If you’re having an unlimited number of rolls prior to your “real” roll, then you would be, in essence, creating a situation that has a statistically lower chance of happening.
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.
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.
Before you roll any dice, the chances of rolling two nat 1s are 1/400. But after you roll your first die, whatever it happened to be, your chances of rolling a nat 1 are 1/20. The chances of the entire scenario have no impact on the probability of the individual rolls
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%.
Ok. I know that this isn’t correct… But isn’t it?
If you’re having an unlimited number of rolls prior to your “real” roll, then you would be, in essence, creating a situation that has a statistically lower chance of happening.
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.
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.
No. You have a five percent chance of rolling any given number on any given roll on a twenty sided die.
Okay, normally, sure. But what if I cross my fingers and kick my heels and rub my lucky clover?
I don’t know but if you rub my lucky clover you’ll get a little squirt of luck.
then it’s 4% each result. you don’t want to know what happens with the missing 20%.
A properly weighted die
You’d end up with a perfectly smooth D20 which would never stop rolling, assuming it was rolled in a vacuum.
Before you roll any dice, the chances of rolling two nat 1s are 1/400. But after you roll your first die, whatever it happened to be, your chances of rolling a nat 1 are 1/20. The chances of the entire scenario have no impact on the probability of the individual rolls
@FearfulSalad @brian
The Gambler’s Fallacy even shows up in humor.
Hitchhiker: “thank you, but aren’t you even a little worried picking up hitchhikers?”
Driver: “nah bro, the odds of a car having TWO serial killers is too tiny to worry about.”
My mother used to tell me there was always one weirdo on every bus. I couldn’t find them.
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%.