First, it should not be all or nothing on dragonite. You should be able to get a little each day if you do all the battles. Non-Perfect run: Random 1 - 6? RNG should not be the only factor in getting dragonite as a reward. If you do the battle, give us a bone. So far, I picked a trap the 1st pick every day. Seems odd.
Second, I am on level 3 and the rewards for winning the battle are the same as level 1.
Third, how did highlighting of selection get past beta testing. You can barely see it on Xbox.
Beta testing is only done on pc because of restrictions the platforms put on the pushing of updates. When beta testing, sometimes 2 or 3 different builds need to be pushed a day to fix bugs testers found or new content. Unfortunately this isn’t possible on console. (although it is sorely needed) I’ve just grabbed some screenshots from Console of the new dungeons and adding it to the bug reports as it 100% needs looking at
This is really important feedback. I believe kafka has passed on feedback to this effect already but starting a feedback thread (if there isn’t one already) would be useful in backing up a big portion of the player bases feelings
Little bit of additional feedback: With the ongoing bug for bonus jewels, it is unclear, if sunday diamonds would get one more per level or would be restricted to one per ten levels.
Obviously I would prefer the former. On sundays, diamonds are the only gems handed out, and going from 20/20/30 to 22/22/32 would feel like a cheap joke.
Somewhere around 80-85. Had an 84 run two days ago.
Well agreed, but exactly how to distribute it is a major discussion. As it stands now it looks like the Perfect Runs pay out 80 Dragonite + 2 per dungeon level, so a good idea could simply be you get 2 Dragonite (x dungeon level) for beating the Gem Dragon boss, plus an extra 80 if it was a perfect run. Note that the Dragonite is awarded immediately upon victory over the Gem Dragon boss, so it kind of makes sense that beating the Gem Boss at all could be worth some Dragonite. But also note that if the hypothesized formula pans out then a high dungeon level would somewhat reduce the importance of getting a Perfect Run (e.g. at Lv.20 this could mean 40 base + 80 if perfect) – not necessarily a bad thing, just noted all the same.
Alternatively, apply a multiplier based on how many traps were encountered during the run – for example:
No traps: 1x multiplier (80-120 Dragonite)
1 trap: 0.25x multiplier (20-30)
2 traps: 0x1. multiplier (8-12)
Finally, maybe one of the random boons for the altar room could be some amount of free Dragonite, or that the next room is guaranteed to not be a trap?
Right. Duh. (Switch player here, we’re still on v6.4 where the bosses are fought in sequence)
Others have suggested the “trap multiplier” already, but keep in mind how this impacts the overall economy of Dragonite payouts (thus, expected grind length for a useful quantity) because it’s most likely someone on the dev team already is…
We know that a perfect run occurs 10% of the time* with a payout of 80+ Dragonite, thus the average payout over time is 8+ Dragonite per run and this can inform us what sound like good multipliers for non-perfect runs.
* I like numbers. Here's hoping these are the correct ones:
Given 6 doors, there are 720 permutations of what order to open them in, but since the 2 traps and 3 bosses are interchangeable there are functionally just 60 ( = 6! / 2! / 3!) different sequences of “runs”, 6 of which are the perfect runs. (ABBB-TT, BABB-TT, BBAB-TT, and BBB- which counts three times because it could be followed by -ATT, -TAT, -TTA)
Now it’s probably easier to calculate two-trap runs than single-trap runs. A two-trap run means that (ignoring the boon room for now) 5 doors were opened, with the last door being a boss. Thus the prior 4 doors were a 2 traps and 2 bosses in any order, which has C(4,2) = 6 combinations (the exact runs being BBTT-B, BTBT-B, BTTB-B, TTBB-B, TBTB-B, TBBT-B). And since the boon room (which has no effect on perfectness) could be added in any position to any run, there are 6 * 6 = 36 runs ending with both traps being sprung before the 3rd boss, right?
Actually, wait, let’s use the same strategy but for single-trap runs. Ignoring the boon room, this means you opened 3 doors before the 3rd boss, which has 3 combinations (sequences TBB-BT, BTB-BT, BBT-BT). Insert the boon room (in any of 6 positions) and there are 3 * 6 = 18 unique runs ending with a single trap sprung before the 3rd boss.
Alternatively, this person helpfully mapped out all 60 unique runs, so with a little bit of analysis there are indeed just 6 perfect runs, 36 runs springing both traps, and 18 runs springing 1 trap. The numbers check out!
Thus the probabilities are:
10%: Perfect Run
30%: Single trap sprung
60%: Both traps sprung
NOW if we divide the average outcome (8+ Dragonite) by these probabilities we get speculative payouts of:
Perfect run: 80+ Dragonite
Single trap: 23+ Dragonite
Double trap: 11+ Dragonite
Which is reasonably close to the arbitrary multipliers I plucked out of the air earlier, right? Cool.
Note that these values functionally triple the total Dragonite payout over time, as each of the three outcomes was calculated from the same expected value a perfect run has now. So maybe if we (or the devs) don’t want to boost the overall Dragonite economy that much, there could be smaller trap multipliers, like:
Perfect Run: 100%
Single trap discovered: 10%
Both traps discovered: 1 Dragonite (or 1% rounded up)