Its 1/6 to roll a 7 with two 6-sided dice. You can get weighted dice that make it more likely to land on a certain number.
Does having one weighted dice change the odds at all? My gut says no but reality is a tricky bitch and I’m convinced im wrong somehow.
If your goal is to roll a 7, then no, weighting one die doesn’t help, because it doesn’t matter which side comes up in the weighted die.
(Another way to look at it is that you can place 1 die on whatever side you like, then roll the 2nd die: you still have a 1/6 chance of rolling a combined 7).
However that is only the case for a 7, because you can roll a 7 with any combination of the first die (and a particular value in the 2nd). If your goal is to roll a 12, then weighting one die towards 6 will affect the odds, because you need a 6 on that first die to roll a 12; any other outcome makes it impossible.
I mean you can get dice weighted differently so you almost always get a 6 and a 1 for example.
There’s no point in weighing one die. The odds are the same as not weighing any die
You’re marginally less likely to roll a 7 with a single weighted die.
My reasoning: if you use your weighted die, whoever you’re gambling against might figure it out. Now they’re pissed, they’re gonna give you the beat down. But you’re packing, and OH SHIT YOU DIDN’T MEAN TO HIT HIM IN THE HEAD YOU WERE AIMING FOR HIS LEG SHIT SHIT and now you’re serving jail time for manslaughter and dice aren’t allowed in jail, thus you won’t be able to roll any 7s for a while.
Don’t worry, there are dice in (at least some) prisons, every cell block had at least one D&D group going with 4-6 people, and they used dice.
Source: Was a prison guard in the ARMY for 4 years at Fort Lewis.
Ah, so the chance of rolling a 7 changes to 1/20?
If you’re rolling D20s, but there are also D4s, D6s, D8s, D10s, D12s, and a D10 Percentage die for 5e, and some spells require multiple of the same dice, so rolling 6d6 is a very real scenario, so you can have plenty of dice to use to find variations to get to 7 with different sided dice.
That just sounds confusing. You’re putting me off murder now.
Once again, D&D improves society.
You’re right! Let’s say we have two dice:
D₁ is fair and has a 1/6 probability of rolling each number from 1-6.
D₂ is weighted, with probabilities P₁, P₂, P₃, P₄, P₅, P₆ to roll each number.We roll D₁, and get a number with the following probability distribution:
1: 1/6
2: 1/6
3: 1/6
4: 1/6
5: 1/6
6: 1/6We roll D₂, and get a number with the following probability distribution:
1: P₁
2: P₂
3: P₃
4: P₄
5: P₅
6: P₆We find the probabilities of every combination of rolls that yields a 7:
1+6: 1/6 P₁
2+5: 1/6 P₂
3+4: 1/6 P₃
4+3: 1/6 P₄
5+2: 1/6 P₅
6+1: 1/6 P₆Adding these together to get the total probability of rolling a 7, we get 1/6 (P₁ + P₂ + P₃ + P₄ + P₅ + P₆). Since the probabilities of rolling each number must sum to 1, we get a probability of 1/6 to roll a 7, and your gut is right. :)
7 is the only number where this property holds. Other numbers will have a probability dependent on the weighting of the die, which could be calculated with a similar method.
If a die is weighted, the first roll is no longer 1/6 probability to get a 7; the roll isn’t random and there isn’t enough info. I think though it’s 3/12 (1/4 for a 7 (6+1, 5+2, 4+3)? Maybe not. I hate this shit.
You need to roll two dice to get a sum of seven. Consider two fair dice: No matter what the first dice lands on, there’s a 1/6 probability that the second dice lands on the number you need to get a total of seven.
Consider now that one dice is weighted such that it always lands on 6. After you’ve thrown this dice, you throw the second dice, which has a 1/6 chance of landing on 1, so the probability of getting seven is still 1/6.
Of course, the order of the dice being thrown is irrelevant, and the same argument holds no matter how the first dice is weighted. Essentially, the probability of getting seven total is unaffected by the “first” dice, so it’s 1/6 no matter what.
That’s if it’s perfectly weighted. If it’s weighted to roll a 6, it might not always land on 6. This would lower the chance of rolling a 7 depending on what the overall probability profile is on the weighted die.E: consider a die weighted to favor 6. Standard dice have opposite faces add up to 7. If this die favors 6 to the extent it never rolls a 1, any time a 6 is rolled on the second die can never result in a total roll of 7.E2: I has the dumb. Apologies.
There is statistically no difference if one is weighted because it’s gone from 6/36 to 1/6.
Just want to expand on this as it’s the most direct explanation.
With two die there are 6 ways to you can roll a seven (each side has one way to add up to seven), and 36 total combinations (6 sides * 6 sides). So the odds are 6 times out of 36 or 6/36.
With one weighted die, you have a set value (say 3 for example). There is only one side on the other die that will equal 7 (4 in our example). So you have 1 out of 6 possibilities, or 1/6 chance.
However, this is only true for 7. If you were targeting 2 for example, the odds can change substantially. Normally you have one way to get 2 (1 and 1) so you’d have 1 out of 36 possible rolls or 1/36. If the weighted die was weighted to 6 though, you’d never be able to get 2, so your odds would be 0.
