There is a way to fix this, to repair my mistake, to determine if my shots really should have missed, and that is fair to both players. It's a good trick, and you don't even have to do any math to use it.

But you might have to roll a lot of dice. ---> More after the fold --->

The trick is to use conditional probability, an application of Bayes Law to correct the mistake. We are not going to do the calculations though, the dice will take care of that for us. To make this work, you need to know two things: The first condition in which you were first rolling the dice (the error), and what the correct and second condition should have been. It works like this: Recreate the first event (the wrong one) by rolling the dice until you get the same result (success, or failure) as the first time. Next apply the second (correct) condition to the same roll, possibly changing the first outcome. The procedure changes the result from the original roll with the probability needed to correct the original error.

Here is a really simple example using a success-roll on a single six-sided die to show how this is done. You might spot some ways to make it even simpler.

Example 1: You roll 1d6 needing 4, 5, or 6 to succeed, and you do succeed. Later you realize that was a mistake, and it should have been just 5 or 6 to succeed. You do not remember what the actual roll was, only that it was successful (therefore a 4, 5, or 6). To correct the mistake roll the die again, re-rolling any results of 1,2,or three. You should now have a die with a 4, 5, or 6 showing. If the die shows a 5 or 6, keep the original result, if it is a 4, change the result to a failure.Now a little math: The first roll had a 50% probability of success (3 out of 6), but should have been ~33% (2/6 or 1/3). After re-rolling the die until you get 4, 5,or 6, the probability of 5 or 6 is ~67% (2/3), and ~33% (1/3) of a 4. The probability that this second roll is a success (2/3) given that you know the first roll succeeded (50%) is 0.5 times 0.67, or ~33%, which gives you the correct probability of success (1/3) as if you had done it right in the first place.

That may have been a little overly-complicated. If you followed the math you know this could have been done with only one additional die roll, which would be much simpler. I did it this way because the same procedure works no matter what sort of dice you are using. Try this on my original example.

Example 2 (my opening example): 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 getting a 10 on the first try, 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.Now the math: The probability of rolling 7+ on 2d6 is 21/36, for 8+ is 15/36, and of rolling exactly 7 is 6/36. Given that my first roll(s) were at least 7 (I know that, because they hit) the conditional probability distribution of my second roll becomes 1/21 chance of a 12, 2/21 chance of 11, 3/21 chance of 10, 4/21 chance of 9, 5/21 chance of 8, and 6/21 chance of 7 and the conditional probability of 8+ is now (adding things up) 15/21. The probability of my first roll (21/36) times the probability of the conditional roll (15/21) is

(21/36) * (15/21) = 15/36 --> The probability I should have used in the first place (Hooray!).

Conditional probability is confusing to a lot of people, usually because they do not consider all the event-probabilities that lead to the final outcome. If you use most obvious way to correct a mistake -

**a simple re-roll**where you ignore the first result and roll again correctly - the player faces a sort of double jeopardy, and has to succeed on two different rolls. The chance of rolling both 7+ and 8+ is (21/36) times (15/36) = (315/1296) or ~24.3%, considerably less that the (15/36) = ~41.7% probability that results if you ignore the result of the first roll.

Conditional dice rolls also work to correct mistakes in failed rolls, and here the unfairness of the

**simple re-roll**is more evident.

Example 3a: Dan rolls to hit at 8+ on 2d6 and misses. Later he realizes the roll should have been at 7+, and talks Scott into giving him a new roll, throwing out the first result (the miss) and replacing it with the result of the second roll. Dan now has a 21/26 chance of hitting with the second roll, but this ignores that Dan already had one chance of hitting which failed. Dan is essentially getting a free re-roll, which is unfair to Scott.Now let's redo Example 3 with conditional probability, and this time just a little math.

Example 3b: Dan rolls to hit at 8+ on 2d6 and misses. Later he realizes the roll should have been at 7+, and Scott agrees to a conditional re-roll. Dan and Scott figure out the difference between the two rolls is just the probability of rolling exactly 7 on 2d6, which is 6/36 or 1/6. Scott agrees that Dan should roll 1d6, and if the results is a 6, to change the miss into a hit.One assumption I am using for all these examples is that there is

*no memory*at all of what the first roll might have been. What if you don't recall exactly what the roll was, but maybe you know it was "more than 5", or "one die was a 6"? You can take this sort of partial information into account too, and probably should. I'll save that for part 2, where I contemplate Conditional Dice Rolls with Partial Information.

[EDIT] - As noted in the comments, there is an additional assumption that there is nothing else that happens in between that cannot be undone. If players make decision based on the outcome, if becomes especially difficult to go back and fix it.