PHY306 Homework 1

Due February 3, 2004

This exercise illustrates the idea of different states that a system can take and gives a feel for the statistics involved. Consider 3 coins being flipped - the 8 possible outcomes are:

 Coin 1 Coin 2 Coin 3 H H H H H T H T H T H H H T T T H T T T H T T T

Each separate outcome is a microstate - a microstate specifies the state of each individual coin (or particle, in a general physical system). If we specify only how MANY heads or tails there are, we specify the macrostate. In this example we have 4 possible macrostates (3 heads, 2 heads, 1 head and no heads). If we know the microstate, we know the macrostate, but not the other way around.

The number of different microstates leading to the same macrostate is called the multiplicity, W. For our 4 macrostates, the multiplicities are (1, 3, 3, 1) - can you see why that is? The probability of finding any given macrostate is the ratio of the number of its microstates to the total number of possible microstates.

Probability of n heads = W (n)/ W (all)

For example, the probability of getting 2 heads is W (n)/ W (all) = 3/8.

Things get more tricky if you have many coins, like 100, for example. Many microstates are possible: 2100 since each coin has two possible states (H or T). The number of macrostates, however, is only 101: 0 heads, 1 head, … up to 100 heads. Now, we investigate how many microstates each macrostate has.

• For 0 heads, each coin falls tails-up, so there is only one microstate possible, and W(0)=1.
• For exactly one head, any one of the coins must fall heads-up, and there are 100 possible ways for this to happen, so W(1)=100. W(1) counts the number of ways of choosing one coin to fall heads-up.
• To find W(2), consider the number of ways of choosing two coins to fall heads-up. You have 100 choices for the first coin, and for each of these you have 99 choices for the second coin. But you could choose any pair in either order, so the number of distinct pairs is W(2) = (100 * 99)/2.
• For W(3), you have 100 choices for the first, 99 for the second and 98 for the third. But any triplet could be chosen in several ways: 3 choices for which one to flip first, and for each of these, 2 choices for which to flip second. Thus the number of distinct triplets is W(3) = (100 * 99 * 98) / (3 * 2). See the pattern?
• W(n) = (100 * 99 * …*(100-n+1))/(n*…* 2)

The denominator is just n! We can write the numerator as (100!)/(100-n)!, and then the general formula becomes W(n) = 100! / (n! * (100-n)!)

• If there are N coins, rather than 100, the multiplicity of a macrostate is

W(N,n) = N! / (n! * (N-n)!)

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Problem 1:

Let's flip 4 coins.

1. list all possible outcomes
2. list all the different macrostates and their probabilities
3. compute the multiplicity of each macrostate using the combinatorial formula above and check that the results agree with wht you got by brute force counting.

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Problem 2:

Now you flip 20 coins

1. how many possible microstates are there?
2. what is the probability of getting the sequence

HTHHTTTHTHHHTHHHHTHT (in exactly that order)?

c) what is the probability of getting 12 heads and 8 tails in any order?