Twelve times 6
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Re: Twelve times 6
If you mean impossible to finish one of those columns in a single turn, then that would be very surprising IMHO and then I would suspect a very weird bug. When on 6-7-8, you can calculate the probability of being able to roll N times like so:
Probability of not busting = (0.92) ^ N
Probability of busting = 1 - (0.92) ^ N
So if you roll say 10 times, the probability of busting would be 1 - 0.92 ^ 10 = 56%. You can play with N to convince yourself that it's definitely possible to finish a column in one go on 6-7-8!
Re: Twelve times 6
It's exceptionally unlikely that anything has changed.ElGera wrote: ↑01 May 2022, 04:49
Hello Ranior.
I am by no means an expert in statistics. I have some notions but nothing more.
However. Before, placing a chip on 6, another on 7, and another on 8 and winning one of those columns was possible. In two player duels, leaving those three columns free could make you lose any game.
But since the new season started, or a month or two before that, it has become impossible for me. I have NEVER made it.
Is it possible that they have fixed something in random and it is no longer possible to run in 6,7,8?
Even when you get on 6, 7, 8 it should be fairly unlikely that you cap a column even if that is your only goal. You'll succeed something like less than 10% of the time. (That's a rough estimate I'm making, I suspect the odds are a little less than that if you're starting from nothing). I'm not sure exactly how many chances you've had in that scenario since you've noticed this strange run, but my suspicion is you overestimate how likely it is to cap a column on rolling 6, 7, 8.
For what it's worth I've not noticed anything different, although I rarely try to cap 6,7,8 in a single turn usually--I think that's often a poor strategy. Personally I think too many players push 6, 7, 8 (8% bust chance) too far while being inconsistent on how far they'll push something like 6, 8, 10 (9% bust chance) which is only slightly wore, or even stuff like 5, 7, 10 (11% bust chance).
Re: Twelve times 6
It looked a fun calculation so I did it.
Let's assume you start the game with 6/7/8 at step 1. So you need 10 steps for 6&8 and 12 steps for 7.
There is 1296 possibilities of dices rolls (6*6*6*6).
104 (8.0%) lead to a loss.
666 (51.4%) have a '6'. 61 (4.7%) have a double 6.
same for 8.
744 (57.4%) have a '7'. 90 (6.9%) have a double 7.
And then I'm kinda stuck, every calcul look really hard.
So let's simplify the scenario and focus on the sixies.
Fails: 104/1296
+1: 666/1296
+2: 61/1296
+0: 465/1296
That give me 28.6% to be able to finish the '6' column while starting with 3 tokens on the first step of the 6, and having others in an infinite column of '7' and '8'.
But we are far from 10%. It's the triple.
Can someone double check? I'm not fully sure of it, but that look good, and that's what i expected.
Here is my document, but i don't advise it, there is no comment: https://docs.google.com/spreadsheets/d/ ... =420911169
Let's assume you start the game with 6/7/8 at step 1. So you need 10 steps for 6&8 and 12 steps for 7.
There is 1296 possibilities of dices rolls (6*6*6*6).
104 (8.0%) lead to a loss.
666 (51.4%) have a '6'. 61 (4.7%) have a double 6.
same for 8.
744 (57.4%) have a '7'. 90 (6.9%) have a double 7.
And then I'm kinda stuck, every calcul look really hard.
So let's simplify the scenario and focus on the sixies.
Fails: 104/1296
+1: 666/1296
+2: 61/1296
+0: 465/1296
That give me 28.6% to be able to finish the '6' column while starting with 3 tokens on the first step of the 6, and having others in an infinite column of '7' and '8'.
So if my maths are correct, it's 28.6% to finish the column 6 only. I suppose most situations where you can finish an 8 you will be able to finsh a 6 too, but not all of them, so this probability should be higher.
But we are far from 10%. It's the triple.
Can someone double check? I'm not fully sure of it, but that look good, and that's what i expected.
