A properly weighted die

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.

How did you manage to spell the same word differently in the same sentence?

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!

Ai probably

This is like a common house fly worrying about the lifespan of Cthulhu.

Right but the way I took the meme was that you would roll *until* you get a 1, then deciding the next roll is the "real" one.

That's what the meme says, but probability doesn't work that way. If you want the result from a roll, what you've roller before has no bearing on the result from this roll. Thus the chance for a single d20 roll is always 1/20, or 5%.

Monty Hall would love this guy

You haven't seen how some of the folks I play with roll.

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!!!

> If you’re having an unlimited number of rolls prior to your “real” roll You'd end up with a perfectly smooth D20 which would never stop rolling, assuming it was rolled in a vacuum.

Okay, normally, sure. But what if I cross my fingers and kick my heels and rub my lucky clover?

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 itm

Are you being facetious, or do you want a non-mathematical explanation?

i assume revenge for stepping on a d4 once?

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.

then it's 4% each result. you don't want to know what happens with the missing 20%.

The standard answer is that the odds of the first roll don't change the odds of the second roll, the second roll still has a 1/20 chance of a 1, no matter what the first roll is. The more thorough answer is that it's a misunderstanding of what probabilities are. Yes, there's a 1/400 chance of rolling 2 1s, but by the time you roll the first die and get a 1, you're not talking about that problem anymore. You've introduced new information to the problem, and thus have to change your calculation. There's a 1/20 chance of rolling 2 1s after you're already rolled one. Let's calculate it... So, there's 400 ways 2 dice can fall, yes, and there's only 1 way that they can both fall on 1. However, there's 20 ways that the first die can fall on 1, one for each possible fall of the second die. So, when we say that that has already happened, we have to eliminate 380 of those 400 die rolls, those are no longer possible. That leaves us with only 20 ways that the second die can fall, and only 1 of those is a 1. So the odds of rolling a on the second die, after already rolling a 1 on the first die is 1/20. We can also calculate it differently. What are the odds of the second die falling on 1? Cause that's the one we care about, really. And there's 20 ways that can happen, one for each possible fall of the first die. So the odds of the second die falling on 1, when rolling 2 dice is 20/400, or 1/20.

D4 is the devil's dice.