We have a die. Success event (S) is 1-4, failure event (F) is 5 and 6 (in other words chances for success are ~0.67 or 2/3 for one roll). We roll the die, say, a billion or n times.
What is the probability of the success events number being higher than the failure events number?
An approximate result, or a clue about the (possible?) tendency for S to go higher/lower as n increases, will suffice!
Extra question if you feel like answering: What is the probability of the number of S being higher than the probability of the number of S in an "alternate universe" (let's call it Sx), where chances of success and failure for the die are equal? (Same n)
The intuitive thing is that if you have one event that is more likely (your $S$) it will overwhelm the other (your $F$) in the long run. In your case, because the probability difference is huge, it will be very soon. For $n$ rolls you should expect $\frac {2n}3$ successes and $\frac n3$ failures. If we use the normal approximation, the standard deviation of the number of successes is $\sqrt {n\frac 13 \cdot \frac 23}=\frac 13\sqrt{2n}$. At the $3\sigma$ level we want $\frac {2n}3-\sqrt {2n} \gt \frac {n}3+\sqrt {2n}$ which requires $\frac n3 \gt 2\sqrt {2n}$ or $n \gt 72$ so if you roll $72$ times (in this approximation) you have $99.7\%$ chance to have more successes than failures. We have ignored the chance of a tie, but for $72$ rolls that is not so likely.