Difference between two Monte-Carlo methods in Ising model

Question

I was working on a Monte-Carlo simulation of the Ising model. It seems that we have two different way to flip a single spin and I didn't quite understand the difference between them.

Say we have $N\times N$ sites $\{\sigma_i\}_{i=1,\ldots N^2}$, each has either spin up or down (denoted by $\sigma_i=+1$ for spin up and $-1$ for spin down). The Hamiltonian reads $$H=-J\sum_{\langle i,j\rangle}\sigma_i\sigma_j$$ A Monte-Carlo simulation, to my understanding, goes like this

1. Randomly choose a site $i=(x,y)$.
2. Calculate some energy and use it to decide how to flip $\sigma_i$ at next step.
3. Flip the spin $\sigma_i$ or not, in a certain probability.
4. Repeat 1.-3.

Now, in Metropolis algorithm steps 2. and 3. goes like

2. Calculate the difference in energy $\Delta E=E_f-E_0$ generated by the spin flip $\sigma_i\to -\sigma_i$.
3. If $\Delta E\leqslant 0$, which means the flipped spin has less total energy, then accept the spin flip $\sigma_i\to -\sigma_i$. If $\Delta E>0$, which means the flipped spin has more energy than the original state, then the spin should be accepted with probability $\mathrm e^{-\beta \Delta E}$ and stay unflipped with probability $1-\mathrm e^{-\beta \Delta E}$.

While another textbook (Section 8.1 by James P. Sethna, Statistical Mechanics:Entropy, Order Parameters, and Complexity) gives so-called Heat Bath Monte-Carlo which says

2. Calculate $E_+$ and $E_-$, the energy given spin $\sigma_i=\pm 1$
3. Set $\sigma_i=+1$ with probability $\mathrm e^{-\beta E_+}/\text{Const.}$ and $\sigma_i=-1$ with probability $\mathrm e^{-\beta E_-}/\text{Const.}$. Here $Const.=\mathrm e^{-\beta E_+}+\mathrm e^{-\beta E_-}$ make sure the probability is normalized.

My question is: Are these two methods in flipping spin effectivly the same? Surely both of them can lead to a thermal equilibrium state, but is it actually identical to flip spin in such ways? If yes, why? If not, does it mean that one method "converges to" thermal equilibrium state quicker than the other?

I have came up with a possible explanation on why they are identical:

Suppose spin up ($\sigma_i=+1$) state has a higher energy, then after the flip in Metropolis algorithm we shall have the spin up with probability $$p'_\text{up}=p_\text{down}\mathrm e^{-\beta(E_\text{up}-E_\text{down})}$$ or we have spin down with probability $$p'_\text{down}=p_\text{up}\cdot 1+p_\text{down}\cdot(1-\mathrm e^{-\beta(E_\text{up}-E_\text{down})})$$ where $p'$ and $p$ stand for the probability $\sigma_i=\pm 1$ after and before the flip manipulation respectivly.

If the system has reached thermal equilibrium state, then $p'=p$ respectivly for spin up and down, thus it can be solved that $$p_\text{up}=e^{-\beta E_\text{up}}/(e^{-\beta E_\text{up}}+e^{-\beta E_\text{down}})\,,\,p_\text{down}=e^{-\beta E_\text{down}}/(e^{-\beta E_\text{up}}+e^{-\beta E_\text{down}})$$

but this is plausible to me, since under non-equilibrium state similar conclusion cannot be derived. Correct me if I am wrong.

Apologize for my poor English. Thanks!