In Fate Core, the game rules for skill resolution involve rolling four six sided dice, and counting faces 1,2 as -1, face 3,4 as 0, faces 5,6 as +1. You then sum up the resulting numbers and add a fixed value from your skill level (typically between 0 and 4) for an overall result. After seeing the result, you can spend a Fate Point to either reroll, or to add +2 to it. The game rules advise you to

Rerolling the dice is a little riskier than just getting the +2 bonus, but has the potential for greater benefit. We recommend you reserve this option for when you’ve rolled a –3 or a –4 on the dice, to maximize the chance that you’ll get a beneficial result from rerolling. The odds are better that way.

Is that the case? That is, is your expected result higher by adding +2 than by rerolling if your original result was a -2? It seems to me that naively, both would have the same expected value, there just would be more variance on the roll than on taking the +2.


7 Answers 7


Yes that is the case.

  • The average score for any given roll will be 0.
  • If you take the +2 bonus and add it to a roll of -1 or higher, you come out better than average.
  • If you take the +2 bonus on a roll of -2 you get 0, which is average. Statistically, the choice between rerolling and taking the bonus is equal here.
  • With a score of -3 or -4 however, your +2 bonus would only get you to -1 or -2 respectively, which is still worse than average. In this case you're better off rerolling.
  • \$\begingroup\$ So, for -2, the expectad value is not better when using +2 over re-roll (which is my question), it is the same. It just may feel safer, as you know what you'll get, rather than risking a worse outcome in some cases, for a better one in others (due to loss aversion bias)? On re-reading I realize this is what the advice claims, you would not get a "beneficial result" from rerolling a -2, you would get a "neutral result". \$\endgroup\$ Aug 15, 2022 at 6:56
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    \$\begingroup\$ @GroodytheHobgoblin Correct! While others will prefer to reroll a -2 because that's more interesting/exciting to them. If what's important to the player is avoiding results below average, then they would take the flat bonus on a -2 every time. \$\endgroup\$
    – Michael
    Aug 15, 2022 at 7:04

Let's test it with AnyDice:

F: d{-1,0,+1}

function: roll SUM:n then reroll as REROLL:d if up to LIMIT:n else add BONUS:n {
  if SUM <= LIMIT { result: REROLL }
  else { result: SUM + BONUS }

output 4dF+2 named "4dF +2"
loop LIMIT over {-4..3} {
  output [roll 4dF then reroll as 4dF if up to LIMIT else add +2]
    named "4dF reroll -4 to [LIMIT] else +2"

AnyDice graph mode screenshot showing the probability distribution of results when rolling 4 Fate dice and either rerolling if the result is below some limit or else adding +2

As you can tell from the graph above (and, in particular, from the averages shown in the legend), rerolling on a -4 or -3 does indeed yield a slightly higher average result (2.07) than just always taking the +2. This makes sense since, as Glazius notes, the expected average of a 4dF roll is zero. Thus, if your first roll happens to be less than -2, rerolling will give a higher average than adding +2. (If you've rolled exactly -2 on your first roll, both rerolling and adding +2 will give the same average result.)

However, if you're trying to roll at least some known target number to succeed, the real question isn't how to maximize the average of your roll, but how to maximize your odds of meeting the target. In that case you should instead look at the probabilities of rolling at least a given number:

AnyDice graph mode screenshot showing the cumulative probability distribution (i.e. the probability of rolling at least a given number) when rolling 4 Fate dice and either rerolling if the result is below some limit or else adding +2

From the graph above we can see that the optimal threshold for rerolling vs. taking the +2 depends on the target number. In fact, the rule that maximizes your success rate turns out to be very simple: take the +2 if that's enough to let you succeed, otherwise reroll. Which makes perfect intuitive sense — you should always take a guaranteed success over an uncertain one, and an uncertain success over a guaranteed failure.

(Of course, you also shouldn't spend a Fate point at all if you'll succeed anyway without it, and it may not be worth spending a point on a reroll if your odds of success will be very low. For example, if you have to roll 4 or more to succeed, then spending a Fate point to add +2 to an initial roll of 2 or 3 may be a very good investment, whereas spending one to reroll an initial roll of 1 or less probably isn't, as it still gives you only a 1.23% chance to succeed.)

