Good old Cyberpunk 2020, 2nd edition, 2nd printing, offers two versions of rolling for statistics to players because point-Buy without rolling is for referees only. They read:

  1. Random: Roll 9d10 and total them. You have this many Character Points.
  2. Fast: Roll 1d10 for each stat (9 in all), rerolling any scores of 2 or less. Place rolls in each stat as desired.

While method one puts the average sum to 49.5 and 5.5 per statistic, the Fast method differs: You can not roll a 1 or 2, because you reroll those. This clearly shifts the average sum and average statistic away from the normal average number.

Thinking about it, the reroll seems to effectively remove the 1 and 2 surfaces on the dice, and thus mimic a d8 running from 3 to 10... At least that's what the intuition claims.

Is the intuition correct that you could replace the Fast method of 1d10 with the rerolls by taking a d8 and adding 2, for fewer dice rolled and rerolled, and ultimately a faster character creation by skipping rerolls?


Cyberpunk 2020 uses only d10 and d6 in its mechanics, but if you own a d8 from other games or use a rolling tool, this would be an option.

The wording above is exactly like that in the book. It means to reroll any 1 or 2, no matter if they appear in the roll or reroll or nth reroll.

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    \$\begingroup\$ There are already two good answers from a simulation and a proof perspective. This comment is just to confirm that your intuition-- that is to say, the informal reasoning that both methods yield a uniform distribution from 3 to 10-- is valid. You're not correct by accident, you're just correct. \$\endgroup\$
    – Novak
    Dec 16, 2023 at 20:31
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    \$\begingroup\$ You could also use a d8 and treat the 1 and 2 as 9 and 10. \$\endgroup\$ Dec 17, 2023 at 2:46
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    \$\begingroup\$ Do you get to keep re-rolling every time it's two or less, or do you only get one re-roll? \$\endgroup\$ Dec 17, 2023 at 3:15
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    \$\begingroup\$ Please note that Cyberpunk 2020 didn't use d8. So while faster, it might also be impossible for players of "just" this game because they might not have the physical dice needed. \$\endgroup\$
    – nvoigt
    Dec 17, 2023 at 9:52
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    \$\begingroup\$ @NickMatteo It was already asked before, and the OP said that you reroll each time you get a 1 or a 2. \$\endgroup\$
    – Eddymage
    Dec 17, 2023 at 20:17

5 Answers 5



From both a practical and theoretical point of view, the two procedures will produce the same results. But... there is a little subtlety, and rolling 1d8+2 is actually the better procedure (for a certain value of "better").

The Details

Over an infinite number of rolls, the probabilities are as follows (I'll do this briefly, as others have already given good answers—I do, however, want this answer to be self-contained and complete).

For any $$n \in \{3,4,5,6,7,8,9,10\},$$ the probability of eventually rolling n on 1d10, with 1s and 2s being rerolled, is given by \begin{align} P(n) &= \sum_{j=1}^{\infty} P(\text{roll $n$ on the $j$-th}\ \mid\ \text{the previous $j-1$ rolls are 1 or 2}) \\ &= \sum_{j=1}^{\infty} \frac{1}{10} \left( \frac{2}{10} \right)^{j-1} \\ &= \frac{1}{10} \frac{1}{1-1/5} \\ &= \frac{1}{8}. \end{align} This is exactly the probability of rolling n on 1d8+2.

However, there is a little subtlety: there is an event which is not being accounted for. Specifically, it is possible to roll nothing but 1s and 2s... forever. This event has probability zero—if you roll 1d10 (rerolling on 1s and 2s) infinitely many times, there is a 0% chance that you will roll forever.

For all intents and purposes, the two probability distributions are the same. In either case, the chance of rolling any particular number between 3 and 10 (inclusive) is the same.

In the real world, however, it is not possible to roll dice infinitely often. At some point, you have to stop rolling (if for no other reason than the eventual heat death of the universe). So the "real world" probability of eventually rolling n on 1d10 (rerolling 1s and 2s) is given by \begin{align} P(n) = \frac{1}{10}\sum_{j=1}^{N} \left( \frac{1}{5} \right)^{j-1} = \frac{1}{10} \frac{1-(1/5)^{N}}{1-1/5} = \frac{1}{8}(1-5^{-N}) < \frac{1}{8}, \end{align} where N is the number of times the die is rolled before you give up. Note that $$ 1-5^{-N} $$ converges to zero very quickly—the probability of rolling 1s and 2s until the heat death of the universe is very small. Indeed, the probability of rolling five or six 1s and 2s in a row is already quite small.

