Explaining a coin toss to a friend problem

[CJ.Be]

I'm currently having trouble trying to explain a coin toss and it's probability to a friend.

The question is

A coin toss has a 50% chance to land heads. You flip the count twice. At least one of the tosses lands a head. What is the probability of both tosses being heads?

Now the correct answer is 1 in 3. For the following reasons.

Ok first coin is flipped, it has 2 options, heads or tails.

We know that we will get at least 1 heads in these 2 coin tosses.

First coin flipped gives us tails, so we know that the next coin has to be heads.

First coin is flipped but it gives us heads. Now we don't know if this was the one that was suppose to be heads or not. We flip the 2nd coin, it can either give us heads or tails. The 2nd coin gives us tails.

First coin is flipped giving us heads, second coin is flipped, it gives us heads.

Note how there are 3 outcomes, if we only know that one of the flips will give us a heads. So we have a 1 in 3 odds that when both coins are flipped, we get heads for both.

This is word for word of how I explain it to him. Now he is thinking that the answer is a 1 in 2 chance. Because we know one coin is going to be 100% heads. So you can just leave that one toss out, and just have the 2nd toss, which would be a 1 in 2 odds.

So I ask that someone might be able to help me to explain this to my friend in some easy more understandable way. Since he seems set in his answer.

 

[dalcde]

This is my way of solving the problem.

When you roll two dice, there are four possible scenarios:
1)HH
2)HT
3)TH
4)TT

Since at least one is a head, scenario 4 is rejected. Hence only 3 scenarios are possible. Only one is both heads (HH), so the probability is 1 in 3.

 

[M-quest]

To me your explanation is simple enough so I think you should tell your friend that a head is also possible on the 1st toss,we are 100% sure of getting atleast 1 head,not where that head will be i.e. 1st or 2nd toss.

 

[jeebs]

I like dalcde's answer, as it is very concise.

ask your friend to list the possible outcomes on a piece of paper, then tell him to get rid of the one that can't happen => 3 choices remain.