$n>2$ participants. Each participant chooses at random. The poll shows the results for the 2 options as percentages rounded to the nearest integer, i.e. $x$ rounded is $\lfloor x+0.5\rfloor$, where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. For example, 1/3 would be shown as 33%. What is the probability that the sum of the 2 shown percentages does not add to 100?
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- Why does
$\lfloor x+0.5\rfloor$round$x$to the nearest integer? Use the ceiling function$\lceil \cdot \rceil$to create a rounding function. - Under what circumstances is
$\lfloor a+1 \rfloor = \lceil a \rceil$true? - Boris flips
$8$fair coins. What is the probability that at least$6$of them are tails?
- Why does
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How are the two percentages related?
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Under what circumstances do the two rounded percentages not add up to 100? Try expressing it in an equation with integer solutions.
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The only scenarios in which the rounded percentages do not add up to 100 are when one of the percentages end in
$.5,$causing both percentages to round up and their sum to be 101. -
Try phrasing the above condition in terms of
$n, k$and$j,$where$k$is the number of votes for the first option and$j$is the integer of that percentage. -
Why not rearrange to get rid of all fractions and decimal points?
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Consider the prime factors of the terms in your equation. What can you deduce about the divisibility of
$n?$ -
$n$must be divisible by$8.$By substituting$n = 8m,$could you derive a new equation in terms of$n, k$and$j?$ -
What are the possible factors of
$m?$How about doing a case split? -
By considering divisibility of
$5,$determine which values of$j$are possible. -
Given the possible values for
$j,$what are the possible values for$k?$ -
What is the probability that the
$n$voters will choose one of the suitable$k?$
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It suffices to consider the answers picking one of the options; suppose there are
$k$of these. Then the rounded percentages of the two answers are$\lfloor \frac{100k}{n} + 0.5\rfloor$and$\lfloor \frac{100(n-k)}{n} + 0.5\rfloor$respectively. Adding these together we get$100 + \lfloor \frac{100k}{n} + 0.5 \rfloor$$+ \lfloor 0.5 - \frac{100k}{n}\rfloor,$which evaluates to 101 when the fractional part of$\frac{100k}{n}$is exactly$0.5.$Thus, we need to solve
$\frac{100k}{n} = j + 0.5$for$j \in \{0, \dots, 99\}.$Rearrange to obtain$200k = n(2j+1).$Since$200 = 2^3 \cdot 5^2$and$2j+1$must be odd, it must be the case that$n$is a multiple of$2^3 = 8.$Writing$n=8m,$our expression becomes$5^2k=m(2j+1)$for$j\in \{0, \ldots, 99\}.$There are three cases:$m$is not a multiple of$5.$This means that$2j+1$must be a multiple of$25,$hence the only possible values for$2j+1$are$S_1 = \{25, 75, 125, 175\}.$$m$is a multiple of$5$but not a multiple of$25.$This means that$2j+1$must be a multiple of$5,$hence the only possible values for$2j+1$are$S_2 = \{5,15,25,35\ldots,195\}.$$m$is a multiple of$25.$There is no restriction on$j$except$j<100,$hence the only possible values for$2j+1$are$S_3 = \{1,3,5,7\ldots,199\}.$
Each
$x \in S_i,$$K(x) = \frac{nx}{200}$will result in the rounded percentages to not add up to$100.$Thus, the probability is:$$ p(n)= \begin{cases} 0, & \text{if } 8\nmid n \\ 2^{-n}\sum_{x \in S_1}\binom{n}{K(x)}, & \text{if } 8\mid n, 5\nmid n \\ 2^{-n}\sum_{x \in S_2}\binom{n}{K(x)}, & \text{if } 40\mid n, 25\nmid n \\ 2^{-n}\sum_{x \in S_3}\binom{n}{K(x)}, & \text{if } 200 \mid n \\ \end{cases} $$