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I think that's actually close, but not exact. In this case, I believe the fastest way to obtain the rate is as the inverse of the average number of chests separating 2 golds (including the last).
So, always after 1 gold chest, the probability that the next gold is at
1 chest = 1/3
2 chests = 2/3 * 1/3
3 chests = 2/3 * 2/3 * 1/3
4 chests = 2/3 * 2/3 * 2/3 * 1
+5 chests = 0
and the average distance is then 1/3 + 2 * 2 / 9 + 3 * 4 / 27 + 4 * 8 / 27 = 2.407...
The inverse of that number is the gold chest average rate = 41.538%
Would that be right?
I'm sure there are a lot of people that play that game and enjoy cracking the maths and probabilities or others that take advantage of some coments in this forum to learn or practice a little.
@Phantom,
If you are interested and you don't mind the explanation, the thing you didn't take into account is that the sequences G, S-G, S-S-G and S-S-S-G are not equiprobable in the series. A quick way to see this is that in your estimation the drop probability for the gold chests doesn't have influence in the result.
For example, if the ratio for gold chest were to be very high (close to 1) the total rate would be also close to 1 no matter the pity timer. If it were to be too low (close to 0) the influence of the pity timer would bring the total rate close to 25%. Your estimation method always gives 40%.
Also, if the pity timer were to be 1 gold for sure after 5 silvers instead of 3, your method would result in 6 / 21 < 1 / 3 (the pity timer would be reducing the total drop rate!! XD )
You were right that probability is favorable cases / total cases, but the tricky part is always adding them in the right way. Even physicist and mathematicians have problems in certain kind of problems. :)
BTW there new automation stop after 1 hour :(
So instead of a 22k gem boost, I got 315 for 7 days.