I'm trying to figure out what is the probability with 3 D10's if I were to roll them one after another."001" is one and "000" is a thousand. Would every number have the same probability or doubles and triples be less common?

And please explain.


Just roll a d1000 in anydice.

The probabilities for rolling 3d10 as the 3 tens places will be exactly the same as rolling a d1000.

These answers shows the math for d100 vs 2d10, it's exactly the same story here just times ten.

The point of using d1000 is that probabilities are easy to calculate: the chance of the number or less is equal to the number in per-mille. The chance of rolling less than 900 is 900-1 in 1000 or 899‰ (89.9%); the chance of rolling equal or higher is the remainder in 1000 or 101‰ (10.1%) SevenSidedDie


If you're using 1 die as a hundreds digit, 1 die as a tens digit, and 1 die as a ones digit, then every number between 1 and 1000 has a 0.1% chance of occurring.

If by doubles you mean 155, 944, etc. and by triples you mean 333, 777, etc. then those have the same probability as any other number in the range. Think about it: each d10 should have a 10% chance of getting any given number, so why would rolling a 3 (or any other number) on one die affect the probability of getting a 3 (or any other number) on another die?

All of this of course assumes your dice are properly balanced and not flawed or being manipulated; using a truly random random number generator is recommended if you're worried about that sort of thing.

  • 1
    \$\begingroup\$ Note that each instance of doubles or triples has an equal 0.1% chance of occurring, but the chance of getting ANY double or triple (when you don't care which one you get) is higher: 1% for triples and something like 31% for doubles if I didn't mess up the math. \$\endgroup\$ – Gregory Avery-Weir Nov 19 '12 at 6:20
  • 3
    \$\begingroup\$ @GregoryWeir 1% for triples and 28% for doubles, according to this precomputed probabilities for the One Roll Engine which looks for exactly that on pools of d10s. \$\endgroup\$ – SevenSidedDie Nov 19 '12 at 7:34

You're reading them as digits, and the dice are in fixed positions. That means that each position has a equal probability (or a "flat" probability curve) of being 0–9, and that flat probability curve remains because you're interpreting them as digits in a number from 1 to 1000. You only start to get non-flat curves if you're adding them together.

If you plug them into Anydice, you'll see the flat curve.


Multiple dice give you a discrete, "blocky" distribution similar to normal when they are added together. This occurs because there are more combinations that sum to results in the average range and fewer combinations as you approach the high and low end of the possible results. For example, with 2 6-sided dice, there is only one combination that produces 2, but there are 6 combinations that produce 7. In the case of percentile or d1000, since each die represents a different place value, even though they are combined to get the final result, the strategy retains the uniform distribution each individual die had (equal probability or flat line), as there are still an equal number of combinations (one) that produce each final result, as if you had rolled a single die.

With a uniform distribution, the chance for any particular result is the same for each possible result. Any particular result is one over the total possible results, so with a d20 for example, any particular result is 1/20 or 5%. Rolling a particular number or under is that number/total possibilities so rolling 15 or under on a d20 is 15/20 or 75%. The chances for rolling a particular number or over is the number plus the remaining possibilities, so rolling a 16 or higher on a d20 is 25% (since there are 5 possibilities, 16-20).

The chance of rolling 950 (exactly) on a d1000 (or 3 d10, where one die represents ones, one tens and one hundreds) is 1/1000 or 0.1%, the same as any other particular number. The chances of rolling 950 or under are 950/1000 or 95%. The chances of rolling higher than 950 are 50/1000 (951-1000 or 50/1000) or 5%.

The slightly odd bit about dpercentile or d100, d1000, et al, is that the highest die represents 1-10 (times the place), while the lower die or dice represent 0-9 (times the place). EDIT- Heh, that is NOT correct, at least not most of the time :) The 0 for the highest place ONLY represents 10 (times the place) when all other dice are 0, in order to make the result 1-100 instead of 0-99 or 1-1000 instead of 0-999, and so on.

  • 2
    \$\begingroup\$ The OP is not talking about 3d10 in the sense of 1d10 + 1d10 + 1d10, but in the sense of 1d10 x 100 + 1d10 x 10 + 1d10, which represents a 1d1000 perfectly and does not approach normal distribution. \$\endgroup\$ – Angew is no longer proud of SO Aug 9 '14 at 21:21
  • \$\begingroup\$ Part of the confusion is likely because you begin your answer with a description of the "normal" method of combining dice. \$\endgroup\$ – thelr Aug 14 '14 at 15:34

It almost seems like you're expecting "exploding dice," which is where when you roll the max value (or certain values) on a dice, you reroll it and add the new value. If you had to get values via exploding dice (eg if you could not roll the second die if you didn't roll a 0 aka 10 on the first die, and couldn't roll the third die if you didn't roll a 0 on the second die), then you'd be in a completely different scenario! You'd have a 90% chance of rolling 1-9, and only a 10% chance of rolling anything above that.

But you're not. You roll all 3 dice every single time, and each digit's place is generated independently of each other's. If the ones-place rolls 1, 7, or 0, it makes no difference whatsoever int he interpretation of the tens-place die nor the hundreds-place die.

  • 1
    \$\begingroup\$ Or a bell curve. \$\endgroup\$ – doppelgreener Sep 10 '14 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.