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=d−t+1dp = \frac{d - t + 1}{d}

Worked example: Eleven or better on a d20 → exactly half, because 11 through 20 is ten faces — press Try an example to run it live, then adjust anything.

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Chance of Meeting a Target Number on One Die explained

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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 d−t+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 (d−1+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 (d−t+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 formula

p=d−t+1dp = \frac{d - t + 1}{d}
Where
  • pp= Chance of success
  • dd= Faces on the die
  • tt= Target number

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