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 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.
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.
I read an article written by an inmate who explained how they’d form dice out of toilet paper (because dice were banned in his case) to play D&D. I assumed that was the case in most facilities.
Yeah, no doubt different facilities run things differently. Depends on where you are, who governs it and it’s security level. Low security gets more privileges than medium, high, or maximum security. Though for us medium security and lower could have small games, dice, dominoes, etc during rec hours.
You had me in the first half for real
There is statistically no difference if one is weighted because it’s gone from 6/36 to 1/6.
Your question is already answered correctly, so I’m just chiming in with thoughts on a similar situation :)
If you weigh both dice, it gets interesting again.
The obvious is to make one die roll always 3 and the other always 4, and get 1/1 chance of 7 but that’s boring and you’ll only get a few throws in before you’re obviously cheating.
Dice are arranged so that opposite sides always add up to 7, meaning you can get 1,2,3 around one corner and 4,5,6 around the opposite corner. So if you weigh opposite corners on each die, you get a 1/3 chance of rolling a 7 by varying combinatons. You might get away with a few more rolls like that.
Tell me the weights and I told you the probability of getting 7.
At the moment you have P(x) = {X=1 : 1/6, X=2: 1/6…X=6: 1/6). Give me the new P(X) and I calculate this for you.
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.
As mentioned by others: No matter how it’s weighed, and no matter what it lands on, there’s a 1/6 probability that the other dice will land on the number you need to get seven. The probability of getting seven is independent of the “first” dice.
Added an example
No, it wouldn’t, as long as only one of the dice is weighted.
If it has a 95% chance to roll a 6, and a 5% chance to roll any other number, or a 100% chance to roll a 6, or a 0% chance to roll a 6, the chance is still 1 in 6 to roll a 7 with two dice (where either zero or one is weighted).
Added an example
Doesn’t actually matter.
A normally weighted die has a weight of 16.67% for each face. No matter what result the first die rolls, the second one has a 16.67% chance of rolling the number needed to total 7. Therefore, the average chance of a (total of) 7 is (16.67 + 16.67 + 16.67 + 16.67 + 16.67 + 16.67) / 6, or, 16.67%, or, 1 in 6.
Consider your example: Die #1 has the following weights:
- 1: 0%
- 2: 20%
- 3: 20%
- 4: 20%
- 5: 20%
- 6: 20%
In your example, if die 2 rolls a 6, there’s a 0% chance of a (total of) 7, instead of the normal 16.67%, but if die 2 rolls a 1, 2, 3, 4, or 5, it has a 20% chance of totaling 7, instead of the normal 16.67%.
The average chance, therefore, is (0 + 20 + 20 + 20 + 20 + 20) / 6, or, 16.67%, or, 1 in 6.
So long as you roll the weighted die first, the odds the unweighted die lands on the number you need is 1/6. If you roll the unweighted die first though, your odds of getting the needed number are no longer 1/6.
The odds that the first die landed on the correct number are 1 in 6, though, so if you’re considering the throw of both dice as a whole, the chance is still 1 in 6 regardless of which die you throw first. (If you’re rolling the unweighted die first and then evaluating the chance of getting a 7 based on that outcome, then you’re correct.)
Added an example
If a die is weighted, the first roll is no longer 1/6 probability to get a 7
Yes, actually, it is. No matter what the first die lands on, there is a 1 in 6 chance that the second die will land on the corresponding value necessary for a “7”. You could glue the first die to the table with “6” (or any other number) showing, and there will be a 1 in 6 chance that the second die will bring the sum to 7.
Weighting one die (to favor “6”) will increase the probability of every outcome over 7, and will decrease the probability of every outcome under 7, but the probability of rolling a 7 will not change.
