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

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?

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:

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

• Great, this makes sense! If we wanted to solve this problem by using a formula, would we use the combinatorics formula? – verkter Jan 21 '17 at 21:43
• I added a solution using formula. – Momo Jan 21 '17 at 21:46
• But why is the P(you 2 head, opponent 0 heads) is not considered in your formula? – verkter Jan 22 '17 at 1:55
• Because you only win if you get one more head than your opponent, not two more. – Momo Jan 22 '17 at 3:39
• But getting two heads, while opponent gets none is still a possibility if we evaluate results after both tosses are complete. – verkter Jan 22 '17 at 3:54

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

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

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) $= p + (1-2p)\cdot\frac 12 = p+\frac 12 -p =\frac 12$
• 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. – true blue anil Jan 22 '17 at 4:00