To do this you need to know 2 things --- You need to know what the probability of success for your first roll, and what that probability should have been. But wait, there is one more thing, an important assumption I rather glossed over last time --- You need to have completely forgotten everything about the first roll except that you succeeded or failed.
And that's harder than you might think. ---> More after the fold --->
This post is perhaps more philosophical than mathematical, because it has more to do with things that you can honestly remember, rather than a clear-cut way to fix a problem. Many players are remarkably good at remembering their dice rolls. Battletech players tend to note when they roll 2's and 12's, which is not surprising since those numbers sometimes lead to a little extra mayhem. But some events are more memorable than others. Consider the following:
Example 4: Dan rolled and hit with two medium lasers needing 7+ on 2d6. Later he realizes the base of the Jenner miniature he was shooting at obscured a light woods (a +1 modifier), and he should have rolled with 8+ to-hit. Dan then approaches Scott to correct his mistake. Scott notes, "I remember that one of the dice was a six."Now we have broken the last assumption, because Scott remembered some information about what was rolled. That's OK though, because we can use this partial information as part of the correction.
Example 4 continued: Dan sets down one die showing a 6, and rolls the other one. If he rolls a 2 or more the result will remain a hit (8+), but will change to a miss if he rolls a 1 (6+1=7).Here the recollection that one of the dice was a 6 mean the conditional re-roll is very likely to succeed (remain a success). If Scott recalled one die was a 2, then the conditional result would stay a success only if the other on a 6 (5 or less changes to a miss). In general, if you remember any bit of information about what the original die roll was, then you can make use of it.
Now here is where things get a little fuzzy for me. If as in the example, one of the dice showed a 6, it seems unlikely that the other dice was also a 6. If I had rolled "12" then I would probably have remembered that. This is true even without Scott's recall of a 6, because if I had rolled a 12 in the first place, I am more likely to remember it. Therefore, I probably did not roll a 12, and maybe I should exclude a 12 result from my re-roll and take this into account. But now we are on a slippery slope, because if I'm pretty sure I didn't roll a 12, then maybe I didn't roll as high as 11 either, or 10 ...
In other words maybe the conditional re-roll I described in my last post isn't quite fair either, because it pretends that you really don't remember anything at all. I may still be getting away with a slightly higher probability of hitting than I should.
This is more of a philosophical problem, but I have a good answer for it in the form of a basic gaming table rule. When it comes to games and mistakes made during play, it's just good sportsmanship it admit the mistake and attempt to make it right with the other player. Further, the person who goofed should expect to pay a bit of a penalty for it. In this case, the conditional dice roll should attempt to partly correct the mistake, but stop short in favor of the other player.
Here is my suggestion. Always assume the original roll was no higher than what was needed, and ignore and numbers above that in the conditional re-roll.
Example 5: Dan rolled and hit with two medium lasers needing 7+ on 2d6. Later he realizes the base of the Jenner miniature he was shooting at obscured a light woods (a +1 modifier), and he should have rolled with 8+ to-hit. Dan then approaches Scott to correct his mistake. He rolls for the first laser until he gets a 7 or an 8 (not a 9, 10, 11, or 12!), and it eventually comes up 8, and this remains a hit. The rolls for the second laser, re-rolling several results less than 7, and finally gets a 7. This result is changed to a miss, and Scott erases the appropriate damage from the record sheet.
This is a little awkward to do with dice, so knowing a little probability will help. We are down to two outcomes, 7 or 8, with probabilities 6/36 and 5/36 on 2d6. We only want those two rolls though, so this reduces to 6/11 and 5/11 if we only consider those two possibilities*. My original probability of hitting on 7+ was too high by 5/36, and given that I rolled at least a 7 but no higher than 8, then the conditional re-roll should have 5/11 chance of remaining a hit, and 6/11 chance of changing to a miss. This is a ~54.5% chance I will now miss, but compare this to my example 2 in the last post, where the chance of missing was 6/21 or ~28.6%. This is almost always an over-correction, but this is more fair to Scott (the injured party), because it gives him the benefit of the doubt for all other partial information - what may or may not be remembered about the original roll.
* [Math note: the two possibilities we are interested in are rolls of 5 or 6, with a total probability of 5/36+6/36 = 11/36. To make the probabilities of the two rolls we are interested in sum to 1.0, we multiply by 36/11, so 5/35 * 36/11 = 5/11, 6/36 * 36/11 = 6/11.]
So we pull out our handy 11-sided die and roll ... 6 or less is a hit and 7 or more is a hit ... and what do you mean you don't have any 11-sided dice?
If you have followed me this far, I'm pretty sure you can figure out the 11- sided die problem. ;-)
In the process of writing this post I thought I was going to need a 36-sided die, so I went and made a table for it. That didn't work out, but here is the table anyway. -->