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.

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; a brand new car! ...so, do you wanna switch to door 3?"

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

I don't know but if you rub my lucky clover you'll get a little squirt of luck.

Maybe the real Cthulhu was the impossibly mind-breaking irrational thought experiments we subjected ourselves to along the way!

Short version, two coin flips. There are 4 options: HH, HT, TH, TT So there's two chances to get one Tails and one Heads, out of 4, so 2/4 = 1/2, half the tosses. Then 1/4 on each of HH and TT. So rolling one Tails is more likely than rolling two. But once you've flipped the first coin, it's "locked in". If it was Heads, the only options left to you are HT and HH. The fact that there *could* have been a T that, if flipped first, would land us in TH is irrelevant fantasy. We've got the H, and all that's left is HT or HH, even odds. Dice are the same. What makes a double 1 rare is that you have to roll 1 specifically and only two times to get there, whereas a single 1 can be first or second, and the other number can be any of the other 19 other numbers. It's the duplication of different results we consider "the same" that make one thing more likely. But once you've already rolled a 1, none of that matters anymore. Now it's just 20 numbers, each equally likely. We're locked in.

Roll my number, roll my number, roll my number, I'm not afraid...

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.

The same logic applies to a nat 20 though

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.