Here is my document, but i don't advise it, there is no comment: https://docs.google.com/spreadsheets/d/ ... =420911169
Re: Twelve times 6
Yeah, looks like I underestimated it, although I'm not quite sure your number is right. (I tried to follow what you were doing but it was hard, I'd try a different approach and might do it when I have time. Think you're missing a few things that lower the odds a little bit though but I can't be sure).
Regardless running some other rough estimates from a different perspective does suggest if you get on 6, 7, 8 you probably do have a 20+% shot of capping a column if that is your sole goal, so my gut feel was wrong. (Likely I got messed up because it's not that likely you can get on 6, 7, 8 so I was probably partially confusing the percentages of how likely it is to be able to do on any turn vs a turn when you actually get on the 6, 7, 8 columns.)
Doesn't change the fact that I highly doubt anything has changed with the code. So many games of Can't Stop are getting played and weird things are going to happen to some players some times.
Regardless running some other rough estimates from a different perspective does suggest if you get on 6, 7, 8 you probably do have a 20+% shot of capping a column if that is your sole goal, so my gut feel was wrong. (Likely I got messed up because it's not that likely you can get on 6, 7, 8 so I was probably partially confusing the percentages of how likely it is to be able to do on any turn vs a turn when you actually get on the 6, 7, 8 columns.)
Doesn't change the fact that I highly doubt anything has changed with the code. So many games of Can't Stop are getting played and weird things are going to happen to some players some times.
Re: Twelve times 6
Thanks!!!Ranior wrote: ↑02 May 2022, 15:01It's exceptionally unlikely that anything has changed.ElGera wrote: ↑01 May 2022, 04:49
Hello Ranior.
I am by no means an expert in statistics. I have some notions but nothing more.
However. Before, placing a chip on 6, another on 7, and another on 8 and winning one of those columns was possible. In two player duels, leaving those three columns free could make you lose any game.
But since the new season started, or a month or two before that, it has become impossible for me. I have NEVER made it.
Is it possible that they have fixed something in random and it is no longer possible to run in 6,7,8?
Even when you get on 6, 7, 8 it should be fairly unlikely that you cap a column even if that is your only goal. You'll succeed something like less than 10% of the time. (That's a rough estimate I'm making, I suspect the odds are a little less than that if you're starting from nothing). I'm not sure exactly how many chances you've had in that scenario since you've noticed this strange run, but my suspicion is you overestimate how likely it is to cap a column on rolling 6, 7, 8.
For what it's worth I've not noticed anything different, although I rarely try to cap 6,7,8 in a single turn usually--I think that's often a poor strategy. Personally I think too many players push 6, 7, 8 (8% bust chance) too far while being inconsistent on how far they'll push something like 6, 8, 10 (9% bust chance) which is only slightly wore, or even stuff like 5, 7, 10 (11% bust chance).
Re: Twelve times 6
I ran some simulations to find the real answer. It turns out that the probability of capping some column in 6-7-8 is about 35%.
Basically I ran millions of "first turns". If it's impossible for the bot to get on 6-7-8, the turn is ignored. When it gets on 6-7-8, it keeps rolling until it busts or finishes a column, prioritizing the columns that are the closest to being finished.
Tweaking the strategy has an influence on the final probability. For instance if the bot simply tries to go for 7s no matter what, the probability drops to 30%.
You can find the code here
https://github.com/simlmx/chickenroll/b ... /cap678.py
Basically I ran millions of "first turns". If it's impossible for the bot to get on 6-7-8, the turn is ignored. When it gets on 6-7-8, it keeps rolling until it busts or finishes a column, prioritizing the columns that are the closest to being finished.
Tweaking the strategy has an influence on the final probability. For instance if the bot simply tries to go for 7s no matter what, the probability drops to 30%.
You can find the code here
https://github.com/simlmx/chickenroll/b ... /cap678.py
Re: Twelve times 6
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Re: Twelve times 6
Look like both my calcul and yours are correct.
28.6% to finish the 'column 6'.
And if you add 'column 7' and 'column 8' you gain 6.4% extra.
28.6% to finish the 'column 6'.
And if you add 'column 7' and 'column 8' you gain 6.4% extra.