Question

Suppose you flip four fair coins.

(a) Make a list of all the possible outcomes, as in Table 2.1.

(b) Make a list of all the different "macrostates" and their probabilities.

(c) Compute the multiplicity of each macrostate using the combinatorial formula $2.6$, and check that these results agree with what you got by bruteforce counting.

Short answer

Determine the multiplicity of each macrostate and ensure that the result is agree with got by brute force counting is

$\begin{array}{r}\mathrm{\Omega }\left(0\text{tails}\right)=\mathrm{\Omega }(4,0)=\left(\begin{array}{c}4\\ 0\end{array}\right)=1\\ \mathrm{\Omega }\left(1\text{tails}\right)=\mathrm{\Omega }(4,1)=\left(\begin{array}{c}4\\ 1\end{array}\right)=4\\ \mathrm{\Omega }\left(2\text{tails}\right)=\mathrm{\Omega }(4,2)=\left(\begin{array}{c}4\\ 2\end{array}\right)=6\\ \mathrm{\Omega }\left(3\text{tails}\right)=\mathrm{\Omega }(4,3)=\left(\begin{array}{c}4\\ 3\end{array}\right)=4\\ \mathrm{\Omega }\left(4\text{tails}\right)=\mathrm{\Omega }(4,4)=\left(\begin{array}{c}4\\ 4\end{array}\right)=1\end{array}$

Step-by-step solution

Step1:Given data

Assume we have four distinguishable fair (equally likely to land heads or tails) coins. There are two if we flip all four. $2^4=16$possible outcomes, because each coin has two possible states and the four coins are distinct from one another Each of these states is referred to as a microstate. On the other hand, we might be more interested in the total number of heads or tails rather than which coins came up in either state. In this case, we'd only be interested in the overall condition of the four coin collection.

Step2:Possible outcomes(part a)

(a)

Step3:Macrostate possibilities(part b)

(b)To make things easier to count, I converted the four coin states into binary numbers using$H=0$ and$T=1$, then simply counted how many times each number appeared. for example:

- if the sum $=0$, so we have$0$tails.

- if the sum$=1$so we have$1$tails.

- if the sum =2, so we have $2$tails.

- if the sum$=3,$so we have$3$tails.

- if the sum$=4$, so we have$3$tails.

Step4:possible outcomes(part b)

Step5:final probability(part b)

Step6:multiplicity macrostate(part c)

(c) In general, we can calculate the probability of getting n heads in N coin flips. The problem can be thought of as the number of ways to choose n coins from a total of N and make these n coins the heads with the remaining N-n tails. We have N coins to choose from for the first of the n coins, so there are N ways to make this first selection. With that first coin selected, there are$N-1$coins from which we can select the next head, and so on, until we reach the final head. for which we have $N-n+1$choices. Thus the number of ways of choosing n coins from N in which the order of the choice does matter is:

$N(N-1)\dots (N-n+1)=\frac{N!}{(N-1)!}$

However, for a given macrostate, the order in which we choose the heads is irrelevant, and because there are n! ways to order each of the macrostates, the actual number of ways to choose the macrostate with n heads is$\mathrm{\Omega }(N,n)=\frac{N!}{(N-1)!n!}=\left(\begin{array}{l}N\\ n\end{array}\right)$

That is, $\Omega (N,n)$is the binomial coefficient$\left(\begin{array}{c}N\\ n\end{array}\right)$. So the multiplicity of macrostates:

$\begin{array}{r}\mathrm{\Omega }\left(0\text{tails}\right)=\mathrm{\Omega }(4,0)=\left(\begin{array}{c}4\\ 0\end{array}\right)=1\\ \mathrm{\Omega }\left(1\text{tails}\right)=\mathrm{\Omega }(4,1)=\left(\begin{array}{c}4\\ 1\end{array}\right)=4\\ \mathrm{\Omega }\left(2\text{tails}\right)=\mathrm{\Omega }(4,2)=\left(\begin{array}{c}4\\ 2\end{array}\right)=6\\ \mathrm{\Omega }\left(3\text{tails}\right)=\mathrm{\Omega }(4,3)=\left(\begin{array}{c}4\\ 3\end{array}\right)=4\\ \mathrm{\Omega }\left(4\text{tails}\right)=\mathrm{\Omega }(4,4)=\left(\begin{array}{c}4\\ 4\end{array}\right)=1\end{array}$