Flight Rising Discussion

How likely is a double? Let’s use a decision tree to illustrate! [img]https://i.ibb.co/9W45gbM/IMG-0001.jpg[/img] Here’s my tree. Now the primary color has a 100% chance of being any color, so we don’t need to worry about that, unless we want a specific double. We’ll say it, whatever it is, is X. The secondary is where we have a decision node. The secondary can be the same as the primary, ie X, or it can be different, Y. There are 177 colors with equal probability of being chosen, so P(X) = 1/177 and P(Y) is everything else = 1 - P(X), or 176/177, since the secondary has to be a color the probability of all the options must sum to 1. Now we have a more complicated set of nodes, but the same rules apply. Each node must have all probabilities add up to 1, because we are accounting for all possible outcomes. Let’s look at secondary = X. The tertiary can either be X or Y, the same as Primary and Secondary or not the same. This is the same as before, P(X) = 1/177 and P(Y) = 176/177. Note they add up to 1. Now if secondary = Y, tertiary can either be X, Y, or neither X nor Y, we’ll call it Z. P(X) = 1/177. P(Y) = 1/177 as well, since we already know Y from the last node. P(Z) is everything else, 175/177. To get P(XXX), all we have to do is multiply the chances associated with each branch we traveled down to get there. 1/177 * 1/177 or 0.00003 (0.003%). These are the odds of a triple. The same applies for a XXY dragon. Go down those branches and multiply 1/177 * 176/177 = 0.00565 (0.565%). This number is the same for XYX and XYY doubles, because you’re multiplying the same numbers in different orders. P(XYZ) is the same logic, but it is much larger, at 0.9831 (98.31%). The sum of all the possibilities is 1 because we are guaranteed a dragon with 3 colors. P(double) = P(XXY) + P(XYX) + P(XYY) because all these outcomes are considered doubles, so any of them would work. This ends up being 0.01685 (1.685%). Decision trees are really useful because they allow you to draw out every possible permutation of decisions and calculate the probability of each outcome. Then you just add all the probabilities counted as “successes” and there’s your probability of success. Hope this helps :) edit: @vacantlyamused beat me to it [emoji=skydancer laughing size=1] here’s my first and only self hatched double, for luck. [s]he’s still unfinished lol[/s] [url=https://www1.flightrising.com/dragon/85974202][img]https://www1.flightrising.com/rendern/350/859743/85974202_350.png[/img][/url]