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The following is from Freedman et al $(2005)$ Statistics.

(a) A die will be rolled some number of times, and you win $1$ dollar if it shows a one more than $20$ percent of the time. Which is better: $60$ rolls, or $600$ rolls?

(d) As in (a); but we win if the percentage of ones is exactly $16$ $2/3$ percent.

According to a key I found online the solution is the following:

(a) $60$ rolls. To win, you need a large percentage error, and that is more likely in $60$ rolls.

(d) $60$ rolls because to get exactly the expected value means getting exactly zero chance error, and that is more likely with fewer rolls.

Aren't these answers contradictory? I think, I understand answer (a): The more I roll the die, the smaller chance I have to reach 20 percent (and the more I approach $16$ $2/3$ percent). But how can I get large percentage error and zero chance error at the same time?

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They are not contradictory, but I don't like the explanation for d. It is true you would rather have $60$ rolls. Intuitively, with $600$ rolls there are too many results very close to right, like $101$ and $99$ that the probability of exactly $100$ gets diluted. You have a higher chance of being between $16 1/3$ and $17$ percent with $600$ rolls, but there are more results in that range so the chance of exactly $16 2/3$ percent is lower.

It is similar for flipping a coin. If you want at least $70\%$ heads and are offered two or ten flips, with two you have $25\%$ chance of winning. With $10$ flips you have about $17\%$ chance. But with two flips you have $50\%$ chance of exactly $1$ but for $10$ flips it is under $25\%$ that you will get exactly $5$.

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    That's very interesting. So the probability of large percentage error and zero chance error are both higher with fewer rolls simply because there is less possible outcomes. I think I get it; and would you agree that the 70 percent heads out of ten flips is 11.7 percent (and not 17 percent)? – 2017-02-09
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    My $17\%$ was for at least $7$ heads. Your $11.7\%$ is for exactly $7$ heads. – 2017-02-10
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    I see, thanks a lot! – 2017-02-10
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They're not contradictory. To see why, calculate the probability of exactly 50% heads for 2 and for 4 rolls, or 25% for 4 rolls and 8.