Amane Kanata, Blue Axolotls, and the Gambler’s Fallacy
How unlucky is Amane Kanata really? Let’s investigate.
At approximately 5 AM on July 12th, 2021, Japan Standard Time, VTuber Amane Kanata ended her latest in a series of streams attempting to spawn a blue Axolotl in Minecraft. Before signing off, she opened her statistics screen to confirm how many attempts she made so far and saw an unthinkable number: 3,275.
But is it really that unthinkable? Let’s take a quick dive into the numbers, and also take the opportunity to talk a bit about statistics in video games.
To provide context, the axolotl is a passive mob first added to Minecraft late 2020, and recently added to the Minecraft server that the Hololive members (including Kanata), play on. Kanata and other members on the server began capturing and breeding axolotls in pursuit of the elusive blue variant.
Pink, brown, yellow, and white axolotl variants can be found in the wild, but the blue variant can only be obtained via breeding. Breeding in Minecraft happens when two adult members of a species are fed the correct food, both within a short time and close in distance to each other. Each time a baby axolotl is spawned, the game does a check with 1/1200 odds (approximately 0.083%), that its color is the desired blue variant. Otherwise, it’s a 50/50 that it’ll be colored the same as one of the wild-type parents.
But back to Kanata’s situation: How unlucky is Kanata? She joked on stream about her bad luck, and her stream chat even joined in by claiming she should apply for a Guinness World Record with her blue axolotl drought.
It’s a simple calculation, actually. In basic statistics we’re taught calculations based on the win chance, but the reverse — the chance of a loss— can be used as well. In our case, it’s 1199/1200 (the probability of losing), raised to the power of 3,275 (the losing streak length). The result is 0.065199436573…, or 6.52%. This means that 6.52% of players that go for a blue axolotl via breeding will require at least 3,275 attempts.
Unlucky, for sure, but there have been much more significant records in games of chance (for example, a 154 streak in craps without rolling a 7, carrying odds of 1 in 1.56 trillion). Anyway, I digress, back on the topic of blue axolotls and statistics.
Question: What is the “midpoint” of the count, the number of attempts at which 50% of the players will have obtained a blue axolotl? Intuition says it should be 600, half of the 1/1200 odds, but that’s incorrect. Solving (1199/1200)^N = 0.5, we get 831.43. Meaning half the players will need to spawn between 831 and 832 axolotls before seeing a blue one.
When it comes to video games, especially where probabilities are involved, human players have a tendency to introduce biases and expectations that lack mathematical support. Games of chance, which is what axolotl breeding effectively is, take advantage of human psychology to increase engagement and extend the time spent with the game. Concepts tried and tested in casinos and gambling houses around the world and across time, find a place in the digital world, in the form of rare spawns, gacha draws, and virtual RPG casinos.
I’ll touch on the most relevant bias related to this topic: the gambler’s fallacy. This is the incorrect belief that previous results of a game of chance have an influence on future results. I’ll break down what I believe are three common statements that connect Kanata’s situation, blue axolotls, and the gambler’s fallacy:
“After 1199 failures, Kanata should be guaranteed a blue axolotl spawn.” As each spawn is a chance independent of all others, no player is ever guaranteed anything. This is like claiming that rolling a 6-sided dice six times should guarantee you one of each result. A deck of cards is a different scenario altogether because if you don’t return the card to the deck and shuffle the cards after each draw, it is no longer an independent probability.
“Fubuki casually took Kanata’s luck by spawning a blue axolotl early for herself.” It sounds pretty silly when said out loud, and is incorrect for the same reasons as above. Fubuki has the same independent 1/1200 chance as Kanata and happened to roll the lucky number sooner than her.
“If the probability of her hitting this losing streak is 6.52%, then she has a 93.48% chance of spawning a blue axolotl soon.” Again, independent probabilities. She has a 1/1200 chance of spawning a blue axolotl on her next try, just as she has a 6.52% probability of going on another 3,275-attempt drought.
As with most games of chance, it’s entertaining to ride the emotional roller coaster as a player or a viewer, collecting serotonin shots with the wins. I hope this short write-up sheds light on the technical side of things. Enjoy the ride and don’t overthink it! (Further reading can be found by searching: “binomial probabilities” and “statistical fallacies”)
My favorite meme related to games of chance has to be the “desire sensor”: the more badly I want an item, the less frequently it seems to show up. What’s yours?
Amane Kanata’s YouTube channel can be found at https://www.youtube.com/channel/UCZlDXzGoo7d44bwdNObFacg
