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Message: <blockquote data-quote="EzekielRaiden" data-source="post: 6877885" data-attributes="member: 6790260">The proportion does not remain the same.When we exclude potentially roll-able options--using your example, 5 or less--we change nothing about how many possible options produce 7, but we do change the total number of "real" possibilities. (The excluded possibilities thus become "virtual"--yes, real dice will make them, but their existence doesn't matter for the real, "recorded" stats.)With an unmodified 2d6, there are 36 possible distinct results, which can be seen by pretending that one die is, say, blue and the other is red. With "2d6, reroll if less than 6 until it's not less than 6" we have removed rolls of 2 (one roll), 3 (two rolls), 4 (three rolls), and 5 (four rolls), for a total of ten rolls removed from the distribution. This, then, means that all the other rolls--the ones not removed--are only possibilities out of 26, rather than 36, since those ten excluded rolls no longer hold sway over the "recorded" results. (Again, using "recorded" as I defined it upthread.)So, we now have our "bell curve," really more a pyramid, as options out of 26. For (say) 1000 actually recorded rolls, this means we should expect 1000*(7/26) = 269 recorded 7s, because the recorded proportion is (7/26) = .2692307692307... approximately. We should expect (6/26) = (3/13) = .230769 repeating as the recorded proportions of both 6 and 8. Etc.This is caused by exactly the same things that make the mean different. We now have (6*5+7*6+8*5+9*4+10*3+11*2+12)/26 = 8.15. The *only* way you can get that 8.15 figure is by dividing by the reduced number of acceptable possibilities (26), which is what causes the proportions in any sufficiently-large sample size to be different from the proportions you would expect of an unmodified 2d6 roll.It doesn't affect the symbols that may be shown, or however one might say that better, but it absolutely does affect what the expected result will be on the whole.To use an even simpler example, let's say we reroll 1s and 2s on a single d6. Then you have (reroll, reroll, 3, 4, 5, 6). Each reroll, itself, is another (reroll, reroll, 3, 4, 5, 6), which makes for an infinite series (as someone else argued previously). When you run that infinite series to its end, the amount of probability that had been present for 1 and 2 becomes proportionally distributed among the other possible results; since all the other results have exactly the same probability, they get the same extra amount, which is (2/6)*(1/4) = (2/24) = 1/12 (the 1/4 is because you're apportioning this into 4 equally-likely acceptable alternatives). You then add that to the original, natural probability of 1/6 for each face: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4. So the proportion of recorded 3s will increase from 1/6 to 1/4.In other words, as I have said twice before, this is a simple and straightforward application of Bayes' Theorem for conditional probability. It has nothing to do with the gambler's fallacy, and asserting that it does is terribly confusing.</blockquote>

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