Is there a difference between rolling 3d6 and rerolling lowest twice, versus rolling 5d6 and dropping the two lowest values?

Question

I was rolling stats for a set of characters (main + backup) with my DM, and he told me I could choose between 3 sets of two rolls. One rolled by him, one rolled by another player, and one rolled by me. Him and the other player use physical dice, rolling three dice, then rerolling the lowest value twice. Myself, I roll electronic dice, and I used a 5d6d2 command.

When the three of us came up with sets of rolls, the results were very different, to the point that you could easily guess which one was mine (the only that was done using electronic dice):

DM's rolls: {14, 17, 14, 15, 15, 18} & {17, 16, 16, 16, 16, 14}

Player's rolls: {15, 17, 17, 16, 18, 17} & {17, 15, 16, 16, 18, 16}

My rolls: {7, 12, 15, 13, 16, 11} & {16, 16, 10, 14, 14, 17}

This isn't that surprising: in my experience, physical dice have a tendency to roll higher (just by comparing what I roll against them, this is far from the first time when this happens). However, we started a friendly argument regarding the statistics we were using for the rolls, and whether we were actually rolling the same/equivalent thing.

My belief is that the two different roll scenarios are equivalent, while my DM believes it's a situation similar to the three doors problem, as he calls it.

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My scenario/rolls

Rolling 5 six-sided dice and taking the three largest values. The 5d6D2 roll, which using this page we can get a neat graph of the possibilities for each roll.

Example roll: {3, 2, 2, 1, 1} Result: 7

My DM's/fellow player's scenario/rolls

Rolling 3 six-sided dice, then rerolling the lowest value twice.

Example roll: Roll: {6, 4, 1} Result: 11

Reroll the 1. Roll: {4} Result: 4

Reroll either 4. Roll: {1} Result: 1

Final roll: {6, 4, 4} Result: 14

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Is there a statistical distribution difference between these two scenarios?

Answer 1

You are both wrong:

• You are wrong because the two rolling procedures do not produce the same distributions of probability.

• Your DM is wrong, because this has nothing to do with the Monty Hall (three doors) problem.

Tackling these in reverse order: in the Monty Hall problem, the host of the game knows where the goats and prize are. He cannot reveal the prize, so when he reveals a goat, you gain extra information. That is, there is a dependence in between the sequence of events which leads to an outcome (winning or losing the game).

In the dice scenario, the die rolls are independent. There is no difference between rolling three dice and rerolling the lowest, and rolling four dice and throwing away the lowest of the first three. You don't get any extra information after rolling three dice, as the fourth roll is independent of the previous three rolls.

Therefore your DM's explanation is incorrect.

Your error is, perhaps, a bit more subtle. Since the question asks whether the probability distributions are different, and not for the probability distributions themselves, let's consider a much simpler case:

• Procedure 1: Flip a coin (a d2, if you will) twice—say tails is $0$ and heads is $1$. Reflip the lowest result. Equivalently, flip three coins, then throw away the lowest result among the first two tosses.

• Procedure 2: Flip a coin three times, and throw away the lowest result.

In either case, you flip the coin $3$ times, leading to $2^3=8$ different events, each corresponding to a sequence of flips. For example, possible events are $$ HHT, \qquad TTT, \qquad\text{and}\qquad HTH. $$ With respect to either procedure, these $8$ events correspond to just three outcomes: you get a sum of $0$, $1$, or $2$ (the total number of heads which "count"). However, the probabilities of these outcomes are not the same. I have chosen this example to be small and tractable, so we can actually show every possible event and outcome, as summarized below:

\begin{array} .\text{Event} & \text{Proc 1} & \text{Proc 2} \\\hline TTT & 0 & 0 \\ TTH & 1 & 1 \\ THT & 1 & 1 \\ THH & 2 & 2 \\ HTT & 1 & 1 \\ HTH & 2 & 2 \\ \color{red}{HHT} & \color{red}{1} & \color{red}{2} \\ HHH & 2 & 2 \end{array}

Notice that these two procedures give very similar results, but that Procedure 2 is, on average, slightly better. The difference occurs in the line which I have colored red—under Procedure 1, you get two heads, but then replace one of those heads with tails on your next toss. Under Procedure 2, you get to keep both heads, so you are better off.

