The question relates to statistics, and how you add up odds.

I'll spare you my function for determining the odds of a die roll. Assume for the sake of simplicity that the foe's AC is 11, and my attack mod is +1, which would mean that a 10 or higher will hit. In this case, the odds of a hit are 0.55

Let's say my first attack is with a dagger; The average damage of a d4 will be 2.5 plus my mod of one. The odds to crit are .05, so I'll subtract that from the odds to hit, calculate it separately, and add the results together. This leads to:

A1: 0.5 * (2.5 + 1) + .05 * (2 * 2.5 + 1) = 1.75 + 0.30 = 2.05 This means my average damage for attack A1 will be just over 2.

My second attack is also with a dagger. It has the same chance to hit, but I don't get the +1 modifier on the damage roll:

A2: 0.5 * 2.5 + .05 * (2*2.5) = 1.25 + .25 = 1.50 This means my average damage for attack A2 will be 1.50

Add these together and I get my average damage per turn of 3.55. Now, ignoring advantage and disadvantage, which I've already figured out how to calculate, this would be my total. However, where I'm having trouble is factoring in my odds to get a Sneak Attack. For the purpose of this experiment, I am positioning myself well and can potentially get a sneak attack on hit every single turn, from either A1 or A2 (Yes; Two handed fighting allows the second strike even if the first misses). The odds of either attack A1 or A2 hitting are .7975 or 79.75% (.55 + .55 - .3025, which is the odds of both attacks hitting). Thus, the average damage of a sneak attack at level one (1d6) is:

S1: .7975 * 3.5 = 2.79125 This means that on average, every turn I will deal 2.79125 in Sneak Attack damage.

My first instinct is to add this figure directly to my figures for A1 and A2, meaning that my average total damage per turn will be 6.34125.

However, And here's where my mind is starting to boggle, A1 and A2 are independent events. Either or neither can hit. S1, the Sneak Attack is...sorta independent??? I guess? Maybe? I'm not sure if I'm properly adding the odds of S1 to the odds of A1 and A2.


  • \$\begingroup\$ Welcome to RPG.SE! Take the tour if you haven't already and see the help center or ask us here in the comments (use @ to ping someone) if you need more guidance. Good Luck and Happy Gaming! \$\endgroup\$
    – Someone_Evil
    May 6, 2022 at 19:24
  • 1
    \$\begingroup\$ I think the title of this question may have a small error. You seem to be asking about two weapon fighting (PHB p195), not attacks with weapons that have the two-handed property (PHB p147). \$\endgroup\$
    – Blckknght
    May 6, 2022 at 21:09

2 Answers 2


My first instinct is to add this figure directly to my figures for A1 and A2, meaning that my average total damage per turn will be 6.34125.

A1 and A2 are independent events. Either or neither can hit. S1, the Sneak Attack is...sorta independent???

The bad news: The variables are dependent. If you miss with both attacks, you're not getting any Sneak Attack damage.

The good news: If you're only interested in the average damage (specifically mean damage, aka expected damage) as opposed to e.g. the entire probability distribution, linearity of expectation (reference 1, reference 2) applies even if the variables are dependent. So you can in fact just add up the expected damages of the first attack, the second attack, and the sneak attack to get the expected total damage.

  • \$\begingroup\$ Here comes the professional stats guru. I'll hand it over to you, @HighDiceRoller \$\endgroup\$ May 6, 2022 at 20:19
  • \$\begingroup\$ I don't actually get it, but I trust you. A correctly calculated series of expectations (which, I guess is what the results of my calculations are) will be accurately summable even if some are dependent on others. \$\endgroup\$
    – Laura Ruby
    May 6, 2022 at 21:24
  • \$\begingroup\$ This example from your second reference helped a lot: If the sum of the numbers rolled on the dice is A and the product of the numbers rolled is B, compute E[A+B]. Solution: We know that E[A]=7, and since the two numbers rolled are independent, we have E[B] = 3.5 ⋅ 3.5=12.25. Thus, despite the fact that A and B are clearly dependent, linearity of expectation tells us that E[A+B] = E[A] + E[B] = 7 + 12.25 = 19.25. \$\endgroup\$
    – Laura Ruby
    May 6, 2022 at 21:27

You are doing it (mostly) right

I'm not a statistician (as some experts around here are, who, if I am wrong will easily explain what is wrong with my answer).

I'm not sure why it bothers you either attack or both can miss. This is already reflected in your probabilities.

One way I learned to calculate the odds for one or more events happening is assume the extreme case where all fail independently (i.e. multiply their probabilities to fail), and substract that from 100%. Here this would be .45 x .45, substract that from 1, then you get the chance that either of them or both hit, which also gets to 0.7975.

This is the chance to hit at all in the round, which you multiply with the expected damage for sneak attack, which you can use once per round. You just can add the resulting sneak damage to the total.

There is one minor flaw: when you hit with a critical, you would also double the sneak attack damage. You only have this chance on whatever of the two rolls you apply the sneak attack to, and on that roll the chance to have a a critical is 5%. So you also have .05 x 3.5 to add to the sneak damage, or multiply that overall sneak damage with 21/20. That is (1 - 0.45 x 0.45) x 3.5 x 21/20 = 2.9308125

Normally, you will always take the sneak attack if the first attack hits. For the case that the second attack also hits and crits, you miss out on the extra crit damage from sneak attack. That scenario has a probability of 0.55 (first hit) * 0.05 (critical on second hit), so we have to deduct 0.55 x 0.05 x 3.5 (the expected critical damage from sneak in this situation), or 0.09625 to correct for it. Deducting this from leaves us with 2.8345625 sneak damage.

Your overall expected damage (barring advantage or other effects) is 2.05 + 1.50 + 2.8345625 = 6.3845625.

You mentioned that your real stat bonus is +3, not +1, and you just use +1 to keep things simple here. You could increase your damage by using short swords instead of daggers, unless you want daggers for other reasons, or you also only use them to keep it simple.

  • \$\begingroup\$ I'm not bothered by A1 or A2 hitting or missing, I'm just thinking that because S1 is dependent, in some weird way, on A1 and A2, that I'm somehow altering the probability in my favor. You make it seem a bit clearer when you point that my .7975 calculation is the chance to hit at all, an entirely separate question as to individual hits and misses, but my poorly remember stats class has me remembering that the world of probability is extremely complicated. The +1 is just for the purpose of this example. My Dex is +3, don't you worry. \$\endgroup\$
    – Laura Ruby
    May 6, 2022 at 19:49
  • \$\begingroup\$ Thats how I remember it, too :). Lets wait if some of the real stats buffs around here have some more light to share on our efforts. \$\endgroup\$ May 6, 2022 at 19:52
  • \$\begingroup\$ Also, to your second edit: The chance to hit of .55 is rolled into the Dex Modifier, which is unchanged between A1 and A2. \$\endgroup\$
    – Laura Ruby
    May 6, 2022 at 19:52
  • \$\begingroup\$ Understood, doing so now. \$\endgroup\$
    – Laura Ruby
    May 6, 2022 at 19:54
  • 1
    \$\begingroup\$ What I mean is you can separately work out your expected average weapon damage per turn (ignoring crits), average extra weapon damage for crits per turn, average sneak damage per turn, and average sneak damage for crits per turn, and then sum them all up to get your total DPS. While they're all tied to the same die rolls, since you're taking averages over many rolls, you can treat them like they're independent. \$\endgroup\$ May 6, 2022 at 21:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .