I have a biased coin $(P(H) \neq P(T))$,
My friend will toss the coin twice, and I have to make bet on either two events :
The two events are:
Event A: both of the tosses will have the same outcome (means HH,TT)
Event B: both of the tosses will have a different outcomes (means HT,TH).
On what event should I bet on to maximize my winning chances.
My answer :
for instance if we assume, for a biased coin, P(H) = 0.4 and P(T) = 0.6
So,
probability of having same outcome on both tosses = P(HH) + P(TT) = 0.40.4 + 0.60.6 = 0.52
And,
probability of having different outcome on tosses = P(HT) + P(TH) = 0.40.6 + 0.40.6 = 0.48
So I should always bet on Event A(i.e. both of the tosses will have the same outcome),
Am I correct?
Best Answer
Think of it generally. Let $p=p_H$ be the probability that your coin comes up $H$. Then $$P(HH)=p^2\quad P(TT)=(1-p)^2\quad P(HT)=P(TH)=p(1-p)$$
so you want to prove that $$p^2+(1-p)^2≥2p(1-p)$$
Expanding terms we see that you want $$2p^2+1-2p≥2p-2p^2$$ which is equivalent to $$4p^2-4p+1≥0\quad \text{or}\quad 4\left(p-\frac 12\right)^2≥0$$
which is true. Note that equality implies that $p=\frac 12$, in which case the coin would in fact be unbiased.