The question regarding dice follows a similar pattern. Rolling three dice, rerolling the lowest, then rerolling the lowest again (or, equivalently, rolling five dice, then dropping the lowest two from the first four) is akin to Procedure 1. Rolling five dice and dropping the lowest two is akin to Procedure 2. Procedure 2 is always going to win out—heuristically, this is because you never replace a high roll with a lower roll.

Answer 2

There is a bit of ambiguity in Scenario 2. If we interpret the procedure literally, it is possible to roll the maximum outcome of $18$ before the two re-rolls occur. If the re-rolls are required no matter the earlier result, then we have what I call Scenario $2$a. If we stop rolling as soon as a total of $18$ is attained, I call this Scenario $2$b.

In any case, the two scenarios indeed do not have the same distribution.

First, it is easy to obtain an empirical distribution with simulation. With $10^6$ simulations per scenario, my results were as follows:

$$ \begin{array}{c|ccc} \text{Total} & \color{darkblue}{\text{Scenario 1}} & \color{orange}{\text{Scenario 2a}} & \color{green}{\text{Scenario 2b}} \\ \hline 3 & 141 & 139 & 134 \\ 4 & 638 & 649 & 690 \\ 5 & 1950 & 2633 & 2567 \\ 6 & 5393 & 6719 & 6671 \\ 7 & 11743 & 15014 & 15304 \\ 8 & 21859 & 28780 & 28813 \\ 9 & 37989 & 50901 & 50688 \\ 10 & 60521 & 79104 & 79556 \\ 11 & 85190 & 110311 & 110929 \\ 12 & 113285 & 137672 & 137669 \\ 13 & 135566 & 151791 & 148344 \\ 14 & 148771 & 137877 & 135410 \\ 15 & 142639 & 115516 & 113169 \\ 16 & 120283 & 87112 & 84080 \\ 17 & 78226 & 53661 & 50417 \\ 18 & 35806 & 22121 & 35559 \\ \end{array}$$ Although they seem reasonably similar, they are clearly not once plotted.

To prove the distributions are not equivalent, we can compute the probability of obtaining a sum of $18$. In scenario $1$, there are $6^5$ total elementary outcomes on five dice. Of these, we require at least three of the dice to show $6$. There is one outcome in which there are exactly $5$ sixes; $5(5) = 25$ outcomes with exactly $4$ sixes; and $\binom{5}{2}(5^2) = 250$ outcomes with exactly $3$ sixes; thus the probability of getting a sum of $18$ is $$\frac{250+25+1}{6^5} = \frac{23}{648} \approx 0.0354938,$$ consistent with the simulation.

To count the same probability for Scenario $2$a, it is important to note that in the initial roll, at least one die must be a six, since there are only two re-rolls allowed. If all three dice are six, we re-roll any die and re-roll it a second time. The first re-roll is irrelevant, since no matter the value, it will be rolled again. So in the unique initial outcome of $3$ sixes, the probability of maintaining it is $1/6$.

If only two of the three dice are six after the initial roll, the same logic applies as in the previous case, and the probability of getting the last die to be a six is $1/6$.

If only one die is a six after the initial roll, we must roll both of the dice that are not six, and each of those re-rolls must result in a six, so the conditional probability of this result is $1/36$.

Now we compute the total probability as $$\frac{1}{6^3}\cdot\frac{1}{6} + \frac{3(5)}{6^3} \cdot \frac{1}{6} + \frac{3(5^2)}{6^3} \cdot \frac{1}{36} = \frac{19}{864} \approx 0.0219907.$$ Again, this is consistent with our simulation.

I leave Scenario $2$b to the reader to consider.