doubles in Castles of burgundy

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mizzou95
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Joined: 05 August 2022, 05:06

Re: doubles in Castles of burgundy

Post by mizzou95 »

“17.22% is very close to this so that is a perfectly reasonable outcome”

I’ll have what you’re smoking. BWAHAHAHAHA. It’s only 5 standard deviations from the mean, which makes it well outside anything that is thought of as reasonable. But hey, that doobie must be good stuff.

lefalaf wrote: 22 August 2026, 05:38 And this here folks is why you should not outsource your brain to generative AI/LLMs, aka autocomplete on steroids.

Did a quick normal web search and checked 3 probability sites to be sure, which all explained the simple math and gave the result. Took no time at all. The probability of doubles with two dice is 16.67%. 17.22% is very close to this so that is a perfectly reasonable outcome. There is nothing to be done.

You'll trust LLM hallucinations, but won't trust the random function in whatever language(s) BGA uses.
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lefalaf
Posts: 179
Joined: 25 October 2022, 02:42

Re: doubles in Castles of burgundy

Post by lefalaf »

mizzou95 wrote: 24 August 2026, 01:11 “17.22% is very close to this so that is a perfectly reasonable outcome”

I’ll have what you’re smoking. BWAHAHAHAHA. It’s only 5 standard deviations from the mean, which makes it well outside anything that is thought of as reasonable. But hey, that doobie must be good stuff.

lefalaf wrote: 22 August 2026, 05:38 And this here folks is why you should not outsource your brain to generative AI/LLMs, aka autocomplete on steroids.

Did a quick normal web search and checked 3 probability sites to be sure, which all explained the simple math and gave the result. Took no time at all. The probability of doubles with two dice is 16.67%. 17.22% is very close to this so that is a perfectly reasonable outcome. There is nothing to be done.

You'll trust LLM hallucinations, but won't trust the random function in whatever language(s) BGA uses.
An extra 55 per 10,000 rolls. You have 3,498 games and 17.21% now (they fixed it and it's correcting! lol). So 54 extra per 10,000 rolls. 25 rolls for you per game = 87,450 rolls. 472.23 "extra" doubles across your games. So an extra double per 7.4 games (3,498/472.23). How is that shocking for something random where each roll is an INDEPENDENT EVENT?
FrankJones
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Joined: 30 June 2024, 00:24

Re: doubles in Castles of burgundy

Post by FrankJones »

54 extra rolls per 10000 rolls is not the same thing as 5400 extra rolls per 1000000 rolls.

Flip a coin ten times and get 8 heads? Happens. I've gone second 8 times in a span of 10 games in games for which going 2nd is a major disadvantage. It happens.

If this happened 80 times in 100 games, that would raise my eyebrows.

As mizzou95 explained, a seemingly small difference can be many standard deviations away from the mean if that small difference is occurring over a very large sample.

5 Standard deviations from the mean is not normal. As previously suggested, I am inclined to blame this on the statistics gathering. several of the games I play have statistics logging method that are clearly making mistakes.
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lefalaf
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Re: doubles in Castles of burgundy

Post by lefalaf »

That was only for the OP’s games. It drops to 6 per 10000, across all CoB games. 16.61 vs 16.67 And no doubt would be even closer to 16.67 with more games. I have fewer games and mine is ~21%. Just 4-5 games there.

All as expected. Larger set, closer. Smaller set, further.

I don’t know where the five std dev, or Barnard’s minuscule odds and “binomial function” come from, but they raise an eyebrow.
FrankJones
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Re: doubles in Castles of burgundy

Post by FrankJones »

lefalaf wrote: 24 August 2026, 02:32

I don’t know where the five std dev, or Barnard’s minuscule odds and “binomial function” come from, but they raise an eyebrow.
These come from a useful thing called "math." :)

Given a data set, we can perform calculations to find the standard deviation, and then calculate how many standard deviations away from the mean a given result is.

This can be done using an excel spreadsheet, or a binomial probability calculator. The "binomial function" (often referred to with other similar names) is a mathematical function, like "multiplication" or "addition".

They all "come from" the same place. The minds and discoveries of mathematicians and scientists.
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lefalaf
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Re: doubles in Castles of burgundy

Post by lefalaf »

FrankJones wrote: 24 August 2026, 02:40
lefalaf wrote: 24 August 2026, 02:32

I don’t know where the five std dev, or Barnard’s minuscule odds and “binomial function” come from, but they raise an eyebrow.
These come from a useful thing called "math." :)

Given a data set, we can perform calculations to find the standard deviation, and then calculate how many standard deviations away from the mean a given result is.

This can be done using an excel spreadsheet, or a binomial probability calculator. The "binomial function" (often referred to with other similar names) is a mathematical function, like "multiplication" or "addition".

They all "come from" the same place. The minds and discoveries of mathematicians and scientists.
I know what they are, but how did the OP arrive at 5 st dev, way, way, way down the tail? AI? Which actual function/formula for Barnard's claim to get odds that are so absurdly low? If there are making serious claims that random isn't random, one needs to fully show their work.
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BarnardsStar
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Re: doubles in Castles of burgundy

Post by BarnardsStar »

lefalaf wrote: 24 August 2026, 02:32 I don’t know where the five std dev, or Barnard’s minuscule odds and “binomial function” come from, but they raise an eyebrow.
https://xkcd.com/1053/

The opening line of the Wikipedia article states,
Wikipedia wrote: In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own Boolean-valued outcome: success (with probability p) or failure (with probability q = 1 − p).
In other words, it's fairly basic statistics (which I think needs to be more widely taught in schools, but I accept most people's eyes glaze over at the very word "binomial"). So, you have this event where "success" has a probablility of p and "failure" has a probability of (1-p). The binomial distribution tells you how likely it is that you will get X successes in the course of N events.

