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 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:
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
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.