#### fuindordm

##### Visitor

In the "low-level wizards suck" thread I didn't think True Strike was ever worth it until someone pointed out that it is only a cantrip (which I had forgotten) and that it can cancel disadvantage. So I decided to do some algebra and work out exactly when casting True Strike before an attack is worth it.

**SUMMARY:**I analyzed the probabilities of hitting and expected damage of two scenarios: casting true strike then attacking once, or attacking twice on subsequent rounds without true strike. I conclude that when the attack has no cost (e.g. a melee attack or cantrip), casting true strike is only worthwhile when you have disadvantage AND you need an 11 or better to hit the target. On the other hand, when the attack has a cost (like a spell slot), then casting True Strike beforehand gives a better damage/slot ratio than casting the spell twice.

**NUANCE:**The spell description for TS says "You point your finger at a target...", perhaps implying that you can't use TS on a target you can't see or pinpoint somehow. However, the most common source of disadvantage on attacks is not being able to see the target. Since the expected damage advantage of TS+Attack over Attack+Attack is small, if your DM rules that you can't TS someone you can't see, then I think the cost of learning this cantrip is not worth the benefit if you intend to use it mainly with free attacks.

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All the following compare two scenarios:

S1 = Cast True Strike in round 1, then make an attack in round 2.

S2 = Make an attack in round 1, make the same attack in round 2.

**MATH: Free attacks, no disadvantage**

Using True Strike with a melee attack or cantrip is never worth it in this case. It's easy to see why: both cases give you two rolls to hit, but in S1 you can never hit more than once. To put it another way, there are 400 possible outcomes of the 2 d20 rolls, which you can visualize on a 20x20 grid. The horizontal axis represents the first roll, and the vertical axis represents the second roll. Let's call the number you need to roll to hit x. In S1, you miss if the two rolls land on a point in the (x-1) x (x-1) square in the lower left corner of the grid; otherwise you hit and do the expected damage from one attack. In S1, the regions of hits and misses look the same. However, there is a (21-x) x (21-x) square of outcomes in the upper right corner of the grid that means you landed two hits, not one, so the expected damage of those outcomes is higher. Therefore, S2 is always better than S1.

**MATH: Free attacks, disadvantage**

In this situation the probability of landing a hit in S1 is easy to calculate: it's just (21-x)/20 where x is the number needed to hit. The expected damage is proportional to the probability. The expected damage in S2 can be calculated by again visualizing the possible outcomes on a grid. Except this time, we're going to look at a 400x400 grid. In effect, I model each roll with disadvantage (2d20) as if it were a d400, since there are 400 possible outcomes of that roll.

If you need x to hit, then on your disadvantaged attack roll there are (21-x)^2 outcomes where you hit and 400-(21-x)^2 outcomes where you miss. If the x-axis is your first attack and the y-axis is your second attack, imagine drawing a cross on the grid with horizontal and vertical lines that cross the axes at position 400-(21-x)^2. This cross divides the grid into four regions: the square in the lower left where you miss both times, the square in the upper right where you hit both times, and the two rectangles left over where you hit once.

The two rectangles each have area [400-(21-x)^2]*(21-x)^2, and the square in the upper right corner has area (21-x)^4.

The expected damage is therefore [2*[400-(21-x)^2]*(21-x)^2 + 2*(21-x)^4]/400^2, equal to the probability of getting one hit plus the probability of getting 2 hits weighted double. Conveniently, the terms proportional to (21-x)^4 in the numerator cancel each other out and we are left with:

S2 exp. damage = 2(21-x)^2/400

S1 exp. damage = (21-x)/20

Equating the two and solving for x gives x=11. So if the number you need to hit the target is 11 or more, the expected damage from using true strike to cancel out your disadvantage, then attacking, is greater. However, the difference is not great. For low values of x, your expected damage in S2 is vastly greater since you have a good chance of hitting twice. For high values of x, the difference is only a few percent. It's worth casting True Strike, but you shouldn't expect a huge difference in the outcome. For example, if x=12 then your E.D. is 0.45 in S1 or 0.405 in S2. If x=20 then your E.D. is 0.05 in S1 or 0.0025 in S2.

**MATH: Costly attacks, no disadvantage**

Let's say you have two choices: you can buy a coupon for 10$ that gives you 10 free pies at your local bakery, or a coupon for 20$ that gives you 15 free pies. Which is better? Obviously the one with the higher pie-to-dollar ratio, the first coupon.

Similarly, if each attack costs a spell slot, then your decision of whether or not to cast True Strike depends on the cost, and you should prefer the strategy with the higher damage-to-slot ratio. I used the same sort of analysis described above, but on a 20x20 grid, and divided the expected damage of S2 by two (because it costs 2 spell slots, or superiority dice, or whatever other limited resource you want to model). We get:

S1 exp. damage / cost = 1 - (x-1)^2/400 (the whole grid minus the lower left square of misses)

S2 exp. damage / cost = (21-x)(x-1)/400 + (21-x)^2/400 (rectangles/2 + 2*upper right square/2)

It turns out that in the range 1-20, S1 always has a higher expected damage than S2. So it is nearly always worthwhile for a wizard to set up their spell requiring an attack roll with True Strike.

**MATH: Costly attacks, disadvantage**

The probabilities are the same as in "Free attacks, disadvantage", but we divide the term for S2 by 2 to get the exp. damage / cost ratio. It turns out that the two ratios are pretty similar over the whole range x=[1,20]. The ratio for S1 is a diagonal line from 1 at x=1 to 0.05 at x=20. The ratio for S2 is a shallow parabola with similar endpoints. The expected damage of S1 is always greater. For example, if you need an 11 to hit, then the damage/cost ratio for S1 is 0.5 and the damage/cost ratio for S2 is 0.25. That's about where the difference is greatest; if you need a high or a low number to hit then the difference is only a few percent.

Note that I am comparing the damage-to-cost ratios, not the expected damage. If you need a high roll to hit, then the expected damage of S1 is slightly greater but the damage-to-cost ratio is also greater, reinforcing the advantage of this scenario. If you need a low roll to hit, then the expected damage of S2 is greater but pursuing this strategy also costs two spell slots instead of one.