PSA: New Version of Hex 3

[tableplay]

But wouldn't it be much easier to catch the horseshoe when there are 6 positions exposed as opposed to just 3 positions exposed?

I think the horseshoe is an overlay symbol Mickey. That is, you have the normal symbols that fall into place after each spin and then there is a random chance you get the horseshoe on any given column (highest for columns 2 and 4, then columns 1 and 5, and then the center as you know). When you get the horseshoe it vacates the cell it appears on (leaving the original symbol in that cell resulting from the spin) and then goes to the top and pushes the bracket up as shown in the youtube excerpt below at the very bottom of this post. If it turns out as you describe and I am mistaken then I will attempt the math under your assumption.
Best, TP.

"https://youtu.be/LRJ_46q5hoY?t=24"

What mickey is saying is there are more spots for the horseshoe to land on the higher the column is, therefore the higher the column is the faster it will progress.

Unless the software "cheats" by treating the masked symbols (i.e. the black area that your column has not expanded into) as potential horseshoe spots and then visually manifests the horseshoe by randomly choosing an unmasked spot to display it on.

But I think everyone assumes it works as Mickey suggests.

Anyway I think the math should be something like:

x + 8/7x+ 8/6x + 8/5x + 8/4x + 8/3x = total cycle length, where x is the number of spins to hit the final horseshoe when the column is at 8

Using the cycle length posted above I get x=1190, so the last horseshoe would take 1190 spins and the first one would take
3172.

Writing this on mobile so not going to double check but feels right to me.

This imbalance is smoothed a little by the fact that you can catch horseshoes that move multiple spots, which will benefit lower columns more (because sometimes the extra spots will be wasted on higher columns).

Yes, I understand what he is saying. I was thinking it works more like Jackpot Ball, where you have a probability of getting a ball (Horseshoe) that is uniform for each spin irrespective of the number of spaces (the same probability for 2 and 4, 1 and 5 have the same probability but different than 2 and 4, etc.). Then the horseshoe appears and bumps up the bracket - you can see regular non-horseshoe symbols at all spots after the horseshoe appears and pushes up the bracket. Otherwise the horseshoe could bump up the bracket and then stay in place as a regular symbol where it initially appeared and you would observe it as such. For those that play for the center progressive, does it seem to move slower at the bottom and faster towards top (I just play the one-bit situations where 2 and 4 are close to the top)? Anyhow, I agree with your math if it works as commonly thought.