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3 years ago ::
Apr 25, 2010 - 5:23PM
#51
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This is mathematically flawed in that there are 2 rolls with brutal as written vs 1 roll with a shift. That is all I am saying. The results are 1/10 and 2/10 for a reroll that would present with probabitity of a higher average. And in regards to the guy having his 1=11 and 2=12 is imo a house rule gone wrong. Just having a number shift as was put makes sense for a house rule but not the same probability or I am sure they would of done this officially in the rules.
Although I am no probability expert I'm pretty sure your probability analysis is the one that is flawed. 1's & 2's don't count, excluding 1's & 2's you have a 1/10 chance of rolling a 3-12, precisely because you exclude rolls of 1 & 2. The other method LoW suggested accomplishes the same thing just with no need to reroll and there's still a 1/10 chance of rolling any given number.
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3 years ago ::
Apr 25, 2010 - 5:43PM
#52
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Date Joined:
Mar 20, 2009
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The probablility of hitting 3-12 and 1-10 is not the same, a 1d12 isn't the same base probability. 1 out of 6 rolls on the die of a d12 means you get a reroll on the d12. When you reroll, say their is a 50% change you roll below the average and 50% change you roll above the average on the die. 1 out of 3 rolls on a d12 below the average has you reroll. So on a d12 if you roll a 1 or a 2, you reroll. Makes sense.
You continue this until you get above a 2. Every time your reroll, consecutive rerolls shifts the probability of what comes up. Flip a coin, call heads, it is heads. Do it again, what do you call? Tails? probability says it should be tails, based on the previous flip, but it is heads again. Do this now a third time. If the first one was heads, the second one was heads, the third as a probability of being tails is even higher then 50%. IT is a fundemental flaw of probability to look at the exact moment of the roll and not base it off the previous roll. Hence why a shift makes perfect sense for min dmg, but brutal is more of a probability scale then a shift and thus has the ability to do rerolls.
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3 years ago ::
Apr 25, 2010 - 5:48PM
#53
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Date Joined:
May 16, 2007
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The probablility of hitting 3-12 and 1-10 is not the same, a 1d12 isn't the same base probability. 1 out of 6 rolls on the die of a d12 means you get a reroll on the d12. When you reroll, say their is a 50% change you roll below the average and 50% change you roll above the average on the die. 1 out of 3 rolls on a d12 below the average has you reroll. So on a d12 if you roll a 1 or a 2, you reroll. Makes sense.
You continue this until you get above a 2. Every time your reroll, consecutive rerolls shifts the probability of what comes up. Flip a coin, call heads, it is heads. Do it again, what do you call? Tails? probability says it should be tails, based on the previous flip, but it is heads again. Do this now a third time. If the first one was heads, the second one was heads, the third as a probability of being tails is even higher then 50%. IT is a fundemental flaw of probability to look at the exact moment of the roll and not base it off the previous roll. Hence why a shift makes perfect sense for min dmg, but brutal is more of a probability scale then a shift and thus has the ability to do rerolls.
1-10 and 3-12 are the same probabilities. They're each 10 possibilities with a 1/10 chance of each possibility. 1's and 2's effectively don't count on 1d12 burtal 2, so on 1d12 brutal 2, you can effectivelyonly get a 3-12, with each possibility being a 1/10. The first roll doesn't matter. It doesn't affect the odds of the 3-12 at all. You can get a 3, a 4, a 5, a 6, a 7, an 8, a 9, a 10, an 11, or a 12, with each one being just as likely to come up up.
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3 years ago ::
Apr 25, 2010 - 5:58PM
#54
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Date Joined:
Mar 20, 2009
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I am not going to teach college level probability and statistics in a forum where the original post I stated was looking for an answer to a question by referencing a RAW that I have yet to see. I never once argued RAI in here other then the fact that this die shift does change the probability.
I have also referenced that there are powers that do state minimum rolls vs rerolling which is what brutal does. I have yet to seen a reference that dictates a timing for brutal other then it is an ability that grants a reroll. I have also shown that the logic behind luckbenders does limit reroll abilities using base logic equtation. And there are powers that let you roll again saying to take the second roll that doesn't have the clause "even if it is lower". I am just finding it hard to believe RAW is RAI without more support then through interpertation which invites RAI.
And yes, stastically speaking the probability of a shift does offer a different outcome then the probability of a reroll.
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3 years ago ::
Apr 25, 2010 - 6:01PM
#55
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Date Joined:
May 16, 2007
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I've taken college level probability, and I think you should look up independence. EDIT: You're right, though, the probability of the dice has nothing to do with whether a reroll from another source uses the brutal property.
