Probability to win the game with coin and dice I was reading a question and answer on this site here,
The problem:

This game is played with a fair coin and a die. First player flips a coin. If it turns out head ($H$), the player proceeds with tossing a die. If it turns out tail ($T$), the player proceeds with flipping a coin for the second time. The player wins if it gets head on the first tossing and 6 on the second or tails on both flips of coin. What is the probability of winning a game?

To solve this problem, I realized there are $8$ possible outcomes in this game: $$(H,1) , (H,2),(H,3),(H,4),(H,5),\color{\green}{(H,6)}, (T,H),\color{green}{(T,T)}$$
And as I demonstrated with the green colors there are only two cases that considered as a win. So I think the probability of winning is $\frac28=\frac14$.
But it has a contradiction
with the answer provided by the user @Ben:

First flip and second toss are independent events. So do first flip
and second flip in the case that first flip is tail.
So use multiplication:
P(head on the first flip and 6 on the second tossing)=P(head on the
first flip)*P(6 on the second
tossing)=$\frac{1}{2}*\frac{1}{6}=\frac{1}{12}$
P(tails on both flip)=$\frac{1}{2}*\frac{1}{2}=\frac{1}{4}$
Win the game if either one of the two events happens, so use addition:
P(winning the game)=P(head on the first flip and 6 on the second
tossing)+P(tails on both flip)=$\frac{1}{12}+\frac{1}{4}=\frac{1}{3}$

I'm confused, is my approach wrong? and why?
 A: The trick is that the various events are not all equally probable: $(H,1)$ happens $1/12$ of the time, and $(T,H)$ happen $1/4$ of the time.
If you do just want to count, you will have to come up with a list of cases that are all equally likely.  In this case, this can be done by replacing the second coin flip with another die roll, and then accepting (say) any item from ${4, 5, 6}$ on the die roll in the tails case.  This gives
$$(H,1), (H, 2), (H, 3), (H, 4), (H, 5), \color{red}{(H, 6)}, (T, 1), (T, 2), (T, 3), \color{red}{(T, 4)}, \color{red}{(T, 5)}, \color{red}{(T, 6)}$$
and it is then obvious that the result is $4/12 = 1/3$.
A: I think the reasoning is very clear by @Dan Uznanski's answer and some comments. To illustrate the result with a simulation, repeating this experiment a million times in the computer:
import random
import numpy as np

def coin():
    if random.randint(0,1) == 0:
        return 'H'
    else:
        return 'T'

def d6():
    return random.randint(1,6)

def iteration():
    if coin() == 'H':
        if d6() == 6:
            return 1
        else:
            return 0
    else:
        if coin() == 'T':
            return 1
        else:
            return 0

# Monte Carlo Results
prob = np.mean([iteration() for i in range(1000000)])

And the output gives prob = 0.332972, very close to the analytical $1/3$ result.
