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A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac{1}{6}$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)

Let $p_i$ be the probability of rolling $i$. Then $sum_{i=1}^6 p_i = 1$.

By Cauchy-Schwarz inequality,

$$begin{align*}
left(sum_{i=1}^6 1^2right) left(sum_{i=1}^6 p_i^2right) &ge
left(sum_{i=1}^6 1p_iright)^2\
6left(sum_{i=1}^6 p_i^2right) &ge 1\
sum_{i=1}^6 p_i^2 &ge frac16end{align*}$$

Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.

Using the hint: For any $ain mathbb{R}$

$$ sum (p_i − a)^2 ge 0$$
$$ sum (a^2 -2ap_i+p_i^2) ge 0$$
$$ 6 a^2 -2asum p_i + sum p_i^2 ge 0$$
$$ sum p_i^2 ge -2a(3a-1)$$

The RHS is a quadratic with roots at $a=0$, $a=1/3$, and maximum at $a_0=1/6$. At that point, the RHS value is $-2 frac 16 (3 frac 16 -1)=1/6$

Hence $$ sum p_i^2 ge frac16$$

Of course, the LHS is the probability of having the same result in both rolls.

For the equality to hold, the first inequality must be an equality, hence $p_i=a=frac16$.

Here’s another way to solve the problem, just for fun. Let $p_i$ be the probability of rolling $i$ when the die is rolled once. Also for simplicity, suppose the faces of the die are numbered $0$ through $5$. This won’t change the probability in question.

For two-roll sequences, consider the event $E_d$ that the second roll is $d$ bigger than the first roll (modulo $6$). So the sequence $3,5$ would be part of event $E_2$; $3,3$ would be part of $E_0$; and $5,3$ would be part of $E_color{red}4$, because $5+color{red}4equiv3$ (mod $6$).

The probability $P(E_d)$ is easy to calculate. It’s $sum_{i=0}^5 p_ip_{(i+d)(!!!!mod!!6)}$, which is the dot product of the vector $vec{p}=langle p_0,p_1,p_2,p_3,p_4,p_5rangle$ and a vector $vec{p_d}$ equal in length to $vec{p}$ having the same coordinates as $vec{p}$ but cyclically permuted. Also, when the die is rolled twice, exactly one of the events $E_d$ for $0le dle5$ occurs, so $sum_{d=0}^5P(E_d)=1$.

Then $P(E_d)=vec vcdotvec v_d={|vec v|^2overcostheta}$, where $theta$ is the angle between $vec v$ and $vec v_d$. The die is biased, so $vec v parallel vec v_d$ only when $d=0$, and $P(E_d)$ has a unique maximum when $d=0$, and $P(E_0)$ is the probability of the die showing the same number on both throws.

The unique maximum value among $6$ numbers whose sum is $1$ must be greater than $1over 6$.