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3 years ago ::
Apr 25, 2010 - 6:29PM
#56
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Date Joined:
Mar 20, 2009
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Independence says that roll 2 doesn't matter what roll 1 was and won't change its outcome. But it only states that the 2 rolls are independent of each other. And this is 100% correct. But it is also known starting with coins as an example that getting a head is 50%. Now the probability of the second coin flip is 50%. But the change of getting 2 heads in a row is not 50% but 25%. So although the probability theory of independence is 100% correct, you can still look at the previous roll in sequence.
Breaking the die roll into top half and bottom half, there is a 50% change to be 7+ and 50% chance to be 6 and lower. So if the first roll was the bottom half (1 or 2) then the second roll is 25% likely to be below or at a 6 on a d12 and 75% likely to be above a 6. Each individual value on a d12 is 1/12 not 1/10. This does matter in the long run.
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3 years ago ::
Apr 25, 2010 - 6:30PM
#57
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Date Joined:
Mar 20, 2009
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Sorry was posting when you updated. but it is good for others to see.
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3 years ago ::
Apr 25, 2010 - 6:31PM
#58
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Date Joined:
May 16, 2007
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for the record, my interpretation is that you reroll the damage dice and the damage dice is 1d12 brutal 2 or 2d4 or whatever the full expression would be for your weapon. But my reasoning isn't anything that hasn't already been said, so i won't bore you with it.
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3 years ago ::
Apr 25, 2010 - 6:47PM
#59
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Date Joined:
Jul 28, 2003
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Independence says that roll 2 doesn't matter what roll 1 was and won't change its outcome. But it only states that the 2 rolls are independent of each other. And this is 100% correct. But it is also known starting with coins as an example that getting a head is 50%. Now the probability of the second coin flip is 50%. But the change of getting 2 heads in a row is not 50% but 25%. So although the probability theory of independence is 100% correct, you can still look at the previous roll in sequence.
Breaking the die roll into top half and bottom half, there is a 50% change to be 7+ and 50% chance to be 6 and lower. So if the first roll was the bottom half (1 or 2) then the second roll is 25% likely to be below or at a 6 on a d12 and 75% likely to be above a 6. Each individual value on a d12 is 1/12 not 1/10. This does matter in the long run.
Getting way off-topic, but:
Probability of heads = .5 (50%)
Probability of heads given the first roll was heads = still 50%. The first roll does not affect the second roll. This despite the fact that the probability of rolling two heads in a row is only 25%. Note that the probability of ANY particular combination is only 25%; H-H, H-T, T-H or T-T.
You cannot look back to the previous result to modify the expectation of the next result.
In fact, the probability of getting two heads in a row given you already rolled heads once is 50%, not 25%.
So one could substitute 1d10 + 2 for 1d12 re-roll 1s and 2s and get the same results - a 1 in 10 chance for any particular result.
Think of it this way. If you cannot count 1s and 2s as valid rolls then, effectively, the only valid results are the numbers 3 -12. Since one roll does not affect the next (independence), you end up with a equal chance for any number from 3 to 12.
It makes to difference that the odds of rolling a 1, then a 2 and then a 3 are not the same as rolling the 3 in the first place. The odds of rolling that "3" have not changed at all.
What is different is the number of rolls you need to make to get your actual result. With a 1d10 + 2 you roll once, with a 1d12 re-roll 1s and 2s, you roll more than once 1 time in 6. So, while the probability for each result has not changed, with the by-the-book method you get top roll more dice.  It's more fun, I think!
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3 years ago ::
Apr 25, 2010 - 7:01PM
#60
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Date Joined:
Mar 20, 2009
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Again off topic, but say you flip a coin and get tails. You flip the coin again and get tails, what was the probability that you would get Tails and Tails. 25%. Just as you stated it. 25% for each possible outcome when there are 2 flips involved. That means you would have 75% chance to not get 2 tails flips in a row. Take a die, divide it in upper half and lower half. If you roll the lower half (Tails) the first time, the chance you will get a lower half the second roll(Tails) although is 50%, the chance for getting in the bottom half 2 times in a row is 25% hence why rolling 2 times doesn't have the same outcome statistically speaking as rolling 1d10 and adding 2. Now if you want to get fancy, start looking into probability and outcomes of groups of series  But there is a difference between 1d12 brutal 2 and 1d10+2. on a 1d10+2 there is a 1/10 chance to get a 3. There is 1/12 to get a 3 on a 1d12 brutal 2. So in the truest sense, 1d10+2 is worse for you.
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