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


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?


1 Answer 1


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.


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