If you chase Reel 3 for the progressive, be warned. One Hex 3 version just surfaced where one player logged in 5,800+ spins and did not get last milestone.
After 5+ hours, player called it quits and took the loss.
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If you chase Reel 3 for the progressive, be warned. One Hex 3 version just surfaced where one player logged in 5,800+ spins and did not get last milestone.
After 5+ hours, player called it quits and took the loss.
Was this some other Twitter dufus? Did you tell him that running a few cycles out on anything is quite possible?
Maybe you should tell him that he needs to understand stochastic calculus to play that machine.....LMAO
1. Nice Ipse Dixit, as usual.Quote:
Originally Posted by MaxPen
Was this some other Twitter dufus? Did you tell him that running a few cycles out on anything is quite possible?
Maybe you should tell him that he needs to understand stochastic calculus to play that machine.....LMAO
2. Hex 3 are getting nerfed in my neck of the woods.
3. Thunder Valley nerfed their Hex 3 a few months ago. Those nerfs at TV made chasing Reel 3 more expensive but DID NOT change the cycle time of getting the last milestone. This nerf makes chasing Reel 3 nearly impossible.
4. The meter movement is 6 cents per 5 spins ($4.00 on $0.80 game); this changes the economics. The progressive resets at $113 (at $0.80). That means the $113 / $2K coin in floor value is no other valid. The $2K number implies 2,500 spins, which was the longest recorded cycle for this “new” version of Hex 3. 5.8K spins > twice previous 2.5K high watermark.
See this tweet: https://mobile.twitter.com/PokerBrad...27721280770048
This person who plays Hex 3 pegs the spins to get a milestone at 1K to 2K. 5.8K is clearly outside the bounds of 1K to 2K, stochastics notwithstanding.
5. This is a PSA since I wrote the original insight on this game on this website. Mickey (make shit up) Crimm has stated on this forum numerous times they don’t change bonus slots.
Here’s one of MC’s many dumb quotes / posts on this subject matter:
The real retard is the one that believes slot machines don’t get changed. Slot machines were being retrofitted / changed as early as 1989 (they got changed to due regulations but don’t tell this to Mickey Crimm).Quote:
Originally Posted by mickeycrimm
“Nearly 10% of Nevada’s 122,000 slot machines are being retrofitted after a decision by gaming authorities that players were being misled by a “near-miss” feature into thinking they were about to hit jackpots.“
Source: https://www.latimes.com/archives/la-...501-story.html
Here is my point: I have been warning on this website / forums that bonus slots get changed. Mickey Crimm tells people they *DON’T* change bonus slots.
This is a PSA about Hex 3. You are more than welcome to ignore this PSA.
Agreed. Nerf term vague.
I stand by what I said.Quote:
Originally Posted by Ex-AP1. Nice Ipse Dixit, as usual.Quote:
Originally Posted by MaxPenWas this some other Twitter dufus? Did you tell him that running a few cycles out on anything is quite possible?
Maybe you should tell him that he needs to understand stochastic calculus to play that machine.....LMAO
2. Hex 3 are getting nerfed in my neck of the woods.
3. Thunder Valley nerfed their Hex 3 a few months ago. Those nerfs at TV made chasing Reel 3 more expensive but DID NOT change the cycle time of getting the last milestone. This nerf makes chasing Reel 3 nearly impossible.
4. The meter movement is 6 cents per 5 spins ($4.00 on $0.80 game); this changes the economics. The progressive resets at $113 (at $0.80). That means the $113 / $2K coin in floor value is no other valid. The $2K number implies 2,500 spins, which was the longest recorded cycle for this “new” version of Hex 3. 5.8K spins > twice previous 2.5K high watermark.
See this tweet: https://mobile.twitter.com/PokerBrad...27721280770048
This person who plays Hex 3 pegs the spins to get a milestone at 1K to 2K. 5.8K is clearly outside the bounds of 1K to 2K, stochastics notwithstanding.
