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.
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?
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.
Many though will land preferentially on one axis or one side. Practice rolls could easily conceal checking whether you have the low rolling die or high rolling die
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
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
Me every time I think about this.
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.
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?
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.
Math
thank you for your thorough in your explanation
You can tell it is by the way it is
The die has no memory of its past roles
Many though will land preferentially on one axis or one side. Practice rolls could easily conceal checking whether you have the low rolling die or high rolling die
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
Jesse, that’s not how probability fucking works.
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