Feynman's Random Walk

Question

In chapter 6 of volume 1 of the Feynman lectures, Feynman explains the random walk. Here's the link for the lecture:

http://www.feynmanlectures.caltech.edu/I_06.html

Feynman says that there is a factor of 2 between D and NH. I dont get that. How? Then he goes on to say that:

We are now in a position to consider a question we have avoided until now. How shall we tell whether a coin is “honest” or “loaded”? We can give now at least a partial answer. For an honest coin, we expect the fraction of the times heads appears to be 0.5, that is, ⟨NH⟩ / N=0.5.(6.13) We also expect an actual NH to deviate from N/2 by about √N/2, or the fraction to deviate by (1/N) * √N / 2=1 / 2√N

Why does it have to be a fraction? And what does the fraction even mean?

Thank you!

Answer 1

Feynman says that there is a factor of 2 between D and NH. I dont get that. How?

He's just saying that starting from $$D=N_H-N_T,\tag{1}$$ where $N_H$ and $N_T$ are the number of head and tails in the analogous example of the coin-tossing game, and taking into account that the total number of steps/tosses is $$N=N_H+N_T, \tag{2}$$ it is possible to combine those two expressions into $$D=2N_H-N,$$ where we see that increasing $N_H$ by one unity increases the distance $D$ by two. That's because of the restriction $(2)$ between $N_H$ and $N_T$: increasing $N_H$ by one means reducing $N_T$ by the same amount, for a fixed $N$, then increasing $D$ by two (see $(1)$).

Why does it have to be a fraction? And what does the fraction even mean?

This fraction is just the probability $P(H)=\frac{<N_H>}{N}$ to obtain heads, as he proceeds to explains at the end of the section. You should however also take into account the deviation from the expected value to match it with the "experimentally determined" probability (eq. (6.14)).