I am trying to find an intuitive and non-rigorous explanation to this math question.
If you have 100 coin flips of a fair coin, P(X = H) = .5, the expected number of heads is 50. Likewise, if you have a weighted coin, where P(X = H) = .3, the expected number of heads would be 30. Now, if you were to win $1,000 dollars if the coin you picked had a true value of heads equal to the expected value after 100 flips, which coin would you choose?
I understand from a pure mathematical perspective that you should choose the weighted coin, as variance is measured by P(1-P), assuming only one flip, and thus having P equal to .5 will maximize your variance. But is there a geometric or simple way to interpret and show this phenomena?