This is a follow up question to my question here: Loading two dice guarantees probability stays same or increases with rolling seven or doubles
Reproducing the linked question:
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.
And here's the actual question this time:
In the case where two dice are loaded alike, so that the chance of throwing doublets-or-seven is unaltered, if ${1\over6} + f$ be the chance of throwing doublets, the chance of throwing triplets with three such dice will be ${1\over{36}} + {1\over2}f$.
So we have $\sum_{i = 1}^6 p_i^2 = {1\over6} + f$, and we want to show that $\sum_{i = 1}^6 p_i^3 = {1\over{36}} + {1\over2}f$. We know that $\sum_{i = 1}^6 p_i = 1$, and we also know that the chance of throwing doublets-or-seven is unaltered means$$p_1 + p_6 = p_2 + p_5 = p_3 + p_4 = {1\over3}.$$(See the answer to my earlier linked question for an explanation.) This implies that$$(p_1 + p_6)^2 + (p_2 + p_5)^2 + (p_3 + p_4)^2 = {1\over3}, \quad (p_1 + p_6)^3 + (p_2 + p_5)^3 + (p_3 + p_4)^3 = {1\over9}.$$I was able to calculate$${1\over6} + f = \left(\sum_{i = 1}^6 p_i^2\right)\left(\sum_{i = 1}^6 p_i\right) = \left(\sum_{i = 1}^6 p_i^3\right) + \left(\sum_{j \neq k} p_j^2p_k\right), \quad1 = \left(\sum_{i = 1}^6 p_i\right)^3 = \left(\sum_{i = 1}^6 p_i^3\right) + 3\left(\sum_{j \neq k} p_j^2p_k\right) + 6\left(\sum_{i \neq j, i \neq k, j \neq k} p_ip_jp_k\right), \quad {1\over3} = \left(\sum_{i = 1}^6 p_i^2\right) + 2(p_1p_6 + p_2p_5 + p_3p_4), \quad {1\over3} = \left((p_1 + p_6)^2 + (p_2 + p_5)^2 + (p_3 + p_4)^2\right)\left(\sum_{i = 1}^6 p_i\right) = \left(\sum_{i = 1}^6 p_i^3\right) + \left(\sum_{j \neq k} p_j^2p_k\right) + 2(p_1^2p_6 + p_2^2p_5 + p_3^2p_4 + p_4^2p_3 + p_5^2p_2 + p_6^2p_1) + 2(p_1p_2p_6 + p_1p_3p_6 + p_1p_4p_6 + p_1p_5p_6 + p_1p_2p_5 + p_2p_3p_5 + p_2p_4p_5 + p_2p_5p_6 + p_1p_3p_4 + p_2p_3p_4 + p_3p_4p_5 + p_3p_4p_6), \quad {1\over9} = \left(\sum_{i = 1}^6 p_i^3\right) + 3(p_1^2p_6 + p_2^2p_5 + p_3^2p_4 + p_4^2p_3 + p_5^2p_2 + p_6^2p_1).$$However, I'm unsure how to combine these into showing that $\sum_{i = 1}^6 p_i^3 = {1\over{36}} + {1\over2}f$. Could anybody help me?
Note that $$(p_1 + p_6)^3 = p_1^3 + p_6^3 + 3p_1 p_6 (p_1 + p_6) = p_1^3 + p_6^3 + p_1 p_6,$$ since $p_1 + p_6 = 1/3$. Then $$\frac{1}{3} = (p_1 + p_6)^2 + (p_2 + p_5)^2 + (p_3 + p_4)^2 = \frac{1}{6} + f + 2(p_1 p_6 + p_2 p_5 + p_3 p_4)$$ implies $$p_1 p_6 + p_2 p_5 + p_3 p_4 = \frac{1}{12} - \frac{f}{2}.$$ Therefore, $$\begin{align} \sum_{i=1}^6 p_i^3 &= (p_1 + p_6)^3 + (p_2 + p_5)^3 + (p_3 + p_4)^3 - (p_1 p_6 + p_2 p_5 + p_3 p_4) \\ &= \frac{1}{9} - \left(\frac{1}{12} - \frac{f}{2}\right) \\ &= \frac{1}{36} + \frac{f}{2}. \end{align}$$