5. This is a PSA since I wrote the original insight on this game on this website. Mickey (make shit up) Crimm has stated on this forum numerous times they don’t change bonus slots.
Here’s one of MC’s many dumb quotes / posts on this subject matter:
The real retard is the one that believes slot machines don’t get changed. Slot machines were being retrofitted / changed as early as 1989 (they got changed to due regulations but don’t tell this to Mickey Crimm).Quote:
Originally Posted by mickeycrimm
“Nearly 10% of Nevada’s 122,000 slot machines are being retrofitted after a decision by gaming authorities that players were being misled by a “near-miss” feature into thinking they were about to hit jackpots.“
Source: https://www.latimes.com/archives/la-...501-story.html
Here is my point: I have been warning on this website / forums that bonus slots get changed. Mickey Crimm tells people they *DON’T* change bonus slots.
This is a PSA about Hex 3. You are more than welcome to ignore this PSA.
Your entire premise about putting a game in then a few months later cutting the payback percentage is not true for the most part. Sure there are always exceptions but its not a routine practice. If they install an 87% game why on earth would they need to cut the payback a few months later. They are already raping the public with the game. Why don't you get that?
Now, this 1K to 2K parameters on the middle column of the Hexes. Are you saying it will move at a minimum of 1000 spins to a maximum of 2000 spins. If this is the case the game would be similar to the old IGT Times Pay games. The hustlers called them "Coming Soon's." An example, a green "Coming Soon" would flash on the screen for 80 coin in. Then it would flash a red "Coming Soon" for an average of 35 coin-in. It could trigger anywhere from the 1st coin-in to the 70th coin-in but the average was 35 coin-in. So the average cycle to trigger the multiplier was 80 + 35 or 115 coin-in. I highly doubt the hexes work that way.
Your buddy saying it trips between 1000 and 2000 spins is not very good information. It's certainly not "pegging it." Telling you what his total spins are with how many times he has gotten movement would be a much more telling statistic. But if you are as sharp as you say you are then you should know that. So I can't see why you are pushing this 1000 to 2000 spins bullshit.
If the observed high water mark was 2500 spins and you were using that as an average number of spins to get movement then I have to say you don't know what the fuck you are doing. With variance being what it is the cycle would be a lot shorter than 2500 spins.
So WTF is the 1000 to 2000 spins bullshit all about? You should be able to tell us total spins versus times you got movement. And if 2.5K was the high water mark you should have seen movement a lot of times where there was far less than 1000 spins made.
So far all I'm seeing from you is incompetence.
Basically these twits don't break the games down themselves. Some of them get ahold of a random par and apply it across the board. When it doesn't contain the frequency of occurrence for the horseshoe in the middle column then they tard out and spout of 1-2k......hurrrdurrr
But Ex-AP is claiming the bonus orbs were changed. I think it's funny that people put this stuff out on public forums then wonder why the shit gets "nerfed"(dumb term). Not necessarily the case in this regard....but??,,,the point is deal with what opening your mouth gets you. Don't act all surprised about it.
A game like Hexbreaker can have a wide range of results. Whereas most slots are in a tight 1-2% of results after 100k spins a game like this a million spins is required.
nn
Not going to post it all. I may consider sending via dm but straight off of one version of the par. The odds of hitting the progressive is 1-11590. There a 6 moves required. You do that math. The meter move is 1.5 percent Those are both facts
I think where ex-ap went wrong is not understanding probability. Whatever the mathematical probability of the event is, say 1000 spins, by the math you will hit the objective within 1000 spins only about 2/3's of the time. There is about a 1 in 7 chance it will take you 2000 spins, and I think 3000 spins is about 1 in 20, something like that. The most cycles I've gone to achieve an objective is 10.4 cycles. That was an extreme case but over a 25 year career and multi-millions of hands and spins it is quite possible to suffer that extreme.Quote:
Originally Posted by Advantageplay
So ex-ap said it was 1000 to 2000 spins then all of a sudden someone went 5800 spins, That doesn't mean the machine got "nerfed." Which is just another way of saying they lowered the payback of the game. And I concede its easy for them to do that these days.
