*and monster kills*for you but does not get chest kills, so you can't close and reopen the game if you want to get an accurate count.

Treasure chests give 10 times as much gold as normal monsters, this is a lot easier to test.

So 1% of the time you meet a monster who gives 10 times as much. That means that an average monsters gives 109% of the gold that an "average monster" gives.

Dora gives you 20% more treasure chests per level up to a maximum level of 50 while Mimzee gives 50% more gold per treasure chest. 20% more treasure chests would mean we get 1.11 normal monsters of gold per monster instead of 1.09, a 1.8% increase. 50% more gold from chests would mean we get 1.14 normal monsters of gold per monster gold instead of 1.09, a 4.6% increase.

But these two ancients combine better than other ancients. Each level of Dora you have makes each level of Mimzee better and vice versa. If I had one ancient that gave me 5% more gold per level and another that gave me 2% more gold per level, then with ten levels of each I would have 150% times 120% = 180% gold. With Dora and Mimzee 10 levels of each means getting treasure chests 3% of the time and getting 60 times the normal gold from those chests, which means getting 254% of the gold I would normally get.

This gets even wackier at higher ranks. If I invested 25 levels into Mammon and 25 levels in Dogcog I would have 125% extra gold from Mammon and effectively 100% extra gold from Dogcog for a total of 4.5 times as much gold. With 25 levels in both Dora and Mimzee I get 7.46 times as much gold. When you get both to 50 a 5% and a 2% ancient would give you 7 times as much gold. Dora and Mimzee gives 27 times as much.

Whats more, while Dora has a maximum level Mimzee does not. Suppose you had maxed out Dora but didn't have Mimzee. That would give you an 11% chance of treasure chests. That gives 199% of "average monster" gold per monster kill. Buying one level of Mimzee would increase that to 254%, at 27.6% increase. Once you have maxed out Dora, you can regard Mimzee as an ancient that gives 27.6% extra gold per. Since I advise buying Mammon at 5% per level, you can imagine what I think of Mimzee. Is it worth paying summoning cost plus 1275 souls to get a supercharged gold providing ancient? Of course it is.

But level 50 Dora may seem a long way off - I'm not particularly close to having level 50 ancients myself. What's the break even point where Mimzee becomes better than Mammon? When Dora's level two of course! Naturally you have to pay the price of the second ancient is a price to pay, but fortunately by the time Mimzee is level 8, Dora is also better than Mammon.

Mammon is a class above both Dogcog and Fortuna, and starting at ancient level 8 both Dora and Mimzee are a class above Mammon as long as you have both. I also think they are a class above all the other ancients as well. Honestly, Mammon is pretty good compared to his brethren.

I made a complicated three dimensional spreadsheet to determine how to level up Dora and Mimzee, but that was just to detect any anomalies. Basically you are increasing the amount of gold you get from treasure chests which is a (1 +

*x*) × (1 +

*y*) type function, so you know you know from my multiplication post that you want to keep

*x*and

*y*a fixed distance apart and in this case that's one. You also want to buy up Dora every time you buy up Mimzee and vice versa because buying one up makes the other one better and makes it worth buying up too.

So I replaced that spreadsheet with a really simple one that treats Dora and Mimzee's levels as a single dimension that goes up together. Here is the sheet. It works the same was as Mammon did. You can change the initial cost - remember to include the cost of summoning both - to figure out when you should buy in, and you can use it to determine when it's a good idea to buy up the next ranks.

My estimate for chest spawn rate was also 1%.

ReplyDeleteThis is one of those things that is really a question of the programmer. It's definitely close to 1% from experimentation. Could be (Math.random()<.01) but we'll never really have enough data to know it's not (Math.random()<Math.pow(Math.PI(),2)/100).

ReplyDeleteIf this were Sandcastle Builder I would think the latter was quite plausible. For nearly any other game, the former is far more likely.

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ReplyDelete