So, from a theoretical point of view, the probability of rolling n is the same for any n between 3 and 10 (inclusive), no matter which procedure is used; and from a practical point of view, it is incredibly unlikely that anyone would roll 1s and 2s forever. On the other hand, rolling 1d8+2 is guaranteed to give a valid result every time the die is rolled, while rolling 1d10 does not.

  • \$\begingroup\$ better as in "avoid the nigh 0 chance to be stuck in a loop"? \$\endgroup\$
    – Trish
    Dec 19, 2023 at 23:28
  • \$\begingroup\$ @Trish Essentially, yes. Over a finite number of rolls, there is a non-zero chance that you will roll nothing but 1s and 2s. This chance is incredibly small over a large number of rolls, but it is nonzero. \$\endgroup\$ Dec 19, 2023 at 23:30
  • \$\begingroup\$ In favor of using a D10, there are some situations where the number of times that a DM secretly rolls 1D8 might convey information to players which would not be conveyed in the same way as if the DM rolled 1D10 with rerolls. For example, if a DM were to roll 1D8 each turn until players reach a certain location, but on one turn they triggered it twice, players might suspect they'd missed a chance to notice something. If, however, the DM sometimes had to reroll 1D10 because the hidden roll was a one or two, the fact that the DM rolled twice on a particular turn wouldn't be noteworthy. \$\endgroup\$
    – supercat
    Dec 19, 2023 at 23:38
  • \$\begingroup\$ On the flip side, one could achieve the desired probabilities precisely without ever need to roll more than three times. If one gets three values from {1,2} in a row, multiply the first number by four and the second by two, add the results together along with the third die (yielding a value from 7 to 14), and subtract 4 to get a value from three to ten. \$\endgroup\$
    – supercat
    Dec 19, 2023 at 23:42
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    \$\begingroup\$ Also note that in the original character point study, the goal isn't to roll a single 3-10 value. A whole series of 3-10 values was needed. So it is not a question of rolling only 1s and 2s forever, but rather how long it takes to roll the required number of 3-10 values. I think the original goal was 10 values? So the terrible eventuality is so much worse than incredibly small - it's ten times incredibly small! Which still probably won't carry you to the heat death of the universe, but hey, something to think about while making sandwiches. \$\endgroup\$
    – adalle
    Dec 19, 2023 at 23:51

Your intuition feels sound to me, but just to be sure, let's check with AnyDice:

function: DIE:n reroll BAD:s {
  if DIE = BAD { result: d{} }
  result: DIE

output [d10 reroll {1,2}] named "d10 reroll 1 and 2"
output d{3..10} named "d{3..10}"
output d8 + 2 named "d8 + 2"

The program above outputs the probability distributions of three ways of rolling a number from 3 to 10:

  1. roll a d10 and reroll any roll of 1 or 2 until you get 3 or higher;
  2. roll an 8 sided die with sides numbered from 3 to 10; or
  3. roll a normal d8 and add 2.

And indeed, all three methods yield a uniform distribution of results from 3 to 10:

AnyDice screenshot

Of course, using the 1d8 + 2 method requires you to have some eight-sided dice available. If you're playing a game that normally only uses ten-sided dice, it may be easier to use the "1d10 and reroll 1 and 2" method than look for a bunch of extra dice.

  • \$\begingroup\$ safe for grabbing the d8 from the game-set^^ \$\endgroup\$
    – Trish
    Dec 16, 2023 at 18:19
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    \$\begingroup\$ Technically there is a nonzero probability that you get 1s and 2s for the rest of your life, so the distributions are technically different. \$\endgroup\$ Dec 17, 2023 at 19:30
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    \$\begingroup\$ @ThomasMarkov: Technically, yes, the probability of rolling only 1s and 2 for the rest of your life is indeed non-zero, but only because "the rest of your life" is not infinite. And I'm pretty sure the Cyberpunk 2020 rules don't actually specify what to do if a player dies in the middle of rerolling, so there's no reason to assume that the last roll would actually stand. More likely the game would just be declared over and the entire roll invalid. (Also, practically, the probability of needing more than 10 rerolls is about one in ten million, and decreases exponentially with each roll.) \$\endgroup\$ Dec 17, 2023 at 21:30
  • \$\begingroup\$ @IlmariKaronen I think the probability of rolling 1 or 2 100 times in a row is substantially lower than the the probability of the player dying of unrelated reasons within the time you need to roll a d10 100 times in a row. That said, if you have a d8 on hand, it does save some annoyance. \$\endgroup\$
    – tomasz
    Dec 19, 2023 at 17:48
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    \$\begingroup\$ @tomasz: I'd say you're right. As a quick ballpack "Fermi estimate", the global age-specific annual death rate for young adults is around 1/1000. Assuming a dice roll takes about 3 seconds, that works out to an average chance of one in 10 billion of dying per roll, which (after some math) puts the break-even point around 13 rolls. Even if you lower the estimated death rate by a factor of 1000 (to account for RPGs generally being a low-risk activity, at least unless you're playing in an ICU hospital bed or while driving a motorcycle), that only raises the break-even point to around 17 rolls. \$\endgroup\$ Dec 19, 2023 at 20:04

Yes, the distribution is the same.

