Is "1d10 rerolling 1&2" equivalent to "1d8+2"?

Question

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?

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Notes

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.

Answer 1

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:

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.

Answer 2

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.

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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.

Answer 3

TL;DR

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.

Answer 4

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.

Answer 5

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.