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Suppose a box contains tickets, each labeled by an integer. Let $X,Y$ and $Z$ be the results of draws at random with replacement from thw box. Show that no matter what the distribution of numbers in the box,

$P(X+Y+Z\text{ is a multiple of }3)\ge 1/4\;.$

Hint: think about remainders when dividing by 3

  • 0
    Hint - restrict Z to {0,1,2}2011-10-05

1 Answers 1

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  1. $X$, $Y$ and $Z$ follow the same distribution.

  2. Let $p_0 = \mathbb{P}( X\mod 3 \equiv 0)$, $p_1 = \mathbb{P}( X\mod 3 \equiv 1)$ and $p_2 = \mathbb{P}( X\mod 3 \equiv 2)$ and $p_0+p_1+p_2=1$.

  3. Now $\mathbb{P}( X + Y + Z \mod 3 \equiv 0) = p_0^3 + p_1^3 + p_2^3 + 6 p_0 p_1 p_2$, because there are 9 triples leading to the desired outcome: $[0,0,0]$, $[1,1,1]$, $[2,2,2]$, and 6 permutations of $[0,1,2]$.

  4. Now minimize $p_0^3 + p_1^3 + p_2^3 + 6 p_0 p_1 p_2$ subject to constraints $p_0, p_1, p_2 \ge 0$ and $p_0+p_1+p_2=1$, which yields exactly $\frac{1}{4}$, that proves the result.

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    Great I get it! Thank you so much!2011-10-05