The answer from Ilmari shows that simulations confirm your intuition: here I'd like to provide a mathematical proof of this. ApexPolent's answer presents another theoretical proof, really elegant and concise. Xander's one focuses on an interesting practical issue.

Rerolling any 1 on a d10 produces a distribution where the possible outcomes are numbers from 2 to 10, each with probability 1/9.

Let's compute the probability to obtain a result of 3 with this rolling technique. This is given by $$ P(3) = P_1 + P_2 + P_3 +\dots $$ where \$P_i\$ stands for the probability of getting the result 3 at the i-th roll, having obtained 1 in the former i-1 rolls: $$ P_i = P(\text{obtain 1 in the }i-1\text{ rolls})\,P(\text{obtain 2 at the } i\text{-th roll}) $$ Then we have $$ P_1 = \frac{1}{10}, \,P_2 = \frac{1}{10}\cdot\frac{1}{10}, \, P_3 = \frac{1}{10}\cdot\frac{1}{10}\cdot\frac{1}{10}, \, \dots, \, P_i = \left(\frac{1}{10}\right)^{i-1}\cdot\frac{1}{10}. $$

Then we can collect everything and write $$ \begin{eqnarray*} P(3) &=& \sum_{i=1}^{\infty} P_i\\ &=& \frac{1}{10}\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^{i-1}\\ &=&\frac{1}{10}\sum_{i=0}^{\infty}\left(\frac{1}{10}\right)^{i}\\ &=& \frac{1}{10}\frac{1}{1-1/10}\\ &=& \frac{1}{10} \, \frac{10}{9} = \frac{1}{9}, \end{eqnarray*} $$ where we used a well-known formula for the geometric series. The same is valid for all the other outcomes.

If you reroll 1 and 2, taking into account that the probability of getting 1 or 2 is 2/10, then you get a set of possible values ranging from 3 to 10, each with probability 1/8: $$ \begin{eqnarray*} P(3) &=& \sum_{i=1}^{\infty} P_i\\ &=& \frac{1}{10}\sum_{i=1}^{\infty}\left(\frac{2}{10}\right)^{i-1}\\ &=&\frac{1}{10}\sum_{i=0}^{\infty}\left(\frac{2}{10}\right)^{i}\\ &=& \frac{1}{10}\frac{1}{1-2/10}\\ &=& \frac{1}{10} \, \frac{10}{8} = \frac{1}{8}. \end{eqnarray*} $$

This corresponds exactly to roll a 1d8+2.

Beside the Anydice outcome, the plot below shows that the simulated rerolling of 10 000 000 d10s confirms the above computation.histogram showing the comparison between the theory and the simulations.

Considering Xander's observation, the table below counts the probability of rolling n times:

# of rolls Frequency (%)
1 80.000851
2 16.000754
3 3.199841
4 0.638720
5 0.127904
6 0.025595
7 0.005111
8 0.000980
9 0.000188
10 0.000053
11 0.000003

Hence, in the 99.2% of the cases one obtains an outcome between 1 and 8 by rerolling just two more times.


Yes, it's the same. I have an easier to follow proof.

Denote the probability of rolling 3 on any given roll as P(3). Then,

$$P(3) = \frac{1}{10} + \frac{2}{10}P(3).$$

In other words, you have a 10% chance to roll 3 immediately, and a 20% chance to reroll, in which case the probability of rolling a 3 is just P(3) again. This is the same as the infinite sum posted by @Eddymage, but packaged recursively.

$$ \therefore \frac{8}{10}P(3)=\frac{1}{10} $$

$$ \therefore P(3)=\frac{1}{8}. $$

Repeat the same argument for 4-10.

  • \$\begingroup\$ clever and intuitive proof! \$\endgroup\$
    – Ben
    Dec 19, 2023 at 19:11

Yes. Assuming fair dice (equal likelihood of rolling each face):

Rolling 1d8, the probability of rolling each result from 1 to 8 is 1/8. Therefore, the chance of 1d8+2 generating each result from 3 to 10 is also 1/8.

Rolling 1d10-drop-1s-and-2s, we can ignore the 1 or 2 results, since they can't happen. Therefore there are 8 events that can happen: 3 to 10, each with equal probability, namely 1/8.

These are the same distribution.


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