What is the probability of getting at least one more head from two coin flips than one coin flip?

Question

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?

Answer 1

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:

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

Answer 2

$\underline{Answer\; for\; you\; getting\; at\; least\; one\; more\; head}$

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) = $\dfrac12 + \dfrac14 + \dfrac14 = \dfrac12$

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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) $= p + (1-2p)\cdot\frac 12 = p+\frac 12 -p =\frac 12$