So imagine flipping three (evenly weighted) coins. You would not be at all surprised if they all came up heads, right? That's a 12.5% likelihood. And if you ask Wolfram Alpha it will tell you exactly that. But, to extend this to 100 coins, you would probably be pretty surprised if they all came up heads. The odds here are 7*10^-31 of that happening. Indeed, 999,999 times out of 1,000,000 you would find the number of heads to be between 26 and 74. Anything outside that range, for 100 coin flips, is very unlikely. (Not impossible.)

If we extend it now to 100*100 or ten thousand coinflips, the range for 99.9999% likelihood falls between 4755 heads and 5245 heads. In other words, with 100 flips, anything below 25% or above 75% was a one in a million chance, but for 10,000 filips, anything below 47.5% or above 52.5% is a one in a million shot. For a million coin flips, the range narrows to 49.75%-50.25% -- outside that range is the one in a million changce. This is the law of large numbers in action -- the more "trials" (coin flips in this case) attempted, the more you expect the long term average to converge to the true expected average.

Now, the numbers look different for rolling doubles because it's a ⅙ chance of "success", but the basic math still applies. For millions of rolls, you expect the overall percentage to be pretty spot on. For 1 million dice rolls, you would have a 99.9999% chance of finding the number of doubles to be in the 16.48%-16.85% range, and when you are looking at 218 million rolls, the range is narrower still (the 99.9999% chance range is between 36306420 and 36360252 doubles rolled, which works out to between 16.654% and 16.679%), and outside that range, even the apparently negligible difference of 16.61% is extraordinarily unlikely.

How unlikely? If you set every computer on the internet to rolling 218 million pairs of dice and they each did it a thousand times a second, the entire lifetime of the universe could go by without seeing as low as 16.61% doubles. Indeed, a million lifetimes of the universe could go by, and you wouldn't expect to see it. At a decillion universe lifespans, it starts to become possible, but still very unlikely. Do you see where I'm going? This is so extraordinarily unlikely, that something is clearly wrong.

It's not the underlying random_int() method, I'm pretty confident of that, but somewhere along the way, something is wrong. Is it the statistics collection? Is it that some doubles are getting disregarded somehow (like the initial roll of the game was suggested)? Is it something else? But something is wrong.
FrankJones
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Joined: 30 June 2024, 00:24

Re: doubles in Castles of burgundy

Post by FrankJones »

lefalaf wrote: 24 August 2026, 03:25

I know what they are, but how did the OP arrive at 5 st dev, way, way, way down the tail? AI? Which actual function/formula for Barnard's claim to get odds that are so absurdly low? If there are making serious claims that random isn't random, one needs to fully show their work.
If you know what they are, then what is the problem? It's a relatively basic math function. Plug in the given values, and the function outputs the result. No AI is needed. Did you miss the part where I listed tools to perform this calculation?

It is not necessary to "show one's work." for a basic function that converts inputs into an output.

If we are using combinatorics, for example, and I write, "5c2 = 10", do I really need to "show my work"? Anyone could open an excel spreadsheet and enter "=COMBIN(5,2)" into a cell, and see that the result is 10.
FrankJones
Posts: 2539
Joined: 30 June 2024, 00:24

Re: doubles in Castles of burgundy

Post by FrankJones »

lefalaf wrote: 24 August 2026, 03:25

I know what they are, but how did the OP arrive at 5 st dev, way, way, way down the tail? AI? Which actual function/formula for Barnard's claim to get odds that are so absurdly low? If there are making serious claims that random isn't random, one needs to fully show their work.
There are various programs that can perform these calculations.

I use Excel, since I am most familiar with excel and I always have a spreadsheet open.

If I type into a cell:

=binom.dist.range(100,.5,26,74) ,

it will return a result showing the probability of flipping a fair coin 100 times and getting between 26 and 74 (inclusive) heads.

The function takes 4 inputs:

Number of trials: 100
Probability of success: .5
range start value: 26
Range end value: 74

The result is 0.9999436. That is the probability that we will have between 26 and 74 heads in 100 coin flips. We call that "p". (In this case, "p" refers to the probability that the number of successful heads results falls between 26 and 74.
(Not to be confused with the p in the original function which, had I listed it that way, would have referred to the probability 0.5 of getting a successful heads result on a single flip.)

Thus, the probability of NOT having between 26 and 74 is equal to 1-p. A very small number. Approximately 1 / 1774219. Approximately 1 out of 1.8 million.

How is the function calculated?
The simpler binomial probability function can tell you the exact probability of getting exactly X successful results in N trials.
For example,
N = 100
p = 0.5
X = 26

would be the inputs to use in the appropriate function to attain the probability of exactly 26 heads in 100 coin flips.

To calculate the probability that X is between 26 and 74, we simply take the summation of the results from performing the above calculation for each number between 26 and 74, inclusive.

Essentially, the excel function is rapidly (nearly instantaneously) returning a result that could be done by hand, since it really involves multiplication and addition, but would be very time consuming and tedious to do so.

The results can be trusted. If these functions in excel were not being calculated correctly, someone would have noticed.

And, BarnardsStar got the same results using an entirely different program.

I hope that clears things up somewhat.
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