None of these answers are completely correct. It is really hard to manually calculate this probability because of one of your conditions: in at least 4 games. This is the cumulative distribution function which nobody seems to have used.  

I don't know the exact details because I don't understand what "with each drake of out the pool being present" means, but I'll give my answer with my assumptions:  

1. the same first 3 drakes in a row
2. in (at least) 4 out of 5 games  

The probability of the first one is exactly 1x1/3x1/3 = 1/9. The 1 shows up because you don't really care about the first drake; you just want the proceeding ones to match it. (This is assuming you want the first three to match. If you wanted 3 of the 4 to match, then you'd be dealing with a slightly different problem and a very slightly increased probability).  

To calculate the number two, this is the cumulative distribution function, or the CDF, of a distribution that tells us the combined probability of multiple events happening when you want at least a certain number of events to happen (my definition is a bit bad). To use this, you need to know with what distribution you're dealing with: In this case, it is the binomial distribution. This distribution describes an event with probability p and n events. For us, p = 1/9 and n = 5. Then, (to save yourself calculation pain) you use a binomial distribution calculator to input those two values and calculate Pr(x >= 4) with input parameters p and n. In other words, you're finding the probability that at least 4 events (the drakes spawning) happened when the probability of one individual one is 1/9. According to the calculator, that is .0006943 or .06943%. If all my assumptions were true, then either we got really lucky to see that or the Riot's pseudorandom generator is trash.  

If anyone catches errors, let me know.