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    \$\begingroup\$ Good point. If you know what you need to succeed, and the +2 doesn't get you there, re-roll. It doesn't really matter what the chance is to succeed, as long as it's better than zero, it's better than taking +2 and not succeeding. \$\endgroup\$
    – Wyrmwood
    Aug 15, 2022 at 17:42

On average, the advice is sound (and I have little to add beyond the existing answers), but strategically (as @AnneAunyme points out), things are more subtle. This is probably why the SRD qualifies the advice with "typically":

Whenever you’re making a skill roll, and you’re in a situation where an aspect might be able to help you, you can spend a fate point to invoke it in order to change the dice result. This allows you to either reroll the dice or add +2 to your roll, whichever is more helpful. (Typically, +2 is a good choice if you rolled –2 or higher, but sometimes you want to risk a reroll to get that +4.)

(Emphasis added to source.)

Players aren't likely making decisions about which benefit to take until after they see their first roll, and context matters a lot (stakes, how many fate points the player has, etc.). In situations where players know (or can reasonably guess) the difficulty on the ladder, there are likely situations where +2 guarantees either failure (rolling a -1 with a difficulty of 4) or success (rolling a -1 with a difficulty of 1), or may be the only way to reach a particular point on the ladder (no amount of re-rerolling will achieve a total of 5 with four fudge dice). Where one option guarantees failure, sane players will not choose to burn a fate point on it, regardless of their initial roll.

Here's a breakdown1 of the statistically "best" situational advice to give after the first roll with chances of a resulting success should the player decide spending a fate point is worth following it:

Difficulty Roll -4 Roll -3 Roll -2 Roll -1 Roll 0 Roll 1 Roll 2 Roll 3 Roll 4
-2 +2 (100%) +2 (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%)
-1 Reroll (81.48%) +2 (100%) +2 (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%)
0 Reroll (61.73%) Reroll (61.73%) +2 (100%) +2 (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%)
1 Reroll (38.27%) Reroll (38.27%) Reroll (38.27%) +2 (100%) +2 (100%) Stop (100%) Stop (100%) Stop (100%) Stop (100%)
2 Reroll (18.52%) Reroll (18.52%) Reroll (18.52%) Reroll (18.52%) +2 (100%) +2 (100%) Stop (100%) Stop (100%) Stop (100%)
3 Reroll (6.17%) Reroll (6.17%) Reroll (6.17%) Reroll (6.17%) Reroll (6.17%) +2 (100%) +2 (100%) Stop (100%) Stop (100%)
4 Reroll (1.23%) Reroll (1.23%) Reroll (1.23%) Reroll (1.23%) Reroll (1.23%) Reroll (1.23%) +2 (100%) +2 (100%) Stop (100%)
5 Stop (0%) Stop (0%) Stop (0%) Stop (0%) Stop (0%) Stop (0%) Stop (0%) +2 (100%) +2 (100%)
6 Stop (0%) Stop (0%) Stop (0%) Stop (0%) Stop (0%) Stop (0%) Stop (0%) Stop (0%) +2 (100%)

1 Generated using dyce (my Python dice probability library) and anydyce (my visualization layer for dyce meant as a rough stand-in for AnyDice). Try it out here: Try dyce [source]

  • \$\begingroup\$ Not sure what I did but I got the table to render right outside of preview mode. \$\endgroup\$ Aug 15, 2022 at 14:34
  • \$\begingroup\$ Look like multiple people editing (or in my case,low rep and suggesting edits) to make it render. 4 spaces in front is a good thing to know,, as are the HTML tags < pre > and < / pre > (but without the spaces). \$\endgroup\$
    – Hennes
    Aug 15, 2022 at 14:37
  • \$\begingroup\$ All good! Thanks for the assist, y'all! It looks like @TomasMarkov's edits did the trick for nice formatting. I wonder if it was a rogue whitespace issue. \$\endgroup\$
    – posita
    Aug 15, 2022 at 14:38
  • \$\begingroup\$ @Hennes My revision made it render properly as an html table, we try to avoid code block formatting as it does not play well with certain alternative viewing devices. \$\endgroup\$ Aug 15, 2022 at 14:40
  • \$\begingroup\$ Surely the "Either" cases should also read "Stop"? There's no point in spending a Fate point on a reroll (or +2) if you're guaranteed to fail anyway. (And the "Reroll" entries on difficulties 4 and 3 are also at least somewhat questionable — would you really want to spend a Fate point just for a 1.23% or 6.17% chance of success?) \$\endgroup\$ Aug 15, 2022 at 15:04

Usually, yes

As pointed out in the other answers, the expected value on average of rolling any number of Fate dice is 0. Thus, adding +2 to a -3 or -4 won't bring it past the average and adding +2 to a -1 or bigger will. Adding +2 to a -2 will bring to the average expected value and nothing more.

That said, sometimes maximizing your average expected value is not what matters. Maybe you know you won't succeed with anything lower than a +2 and thus should re-roll a -1 despite it lowering the expected value. Maybe you know that a -1 on the dice is enough and thus should get the +2 on a -3.