Now, Advantageplay, who obviously has a par sheet, tells us its 11,590 spins average to collect the black cat progressive. Thanks, AP. With there being 6 levels to clear that would be an average of 1931 spins per level.
But wouldn't it be much easier to catch the horseshoe when there are 6 positions exposed as opposed to just 3 positions exposed?
If it takes 11,590 spins to move thru the six levels then level 3 (3 positions exposed) should be the hardest to hit. Level 4 should be slightly easier. And level 5 slightly easier than level 4. Level 8 should be the easiest to hit.
With 11,590 spins overall to move it thru six levels, the frequency of each level is the question. The 3rd level should have the longest frequency, then the 4th level, etc. I'm trying to think of a way to do the math. Meanwhile, if anyone else knows how to do it please don't wait on me and go ahead and show us.
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.Quote:
Originally Posted by mickeycrimm
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.Quote:
Originally Posted by tableplayI 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.Quote:
Originally Posted by mickeycrimm
Best, TP.
"https://youtu.be/LRJ_46q5hoY?t=24"
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.Quote:
Originally Posted by smurgerburgerWhat 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.Quote:
Originally Posted by tableplayI 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.Quote:
Originally Posted by mickeycrimm
Best, TP.
"https://youtu.be/LRJ_46q5hoY?t=24"
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).
I agree it could work that way and probably should have been designed that way if it doesn't.
To your question I think it's particularly noticeable from the behavior or the 2nd and 5th columns during the course of a center column play that they tend to stick longer at 3 and then progress faster as they grow.
The acid test is if folks that play the center could chime in here (I doubt there is enough youtube footage to gather the required data).Quote:
Originally Posted by smurgerburger
Yes, that's the way I'm thinking it works. Thanks.Quote:
Originally Posted by smurgerburgerWhat 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.Quote:
Originally Posted by tableplayI 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.Quote:
Originally Posted by mickeycrimm
Best, TP.
"https://youtu.be/LRJ_46q5hoY?t=24"
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).
I've only done maybe a dozen plays. Although I'm in an area with a lot of hexes they sit there forever on the 3rd of 4th level. Once they reach the 5th level I usually find them at reset a few days later. It's rare to find one on the 6th level which is where I will consider playing. I will automatically grab a seat if I find one on the 7th or 8th level.Quote:
Originally Posted by tableplayThe acid test is if folks that play the center could chime in here (I doubt there is enough youtube footage to gather the required data).Quote:
Originally Posted by smurgerburger
From a limited number of plays, and I didn't keep stats on it, it's just an observation. I spend more time on the 6th level than the 7th level, and spend more time on the 7th level than the 8th level.
I'm pretty bad these days about collecting stats. To lazy to do it anymore.
Thanks guys. So it sounds like it's just like a regular hex3 symbol then except with the additional ability to push the bracket up 1 or 2 spots (usually one spot) just as Mickey originally stated. So we can apply Smurgerburger's equation for columns 2,4 and columns 1,5 using the stats Mickey had collected on these and posted in an earlier post (and different thread IIRC) for more accurate cost to hit analyses for the luck zones above these columns.
My sample space on the 1/5 and 2/4 columms is probably not big enough.Quote:
Originally Posted by tableplay
Can't say how accurate it is but I picked up a little more information. There supposedly are 6 different paybacks for the game with the lowest being 85%. So we can guess there is about 2% difference between each payback. 85, 87, 89, 91, 93, 95%. One hex player told me the 85%'ers are found on the CZR properties.
