The trick is to say "this is just a practice roll" where the die can hear you, but wink at the GM so they know it's the real roll. That way, the die will be a spiteful little punk and throw out the nat20 for the "practice".
But don't do that too often, or the die will figure out the trick.
And when the Nat 1 shows up, rub your eye because you had sand in it.
This kind of thinking is wasteful. Every d20 has a finite lifespan. It was created, and it will, at some time in the future be destroyed, as all things are. That means it has a finite number of rolls in its lifetime, with an equal distribution of all possible outcomes. When you "practice roll" and get a nat 20, you have wasted one of the limited number of nat 20s that die has in it. Think of the 20s. Don't practice roll.
This is like a common house fly worrying about the lifespan of Cthulhu.
You haven't seen how some of the folks I play with roll.
And of course the traditional sentence for dice which misbehave one too many times.
Maybe the real Cthulhu was the impossibly mind-breaking irrational thought experiments we subjected ourselves to along the way! :D
On the contrary, it will not be the number of rolls that destroys it, but being thrown away. You should roll it as much as you can before then, any time spent not rolling is time wasted!
🎶These dice are spinning around me
🎶The whole table's spinning without me
🎶Every sesh sends future to past
🎶Every roll leaves me one less to my last
Besides, everyone knows you play the long game of training your dice by always resting them with the high value up.
It probably does nothing, but maybe the atoms shift over time and it warps just a bit and rolls better.
That’s stupid. But obviously how the dice strikes the table impacts its balance and therefore the probability of rolling specific numbers. So we must figure out what side need to strike the table first to decrease the probability of getting an undesirable roll. Boom, I out physicsed you’re probabilities.
I did this with d6. I am a master at rolling low, cursed you may say but if I pick up a 1 the chance of rolling another one is lower. Or at least in my mind and memory.
The funny thing is that this logic assumes the rolls are independent (so you can just multiply probabilities), but the definition of independence is that past rolls can't affect future ones. So basically it's saying that past rolls can't affect future ones and therefore they must.
Thats the same argument to use taking a bomb on a plane. What are the odds of having 2 bombs on board?
Monty Hall would love this guy
It literally doesn't matter whether you stick with your door or switch.
Takes mathematical model and shoves it in the trash
No! I won't listen! It doesn't matter, I tell you!!!
Man there's something about the monty hall problem that just messes with human reasoning. I get it now and it's really not even complicated at all but when you first learn about it you tend to overthink it. Now I don't even understand how I was ever confused.
I think the problem is that people forget Monty Hall has information that the contestant does not. The naive assumption is that he's just picking a door and you're just picking a door. The unsophisticated viewer never really stops to think about why Monty Hall never points to a door and reveals a prize by mistake.
One way I've had success explaining it is to expand the problem to more than three doors. Assume 100 doors. Monty Hall then says "Open 98 doors" and fails to reveal a prize behind any of them. Now its a bit more clear that he knows something you don't.
Yes, it is more like a sleigh of hand or a magic trick. When the presenter discards an option, they are acting as a hand of god that skews the probability.
Maybe? I don't think that was my issue. I think I was overthinking it and using the second "choice" as an event with separate odds.
The thing you're getting by switching is the benefit of the information provided by the person who revealed an empty door.
Before a door is open, you have a 1/3 chance of selecting correctly.
After you select a door, the host picks from the other two doors. This provides extra information you didn't have during your initial selection. The host points to a door they know is a dud and asks for it to open. So now you're left with the question "Did I pick the correct door on the first go? Or did the host skip the door that had the prize?" There's a 1/3 chance you picked the right door initially and a 2/3 chance the host had to avoid the prize-door.
Yeah I think the easiest way of understanding how monty affects the choice is to imagine 100 doors, and after you pick one monty opens 97 other ones. Wouldn't you want to change after that?
tigeruppercut@lemmy.zip I interviewed someone for a jr data analyst roll once, and one of the questions I asked was to explain the Monty Hall problem.
They told me the odds were 50:50, so I rolled out the 100 door version.
They told me the odds were 50:50.
Management decided to hire him, anyway.
Are you being facetious, or do you want a non-mathematical explanation?
Imagine if he didn't always show the other goat. "So you picked door number one. Let's see what's behind door number 2!"
Door 2 reveals a brand new car
"... So, do you wanna switch to door 3?"
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.
The die has no memory of its past roles
How did you manage to spell the same word differently in the same sentence?
The same logic applies to a nat 20 though
The die need to warm up. I have to practice my release to make sure of a good number. Don't take this from me.
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