## Thursday, April 17, 2014

### The Two Strategies

 Follow @ProbabilityPuz
Q: You are in a game where you get to toss a pair of coins once. There are two boxes (A & B) holding a pair each. Box A's coins are fair however B's coins are biased with probability of heads being $$0.6$$ and $$0.4$$ respectively. You are paid for the expected number of heads you will win. Which of the boxes should you pick?

Machine Learning: The Art and Science of Algorithms that Make Sense of Data

A: The expected number of heads if you chose box A is easy to calculate as
$$E(\text{heads}| A) = \frac{1}{2} + \frac{1}{2} = 1$$
However the expected number of heads if you chose box B is also the same
$$E(\text{heads}| B) = \frac{4}{10} + \frac{6}{10} = 1$$
The average yield being the same could make one think that both boxes yield the same. However there is one difference, its the variance. The variance of a distribution of a random variable $$X$$ is defined as
$$Var(X) = \sum_{i=0}^{N} (x_i - \bar{x})^{2}p_i$$
where $$p_i$$ is the probability of $$x_i$$. Given this, lets compute the variance of each strategy
$$Var(X | \text{Box=A}) = (\frac{1}{2} - 1)^2 \frac{1}{2} + (\frac{1}{2} - 1)^2 \frac{1}{2} = \frac{1}{4} = 0.25 \\ Var(X | \text{Box=B}) = (\frac{2}{5} - 1)^2 \frac{2}{5} + (\frac{3}{5} - 1)^2 \frac{3}{5} = \frac{6}{25} = 0.24$$
The variance for box B is slightly tighter than box A which makes choosing the coins in box B a better strategy.

If you are looking to buy some books on probability theory here is a good list.