Dice Probability

[frogstar_A]

I agree it’s more likely that the % of doubles has been misread as the number.

But in general it is tiresome to have all these threads about the RNG, which seems fine. Amongst the millions and millions of games played on BGA of course all sorts of funny old things are bound to happen; but of course you don’t get 99,999 players posting “I played a game today and the RNG seemed fine” you get the 1 person who posts “my opponent rolled three Yahtzee’s in a row so the RNG is clearly flawed”

[Nekojin]

Unless and until the BGA moderation decide to put a ban on this topic, it's gonna keep coming up, no matter what we do. If we somehow manage to prove to one doubter that the randomness is working just fine, along comes another one a day later to revive the conversation all over again. Or make a new thread complaining about it, often right above the last randomness complaint thread.

Probability is hard to understand. Even people who are strong in math can still completely misunderstand it, because it's not just math. I had a friend with a CompSci degree, very good at advanced mathematics like Calculus, who still believed in the Law of Averages and was quick to recognize patterns and try to plan for when the "randomness" came back to balance the "average." Sometimes things worked out the way he expected, and he smirked about how smart he was. But just as often, the streak that he identified continued on its merry way, leaving his plans (and his battlemechs) in shambles.

The dice do not have memory. The last ten rolls have zero impact on the next roll. Likewise for the last hundred, thousand, and million rolls. The dice have no obligation to bring the sequence back to an average, and no ability to do so. A few lucky (or unlucky, depending on your perspective) can skew the numbers far to one side of the average, and it may take millions of rolls for it to come back to a perfect average across the rolls. Or it may never reach a perfect average, because all of those past rolls are now irrelevant with respect to the next roll.

If dice rolls and probability were as predictable as some of these people think it is (or think it should be), casinos would be going broke. The casinos live on streaks. And the casinos live on people who misunderstand probability.

[RobertBr]

I am assuming good faith and accurate information.

Why? Almost every poster like this has lacked both.

I've only ever seen one problem on BGA (and it wasn't with the random number generation) and people like the OP made it much harder for sensible people to gain traction with the site. Giving such people a credence they don't deserve is not productive.

[PhilipK_]

the prob came from one of my prob stat books. Chat Gpt confirmed the figure and the formula (for what that;s worth). I trust that it is sold but could ask a PhD prof.

i initially thought it would be like a lotta factorial calculation. for example, 50 balls to draw 6 numbers where the balls do not replenish:

(50 * 49 * 48 * 47 * 46 * 45) / ( 1 * 2 * 3 * 4 * 5 * 6) = prob of your ticket having all 6.

Also, no, not a bug, just a comment on "random: # generators. Our computer science profs all scoff at the notion that computers can truly generate "random" #'s, and they've shown us the proofs of the fallacy.

lastly, these were dice from Castles of Burgundy. 5 rounds x 5 turns per rd. 25 pair of d6 die.

[PhilipK_]

"Why? Almost every poster like this has lacked both."

I was in no way questioning BGA; just commenting on the "random" dice, and used the word in quotes.

care to tell us the lotto nmbrs for tmw? I mean, since you know my education and intent without having met me. Bravo "well played:. Note the use of quotation marks again

[PhilipK_]

":I agree it’s more likely that the % of doubles has been misread as the number."

the bottom of the Game Result page for Castles of Burgundy displays the % doubles rolled at game end.

[Mathew5000]

lastly, these were dice from Castles of Burgundy. 5 rounds x 5 turns per rd. 25 pair of d6 die.

I'm a little confused. I've never played Castles of Burgundy. You're saying there are are 5 rounds of 5 turns in one game, so isn't that 25 dice rolls in one game? But in your original post you mentioned that 24 of the 50 rolls were doubles, on two separate days. Could you please clarify and, most importantly, provide links (or just the table numbers) for where this took place. Try navigating to https://boardgamearena.com/gamestats?pl ... me_id=1390 and from there you can get links to the particular tables of interest.

[robinzig]

lastly, these were dice from Castles of Burgundy. 5 rounds x 5 turns per rd. 25 pair of d6 die.

I'm a little confused. I've never played Castles of Burgundy. You're saying there are are 5 rounds of 5 turns in one game, so isn't that 25 dice rolls in one game? But in your original post you mentioned that 24 of the 50 rolls were doubles, on two separate days. Could you please clarify and, most importantly, provide links (or just the table numbers) for where this took place. Try navigating to https://boardgamearena.com/gamestats?pl ... me_id=1390 and from there you can get links to the particular tables of interest.

I don't know for sure what they meant, but that 25 pairs of dice per game is *per player* per game. In a 2p game (which is probably the most popular player count for CoB), that would be 50 rolls in total.

[Gulchen]

I otherwise would've guessed that, but all of the OP's CoB games since August 8 have been 3-player, rather than 2-player.

[gmrsmrt]

the prob came from one of my prob stat books. Chat Gpt confirmed the figure and the formula (for what that;s worth). I trust that it is sold but could ask a PhD prof.

i initially thought it would be like a lotta factorial calculation. for example, 50 balls to draw 6 numbers where the balls do not replenish:

(50 * 49 * 48 * 47 * 46 * 45) / ( 1 * 2 * 3 * 4 * 5 * 6) = prob of your ticket having all 6.

Also, no, not a bug, just a comment on "random: # generators. Our computer science profs all scoff at the notion that computers can truly generate "random" #'s, and they've shown us the proofs of the fallacy.

lastly, these were dice from Castles of Burgundy. 5 rounds x 5 turns per rd. 25 pair of d6 die.

Math is not ok. If the probability of an event is P the probability of having NP times this event out of N trials is

P**NP * (1-P)**(1-NP) * N! /(NP!)/[(N-NP)!]
However this is not the main problem.
The occurrence you reported is too rare so if it is true there is certainly a bug.
To sort this out you have to post the link to the two tables you mentioned. If what you claim is confirmed I am pretty confident it will be investigated.