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?

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.
