this post was submitted on 02 Oct 2025
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[–] TehBamski@lemmy.world 9 points 17 hours ago (4 children)

Me every time I think about this.

[–] __nobodynowhere@sh.itjust.works 1 points 3 minutes ago

The die has no memory of its past roles

[–] thatKamGuy@sh.itjust.works 20 points 14 hours ago (1 children)

Weirdly enough, it’s just the way probability works.

Once something stops being a possibility, and becomes a fact (ie. dice are rolled, numbers known) - future probability is no longer affected (assuming independent events like die rolls).

e.g. you have a 1/400 chance of rolling two 1s on a D20 back-to-back. But if your first roll is a 1, you’re back down to the standard 1/20 chance of doing it again - because one of the conditions has already been met.

[–] LoreleiSankTheShip@lemmy.ml 2 points 2 hours ago (1 children)

That's very interesting to me (I am a bit mathematically illiterate when it comes to probability). Wouldn't it still have a lower chance of being a 1 if you said you want your second roll to be the one that counts beforehand? Or would different permutations screw with the odds, say rolling a 12 then a 1, rolling a 15 and a 1, etc, counting towards unfavourable possibilities and bringing it back to 1/20?

[–] thatKamGuy@sh.itjust.works 1 points 1 hour ago

Because the outcome of a dice roll is an independent event (ie. the outcome of any given event does not impact subsequent events), it doesn’t matter if you said only your 2nd/3rd/4th etc. roll counted. Every roll has a 1/20 chance of rolling a 1 on a D20 die.

Consider this thought experiment, there are ~60.5m people, each rolling a 6-sided die. Only the people who roll a 6 can continue to the next round, and the game continues until there is only 1 winner.

After the first roll, only ~10m people remain in the game. After the second roll, ~1.7m people remain After the third roll, ~280K After the fourth, ~46.5K 5th, ~7.8K 6th, ~1.3K 7th, ~216 8th, ~36 9th, ~6 After the 10th and final roll, there should only be ~1 player remaining.

So even though initially there is only a 1-in-65m chance of rolling 10 6s back-to-back initially, each attempt still has a 1/6 chance of succeeding. By the time we get down to the final six contestants, they have each rolled a 6 nine times in a row - yet their chances of rolling it another time is still 1/6.

[–] DragonTypeWyvern@midwest.social 5 points 17 hours ago (1 children)
[–] festnt@sh.itjust.works 2 points 12 hours ago (1 children)

thank you for your thorough in your explanation

[–] DragonTypeWyvern@midwest.social 2 points 12 hours ago

You can tell it is by the way it is

[–] starman2112@sh.itjust.works 0 points 14 hours ago (1 children)

The math checks out, but the problem is the danger of rolling a nat 20 on your practice roll. The odds of getting two nat 20s in a row are almost as low as the odds of getting two nat 1s, so you may be screwing yourself out of a crit

[–] Cliff@lemmy.world 4 points 9 hours ago (1 children)

Jesse, that's not how probability fucking works.

[–] starman2112@sh.itjust.works 3 points 8 hours ago* (last edited 8 hours ago)

Gosh it's almost like I was joking by coming to a correct conclusion through faulty reasoning

I mean I could have just been a complete dweeb and explain that the outcome of the second roll is unaffected by the outcome of the first, and you are just as likely to roll two ones in a row as you are to roll any two numbers, but then I'd have to find a locker to shove myself in