Comment on Odds of rolling a 7 with a weighted die
thebestaquaman@lemmy.world 1 day agoYou 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.
meco03211@lemmy.world 1 day ago
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.
KoboldCoterie@pawb.social 1 day ago
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).
meco03211@lemmy.world 22 hours ago
Added an example
KoboldCoterie@pawb.social 22 hours ago
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 same number. 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 2, 3, 4, 5 or 6, it has a 20% chance of totaling 7, instead of the normal 16.67%.
The average chance, therefore, (0 + 20 + 20 + 20 + 20 + 20) / 6, or, 16.67%, or, 1 in 6.
thebestaquaman@lemmy.world 1 day ago
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.
meco03211@lemmy.world 22 hours ago
Added an example