To sum up, this is a good rule of thumb for when you don't target a specific result and just want to maximize the expected result. On the contrary, if a specific result is important to you, you should take the +2 if it is enough to reach it and the re-roll otherwise.


You're not likely to beat zero, but you're not likely to do worse, either.

Here's a simple Fate die roller done up in Anydice. Inspecting the results in "at least" mode will show that your chance of getting a 0 or better, beating (-3 + 2), is 62%, and complimentarily (because the result curve is symmetric), your chance of getting a +1 or better, beating (-2 + 2), is 38%. So you've got about 24% odds of breaking even.

Unlike simple polyhedral dice with an even number of sides, having a big slice of probability at the center weight means that 4dF's expected value of 0 actually happens fairly frequently. Rerolling all the dice when they've come up -3 or -4 gives you an expected gain of 3 and 4, and rerolling the dice when they've come up -2 is an expected gain of 2 exactly. Rerolling versus taking the +2 is kind of down to psychology at that point: are you tempted by gains, do you want to avoid losses? You can't count on either but you can't count either out.

Conditions and variants

Note that it's a usual assumption of Fate that you're not just rolling dice into a void hoping for a big number; rather, because the GM is usually setting the opposition, they can provide you with a number you're expected to hit, whether that's passive opposition or what the active opposition rolled. (In general this is good design practice when players have a limited pool of resources to spend to influence rolls.) It's also likely that you'll know your total Fate Point "budget" for the roll: since you can't use an Aspect more than once to change the result of the dice, whether that's a reroll or a +2, you know how much you could possibly spend based on how many Aspects are actually relevant to what you're trying.

This can obviously create situations where you have to hope for a lucky reroll as the only way to hit the number you're after. However, the GM also has their own scene pool of Fate Points that they can spend on invokes to beef up passive opposition or boost or reroll active opposition, so a GM number is truly final only when the GM pool has run dry. More than that, your own ration of Fate Points has to last you the entire night, so even if you could spend two Fate Points to guarantee something, gambling one on a relatively likely reroll can save you one for later. It's always useful to know how you can spend your Fate Points efficiently.

The odds also assume you're resolving under basic Fate conditions, where nothing matters except the end roll of the dice. One of the common tricks the various Fate drifts pull is to do something more with the randomizer, whether that's introducing the idea of a "Fate critical" when three or more dice have matching faces (often unlocked by stunts for certain skills) or if it's something more esoteric, like using Evil Hat's Deck of Fate and auguring something from the extra card greebles. The odds calculation gets a bit thrown out of whack in those cases, making a reroll potentially more overall beneficial even at higher numbers.

  • \$\begingroup\$ The heading here buries the lede a bit. The question isn't about the expected value, but about whether rerolling only on a -3/-4 is advisable and accurate. I suggest this answer should lead with a statement about whether that's the case. For example, use a heading of "Yes, reroll only on a -3 or worse", and explain in that first paragraph that the expected value of a roll is 0. \$\endgroup\$ Aug 15, 2022 at 15:53
  • \$\begingroup\$ @doppelgreener something like this, maybe? \$\endgroup\$
    – Glazius
    Aug 15, 2022 at 17:25

Fate dice can be represented by this polynomial:

$$\frac{x^{-1} + 1 + x^1}{3}$$ or

$$\frac{1 + x^1 + x^2}{3x}$$

When you roll 4 dice you get:

$$\frac{(1 + x^1 + x^2)^4}{81x^4}$$ or

$$\frac{(1 + x^1 + x^2)^4}{81x^4}$$


$$\frac{{(1 + x^1 + x^2)^2}^2}{81x^4}$$

$$\frac{(1 + 2 x^1 + 3 x^2 + 2x ^3 + x^4)^2}{81x^4}$$

$$\frac{1 + 4 x^1 + 10 x^2 + 16 x ^3 + 19 x^4 + 16 x^5 + 10 x^6 + 4 x^7 + x^8}{81x^4}$$

$$\frac{1x^{-4} + 4 x^{-3} + 10 x^{-2} + 16 x ^{-1} + 19 x^0 + 16 x^1 + 10 x^2 + 4 x^3 + x^4}{81}$$

where the \$x^n\$ coefficient is the chance you'll get the result \$n\$ after the roll.

Adding +2 had no variance as you know exactly what the result will be; you are adding after you see the result.

Rerolling compared to just a +2 when you are at -2:

It has a \$\frac{5}{81}\$ chance of being worse, a \$\frac{10}{81}\$ chance of not doing anything, and a \$\frac{31}{81}\$ chance of being worse than adding +2.

