This is is a probability problem I'm having trouble solving.
You have 2 special events that can happen every second while playing a game.
$P(Event A)$ is $\frac{1}{1,000,000}$ chance per second
$P(Event B)$ is $25\%$ more likely to happen than $P(Event A)$
What is the chance that in a period of 8 hours, you get Event A exactly twice, and Event B exactly once?
I know that:
Event B is $1.25/1,000,000=1/800,000$
I probably need to use binomial distribution.
Can you help me?
Note: my intuition (as expressed in the comments) was off. The event you want, while unlikely, is not "astronomically unlikely".
Let $p_A$ (resp. $p_B$) be the probability of seeing event $A$ (resp. $B$) in a given second. The problem isn't clear, but let's assume that each second is an independent trial. That is, in any given second I might see either, both or neither of $A,B$ but I can't see $A$ or $B$ twice and seeing $A$ is independent of seeing $B$.
We compute. The probability of seeing $A$ exactly twice in $8$ hours is $$\psi_1=\binom{28800}{2}\times p_A^2\times (1-p_A)^{28798}$$
Using $p_A=10^{-6}$ we get $\psi_1=\fbox {.000402933}$.
The probability of seeing $B$ exactly once in $8$ hours is $$\psi_2=\binom{28800}{1}\times p_B\times (1-p_B)^{28799}$$
Using $p_B=1.25*10^{-6}$ we get $\psi_2=\fbox {.0347271}$
Independence tells us that the probability you want is just the product $\psi_1\times\psi_2$. Using your values we get about $\fbox {.000014}$