E(x) but not in our case.