Editing Probability (section)

===Combination of Two/What are the odds of at least one thing happening in a set of two?===

You're trying out DnD 5e, and you learn about its fancy new Advantage mechanic, where you can roll two d20s and take the higher of the two. If your rogue has advantage and needs a 9+ on a d20 to hit, what is the chance you're going to need to test out your new boots of frantically running away?

Calculating this rogue's chances is a matter of combining the rules we've looked at above. To get the expected number of hits we can just add the chances from each d20: <nowiki> P({rolling 9+ on a d20})=.6</nowiki>, so we would expect <nowiki> 2*.6=1.2 </nowiki> successes, and if we wanted expected damage we could stop there.

However, this is clearly bullshit, because the rogue can't hit twice on one attack. To compensate for this, we subtract the chance of getting 9+ on ''both'' rolls. This, known as the inclusion-exclusion principle, gives us the real formula we want: [[File:Probability Eq3.png|x30px]]. This rogue, then, is looking at a .6+.6-.36='''.84 chance to hit''', or only a 16% chance to miss that crucial backstab.

We can get to this probability in a number of other ways, too.  The simplest might be one minus the chance that neither attack hits. We know that's the product of the two miss rates, so this works out to [[File:Probability Eq4.png|x30px]], or 1-(.4*.4)=.84. Convenient.

To check, we could alternatively look at the three mutually exclusive ways we could succeed this attack. The rogue could A: hit on both dice, B: hit on the first but miss the second, or C: hit on the second but miss the first. For the numbers:

[[File:Probability Eq5.png|x60px]]

We're now pretty sure we know the rogue's success rate, and are ready with the fleeing boots for that 16% failure chance.