Here's a question from my probability textbook:
If two dice are loaded alike in any way whatever, show that the chance of throwing doublets-or-seven is increased by the loading, except in one particular case in which it is unaltered.
Here's what I did. Let's say the probability of rolling $k$ is ${{a_k}\over{\sum_{i = 1}^6 a_i}}$. So we want to show that$${{\left(\sum_{i = 1}^6 a_i^2\right) + 2(a_1a_6 + a_2a_5 + a_3a_4)}\over{\left(\sum_{i = 1}^6 a_i \right)^2}} \ge {1\over6} + {1\over6} = {1\over3}.$$It's clear that$$(1 + 1 + 1 + 1 + 1 + 1)\left(\sum_{i = 1}^6 a_i^2\right) \ge \left(\sum_{i = 1}^6 a_i\right)^2$$by Cauchy-Schwarz. However, I don't know how to show$$12(a_1a_6 + a_2a_5 + a_3a_4) \ge \left(\sum_{i = 1}^6 a_i\right)^2.$$Could anyone help?
To make it easier, we assume probability of $i$ showing up is $p_i$ on both dice (as they are loaded alike).
So we have,
$ \displaystyle \left(\sum_{i=1}^6 p_i^2 \right) + 2 (p_1 p_6 + p_2 p_5 + p_3 p_4) \geq \frac 16 + \frac 16 = \frac{1}{3}$
Or, $(p_1 + p_6)^2 + (p_2 + p_5)^2 + (p_3 + p_4)^2 \geq \frac{1}{3}$
Say, $p_1 + p_6 = a, p_2 + p_5 = b, p_3 + p_4 = c$
Then we need to show that,
$a^2 + b^2 + c^2 \geq \frac{1}{3}$ where $a+b+c = 1, a, b, c \ge 0$
Using Root-Mean Square and Arithmetic Mean,
$\sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{a+b+c}{3}$
$ \implies a^2 + b^2 + c^2 \geq \frac{1}{3}$
Equality occurs when $a = b = c = \frac{1}{3}$
So the desired probability is always greater than $\frac 13$, except when they are loaded such that $p_1 + p_6 = p_2 + p_5 = p_3 + p_4 = \frac 13$. Then the probability remains $\frac 13$.