How to wire MULTIPLE LED driver dimming leads?

I have 2xHLG185-1400 (COBs) and 1xHLG240-1050 (QBs), each with it's own pot & 10k resistor on the negative side.
I tried wiring the positive dimming leads of all 3 drivers to one wago and from there to the pot.
Tied the negative leads to a second wago and put a 47k resistor between it and the pot.
I read everywhere I need a 33k pot. It's in the mail. While I wait I thought I'll try a 47k like stardustsailor suggested here: https://www.rollitup.org/t/meanwell-led-drivers-3-in-1-dimming-function.838760/#post-10725098
I figured I get something..
What I got is probably a short of the dimming leads.
The LEDs were on 100% (power monitor connected on the AC side of the QBs) and the pot did not do anything.
I also messed around with a small electricity simulator and hooking up the leads that way got them shorted.
So what am I doing wrong? How do I wire it the right way?

That's exactly what I did.
I must have shorted the wires somehow. I'll give it another go.
Thanks Airwalker16!

I tried it again with the same result so I tried it without the resistor.
I have the power monitor attached on the AC side of the HLG240 so I plugged this one in first.
It measured 30w and the QBs were very dim.
I plugged in the a HLG185 and the meter showed 50w (that's only for the HLG240) and QBs and now COBs stayed dim.
I plugged in a second HLG185 (now all 3 drivers are plugged in) and brightness jumped to max, so did the meter showing 250w.
Maximum value for the HLG240 on a single potentiometer is 248w. Minimum is 27w.
I'm struggling to understand the results.
I understand I gave a lot of headroom with a 47k resistor (as I do with a 10k resistor currently to reach 100k) having me stay at max power regardless of the pot position (I think).
I then tried 3x10k resistors in series (it might be a stupid idea - I don't have any formal education) thinking I'll at least be close to the 33k I need and I'll see something change when I turn the knob.
It did. It dims from around 10w to maximum of 250w but only taking about a third of turn of the knob. I guess resistance is not additive.
I'm waiting for the 33k resistor to arrive now.
Could someone explain the results to me?
Am I putting my drivers at risk playing around like this?

Try the pot without the resistors. You want a (10k) resistor for two things: to ensure you don´t dim below 10% (100k resistance is 100%, 10k 10%) and in case your pot does not have true 100k resistance but less than that (e.g. 95k of true resistance, so you add the 10k to make sure you are able to go up to 100k). A 33k resistor would mean you are not able to dim below 33%. If you are wiring 2 drivers together, go with a 50k pot and a 5k resistor (or if you use a 10k, you would not be able to dim below 20%). I hope this helps.

I did try it without the resistors and got results I didn't fully understand (it's a 100k pot, 90k really).
At the moment, I do have a 10k resistor on each of pots because I measured them at around 90k. One pot for each driver. I understand the theory.
I am trying to tie up 3 drivers. I need a 33k but I didn't have one but I do have 100K pots so I ordered 33k resistors. That should work right?
If I wanted to go this way and my pot is 90k, what resistor would I need to get even dimming on all 3 drivers and make sure I can't dim to minimum?

With a 33k resistor and 3 drivers I understand you would not be able to dim. It will always be at least at 33k resistance (which is 100% output for 3 drivers put together). Try a without a resistor, or use a 3k one (which is about 10% of 33k), or just try to look for a pot that measures 33k.

I tried it without a resistor. The brightness indeed maxes out a third of the way.
I ordered a 33k 2W pot. That should do it, right? Just add a 3.3K resistor..
Thanks