We have a biased coin with probability of heads p. We win $1 if we predict the number of heads correctly. Otherwise, nothing.
How would one go to develop the optimal betting strategy to maximize expected earnings in a two coin toss game?
That is, what is the number of heads we should predict in two coin tosses to maximize expected winnings?
Best Answer
Since you win \$1 if you guess correctly, the expected winning is equal to the probability that you are correct. Let the number of heads be denoted by $X$. $$ P(X=0)=(1-p)^2\\ P(X=1)=2p(1-p)\\ P(X=2)=p^2 $$ To maximize the probability of being right, you need to pick the value of $X$ that has the highest probability.
You can show that if $0\le p\le 1/3$, $X=0$ has the highest probability, if $1/3\le p\le 2/3$, $X=2$ has the highest probability, and if $1/3\le p\le 1$, $X=2$ has the highest probability.