Let $$B_i$$ ($$i = 1, 2, 3$$) be three independent events in a sample space. The probability that only $$B_1$$ occur is $$\alpha$$, only $$B_2$$ occurs is $$\beta$$ and only $$B_3$$ occurs is $$\gamma$$. Let $$p$$ be the probability that none of the events $$B_i$$ occurs and these 4 probabilities satisfy the equations $$(\alpha - 2\beta)p = \alpha\beta$$ and $$(\beta - 3\gamma)p = 2\beta\gamma$$ (All the probabilities are assumed to lie in the interval (0, 1)). Then $$\frac{P(B_1)}{P(B_3)}$$ is equal to ______.

Let $$P(B_i) = p_i$$ and $$P(\overline{B_i}) = 1 - p_i = q_i$$ for $$i = 1, 2, 3$$. Since the events are independent, we have:

$$\alpha = p_1 q_2 q_3$$ (only $$B_1$$ occurs), $$\beta = q_1 p_2 q_3$$ (only $$B_2$$ occurs), $$\gamma = q_1 q_2 p_3$$ (only $$B_3$$ occurs), and $$p = q_1 q_2 q_3$$ (none occurs).

We note that $$\frac{\alpha}{p} = \frac{p_1}{q_1}$$, $$\frac{\beta}{p} = \frac{p_2}{q_2}$$, and $$\frac{\gamma}{p} = \frac{p_3}{q_3}$$.

Let $$a = \frac{p_1}{q_1}$$, $$b = \frac{p_2}{q_2}$$, $$c = \frac{p_3}{q_3}$$. Then $$\alpha = ap$$, $$\beta = bp$$, $$\gamma = cp$$.

Substituting into the first equation $$(\alpha - 2\beta)p = \alpha\beta$$, we get $$(ap - 2bp)p = ap \cdot bp$$, which simplifies to $$p^2(a - 2b) = abp^2$$. Since $$p \neq 0$$, we get $$a - 2b = ab$$, so $$a(1 - b) = 2b$$, giving $$a = \frac{2b}{1 - b}$$.

Substituting into the second equation $$(\beta - 3\gamma)p = 2\beta\gamma$$, we get $$(bp - 3cp)p = 2bp \cdot cp$$, which simplifies to $$b - 3c = 2bc$$, so $$b(1 - 2c) = 3c$$, giving $$b = \frac{3c}{1 - 2c}$$.

Now $$1 - b = 1 - \frac{3c}{1 - 2c} = \frac{1 - 5c}{1 - 2c}$$. So $$a = \frac{2 \cdot \frac{3c}{1 - 2c}}{\frac{1 - 5c}{1 - 2c}} = \frac{6c}{1 - 5c}$$.

We need $$\frac{P(B_1)}{P(B_3)} = \frac{p_1}{p_3}$$. Since $$p_i = \frac{a_i}{1 + a_i}$$ (where $$a_1 = a$$, $$a_3 = c$$), we get:

$$\frac{p_1}{p_3} = \frac{a}{1 + a} \cdot \frac{1 + c}{c} = \frac{a(1 + c)}{c(1 + a)}$$

Now $$1 + a = 1 + \frac{6c}{1 - 5c} = \frac{1 + c}{1 - 5c}$$.

So $$\frac{p_1}{p_3} = \frac{\frac{6c}{1 - 5c} \cdot (1 + c)}{c \cdot \frac{1 + c}{1 - 5c}} = \frac{6c(1 + c)(1 - 5c)}{c(1 + c)(1 - 5c)} = 6$$.

Hence, the answer is $$6$$.