With smurgerburger's formula I come up with this number of spins per each level:
3rd level: 3172
4th level: 2775
5th level: 2379
6th level: 1982
7th level: 1586
8th level: 1190
So playing from the 6th level there would be an average of 4758 spins to complete the play. I use the speed stop and average 1800 spins an hour so average seat time is 2.6 hours.
Another thing I was told. The EV playing from the 6th level is the same as playing from the 8th level. It's worth the same amount of money. So the only difference should be the seat time involved. And it should mean there is no cost to play from the 6th level.
Another clue I was given. Some hex players will play from the 5th level if the meter is twice the reset. The meter resets to 141.25 units.
Also, some hex players are playing high denom and running multiple cards for freeplay.
Thank you for the convenient formula Smurgerburger and great stuff Mickey. So the main value in taking it at the 6th spot (or 7th) rather than the 8th would be the additional card points for those that run a card(s) since the EV from the play itself is the same.Quote:
Originally Posted by mickeycrimm
I forgot another clue I was given. Hex players even play on CZR properties because of the big freeplay they send. At least they used to. I'm sure there was multi-carding involved.Quote:
Originally Posted by tableplayThank you for the convenient formula Smurgerburger and great stuff Mickey. So the main value in taking it at the 6th spot (or 7th) rather than the 8th would be the additional card points for those that run a card(s) since the EV from the play itself is the same.Quote:
Originally Posted by mickeycrimm
I lost 400 taking a .80 Hex at 6 at a CZR property, for what it's worth. I would suspect they crank it way down there, but could just be terrible variance.Quote:
Originally Posted by mickeycrimmI forgot another clue I was given. Hex players even play on CZR properties because of the big freeplay they send. At least they used to. I'm sure there was multi-carding involved.Quote:
Originally Posted by tableplayThank you for the convenient formula Smurgerburger and great stuff Mickey. So the main value in taking it at the 6th spot (or 7th) rather than the 8th would be the additional card points for those that run a card(s) since the EV from the play itself is the same.Quote:
Originally Posted by mickeycrimm
Mickey, if I just signed up for a card at a new property and coin on opportunities were scarce that day to generate mail, how bad of a bath do think I would take getting on a Hex at 5 at your average non-CZR property, assumingthe progressive was near reset? Any speculation on what the RTP to player could be there?
I saw the bit about the average number of spins to get it to 6 from 5 is 2379, but is there a way to factor in the value of the middle column being two above reset? Or, is the RTP at 3 just insanely low because the RTP as a whole factors in all play positions during the entire course of the machines play?
All I have is clues to go by. If hex players take a hex on the 5th level when the meter is double the reset value it should mean the cost to spin off the 5th level would be about 141 units. That would be about a 94% return. And I assume they are talking about machines that have a reasonable overall payback like 89% and higher.Quote:
Originally Posted by RounderI lost 400 taking a .80 Hex at 6 at a CZR property, for what it's worth. I would suspect they crank it way down there, but could just be terrible variance.Quote:
Originally Posted by mickeycrimmI forgot another clue I was given. Hex players even play on CZR properties because of the big freeplay they send. At least they used to. I'm sure there was multi-carding involved.Quote:
Originally Posted by tableplay
Mickey, if I just signed up for a card at a new property and coin on opportunities were scarce that day to generate mail, how bad of a bath do think I would take getting on a Hex at 5 at your average non-CZR property, assumingthe progressive was near reset? Any speculation on what the RTP to player could be there?
I saw the bit about the average number of spins to get it to 6 from 5 is 2379, but is there a way to factor in the value of the middle column being two above reset? Or, is the RTP at 3 just insanely low because the RTP as a whole factors in all play positions during the entire course of the machines play?
I suspect that hex players pull the same tricks as the 10K must hit players and get them to work on the machine for some reason so they can get a glimpse of the payback.
PS You were playing an 85% hex on the CZR property.
Bump.
Since we're bumping, anyone have an opinion about the EV of a center column 5 at the 14.x% rtp?