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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?

Happy Holidays.

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To make this easier to make sense of, try looking at a more extreme version of what's happening.

Start off by considering a two-tailed coin where $P(X\! =\! H) = 0$. In that case, the expected number of heads is $0$ and there is no variance at all. If this is an option, it's clearly the best choice: you're guaranteed to win the competition by picking it because the only possibility is for the outcome to match the expected value.

Once that makes sense, we can back off a bit and consider an extremely weighted coin where $P(X\! =\! H) = 0.01$. In that case, the expected number of heads is $1$. But because it's so heavily weighted, the chance of getting a head on any of the 100 flips is very low, so it's still a much better bet that we'll win picking this coin rather than a fair coin. From this, we can start to gain an intuitive grasp that low/high probabilities imply small variance which means that we're more likely to land "near" the expected value.

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