Chance of Meeting a Target Number on One Die

Also known as roll target number probability · chance to roll t or higher · difficulty number probability · d20 target number odds · probability of rolling at least

p=dt+1dp = \frac{d - t + 1}{d}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

This is classical probability at its most direct: count the faces that succeed and divide by the faces there are. If the target is tt and the die has dd faces, the successes are t,t+1,,dt, t+1, \ldots, d, and there are dt+1d - t + 1 of them.

The plus one is the whole difficulty, and it is where nearly every error in this arithmetic comes from. Counting from 11 to 20 on a twenty-sided die gives ten faces, not nine: both ends are included, because the target itself succeeds. The general name for this is the fencepost error — a fence a hundred metres long with posts every ten metres needs eleven posts, not ten — and it is worth checking by hand the first few times. If the answer feels one face short, it is.

Two boundary cases confirm the formula behaves. A target of 1 gives (d1+1)/d=1(d - 1 + 1)/d = 1: every face clears it, which is right. A target of d+1d + 1 gives zero: nothing on the die reaches it, which is also right, and the calculator flags it as an error rather than returning a bare zero, because a target above the die is far more often a typing mistake than a deliberate impossibility.

What the formula assumes is a FAIR die, and that assumption is doing quiet work. Every face must be equally likely. Cheap injection-moulded dice with hollowed pips are measurably not fair, casino dice are made to tolerances precisely because it matters, and a die that has been dropped on a corner may no longer be the die it was. None of this changes the arithmetic; it changes whether the arithmetic applies.

And once again: the die has no memory. Four failures in a row do not improve the fifth attempt by any amount whatever. This is the page where that error does the most damage, because a target roll is usually attempted repeatedly and it is exactly the setting in which "I'm due" feels most reasonable. The chance is (dt+1)/d(d - t + 1)/d on the first attempt and on the thousandth, unchanged and unchanging. If you want the chance that at least one of several attempts succeeds, that is a different and answerable question — it is one minus the chance they all fail — and it lives on its own page. What it is not, ever, is a rising chance on any individual roll.

Chance of Meeting a Target Number on One Die
p=dt+1dp = \frac{d - t + 1}{d}
tdp
Where
  • pp= Chance of success
  • dd= Faces on the die
  • tt= Target number
Missing one of these? Work it out first, then come back