law of large numbers: zero chance error 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? 

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.
