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