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Let's say that I have two fair coins and my opponent has one fair coin. Me and my opponent can flip each of our coins once. I win only if I get at least one more head than my opponent. What is the probability of me winning?

I don't understand how to think about this problem in terms of the probabilities. The event of me getting at least one head from the two coin flips is $$P(TH) + P(HH) = 0.5 + 0.25 = 0.75 $$ while the probability of my opponent getting at least one head is $$ P(H) = 0.5.$$ I am confused by contrasting my opponents probability of winning with mine. How would I go about solving this problem?

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If the first two flips are yours and the third is your opponent's you get 8 possible outcomes: $HH\ H, HH\ T,\ldots, TT\ T$, each with probability $1/8$

To have at least one head more than your opponent, you need either 2 heads vs opponent's 0 or 1 heads or 1 head vs opponent's 0 heads, so the favorable outcomes are:

$HH\ H, HH\ T, HT\ T, TH\ T$.

So probability of winning is $4/8=1/2$

Or you can use independence:

$$P(\text{winning})=P(\text{you 2 heads, opponent 1 head})+P(\text{you 2 heads, opponent 0 heads})+P(\text{you 1 head, opponent 0 heads})\\=P(\text{you 2 heads})P(\text{opponent 1 head})+P(\text{you 2 heads})P(\text{opponent 0 heads})+P(\text{you 1 head})P(\text{opponent 0 heads})\\=\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{2}+\frac{2}{4}\cdot\frac{1}{2}=\frac{4}{8}=\frac{1}{2}$$

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    Great, this makes sense! If we wanted to solve this problem by using a formula, would we use the combinatorics formula? – 2017-01-21
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    I added a solution using formula. – 2017-01-21
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    But why is the P(you 2 head, opponent 0 heads) is not considered in your formula? – 2017-01-22
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    Because you only win if you get one more head than your opponent, not two more. – 2017-01-22
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    But getting two heads, while opponent gets none is still a possibility if we evaluate results after both tosses are complete. – 2017-01-22
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    ... so make it clear: does two heads against none count as a win? Your problem says "win only if I get one more head than my opponent", so I assumed it doesn't. – 2017-01-22
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    You helped me realize that wording matters, my fault. I meant to say "win by at least one head". I will edit the question... This changes will change the answer. I wish @true blue anil did not delete his answer! – 2017-01-22
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    I updated my answer. You can also compress the first two probabilities to $P(\text{you 2 heads})$, since in this case you win regardless of what opponent flips. – 2017-01-22
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$\underline{Answer\; for\; you\; getting\; at\; least\; one\; more\; head}$

Consider that initially, you both toss only one coin.

You can win only if you have already won, or are equal and win with your second toss.

Denoting your results in caps, and opponents in lowercase for clarity,

P(you win) = P(Ht) + P(HhH) + P(TtH) = $\dfrac12 + \dfrac14 + \dfrac14 = \dfrac12$

Added material

Interestingly, if you toss (n+1) coins against n tossed by your opponent, P(You win) is still $\dfrac12$

After tossing $n$ coins each, let $p$ be the probability that you are ahead. By symmetry, $p$ is also the probability that your opponent is ahead, and the probability of a tie is $1-2p$. You have just two ways to win: either you are ahead before the last toss, or there is a tie and you then get $H$.

Thus P(You win) $= p + (1-2p)\cdot\frac 12 = p+\frac 12 -p =\frac 12$

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    But what if you don't care about the order and just look at the results after all tosses are complete? – 2017-01-22
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    We are **not** caring about order, we are just delaying tossing your coin 2 (or delaying declaring its result) , to enable the situation to be seen clearly. – 2017-01-22
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    I am adding some more material . – 2017-01-22
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    OK, but what if we **did not** delay the tossing and evaluated the results at the end? Then we would have to consider P(HHt), P(HHh) and P(THt)? You brought up a very good point on timing when we evaluate the win. – 2017-01-22
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    Will it affect the results if we label one of your coins **1** and one of them **2** ? I have added material that explains your win probability remains 1/2 even if you have $(n+1)$ coins against your opponent's $n$. I think it should make it very clear. – 2017-01-22