It also has a \$\frac{31}{81}\$ chance of being better than your +2.

In comparison, when at -3:

It has a \$\frac{1}{81}\$ chance of being worse, a \$\frac{4}{81}\$ chance of not doing anything, and a \$\frac{15}{81}\$ chance of being worse than adding +2.

It also has a \$\frac{50}{81}\$ chance of being better than your +2.

When at -4:

It has no chance of being worse, a \$\frac{1}{81}\$ chance of not doing anything, and a \$\frac{4}{81}\$ chance of being worse than adding +2.

It also has a \$\frac{66}{81}\$ chance of being better than your +2.


But really, the player should calculate the expected difficulty distribution and subtract it from their roll. Then work out if +2 is better than a reroll.

A cute way to model this is generate a distribution of the expected difficulty, then subtract it from your skill bonus.

So if your skill bonus is +3 and you expect the difficulty to be 50% 0 and 50% 1, then this is the polynomial \$\frac{x^1 + x^2}{2}\$. Multiply that by the above FATE roll polynomial to get the full distribution of your chance of success on a reroll.

For the +2 case, you just apply your known roll (say \$x^{-3}\$) to the difficulty distribution and see what your odds are of success (net of \$x^0\$ or higher).


You could also think about the scenarios. If the difficulty is D, what are the odds that +2 wins? A reroll wins? A reroll with a +2?

Then work out the relative odds of each possible difficulty.


Naturally, doing this live at the table is not that polite, unless you are playing by snail mail post. So you end up having to wing it.

You'll use a reroll if you don't think +2 is likely to succeed, but a reroll (and possibly a +2 after that) is better odds for the cost.

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    \$\begingroup\$ Math pedant here, the definition of a polynomial requires non-negative exponents, so an expression with \$x^{-1}\$ anywhere is not a polynomial. \$\endgroup\$ Aug 15, 2022 at 15:49
  • \$\begingroup\$ @ThomasMarkov Sure, pretend I wrote \$\mathbb{R}[x,y]/(xy)\$ and wrote \$y\$ as \$x^{-1}\$? I'm messing around with counting polynomials, and honestly they work with rational functions; but full rational functions is only needed for stuff like exploding die rolls. \$\endgroup\$
    – Yakk
    Aug 15, 2022 at 15:51
  • \$\begingroup\$ Like I said, it's a pedantic observation. Everything you do with it seems correct. \$\endgroup\$ Aug 15, 2022 at 15:58
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    \$\begingroup\$ You could just say it’s a Laurent polynomial and it could be technically correct. \$\endgroup\$ Aug 15, 2022 at 16:16
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    \$\begingroup\$ FWIW I've usually seen this approach referred to using the term "generating functions", e.g. stats.stackexchange.com/a/242857/351712 and math.stackexchange.com/a/3792882/1035142 . \$\endgroup\$ Aug 18, 2022 at 0:04

This is an interesting probability question!

So the key question is: What is the cutoff point in the possible outcomes of the first roll where the expected outcome of a re-roll is at least "3 better?"

In order to do that, we need to calculate the probabilities of each outcome in the set {-4, -3, ... , 3, 4}.

They are:

$$p(4) = \frac{1}{81}$$ $$p(3) = \frac{4}{81}$$ $$p(2) = \frac{10}{81}$$ $$p(1) = \frac{16}{81}$$ $$p(0) = \frac{19}{81}$$

(see example calculation at bottom of post)

...others by symmetry

So, if after the first roll you are at a -3 value, you would need to roll a value of 0 or more to be "better" than just taking the +2. The odds of rolling 0 or more is:

$$p(0) + p(1) + p(2) + p(3) + p(4) = \frac{50}{81} > 50\%$$

And even higher if we started out at -4

But if our first rolls sum to a -2 then the odds of rolling a 1 or better in the second roll diminish to:

$$\frac{16 + 10 + 4 + 1}{81} = \frac{31}{81} < 50\%$$

So the advice is correct!!

Example for calculating \$p(2)\$:

You can consider the die to be equivalent to a "3 sided" die with {-1, 0, 1} on the faces. The math is easier.

So then there are several ways to sum to a "2"

case a: all arrangements of 1, 1, 0, 0

case b: all arrangements of 1, 1, 1, -1

The odds of any particular roll is: $$\frac{1}{3}^4$$ as each side has an equal chance of showing up. Then we just look at the arrangements.

In case a we have a \$4\choose2\$ or 6 arrangements and in case b we have \$4\choose1\$ or 4 arrangements for a total of 10 arrangements or $$p(2)=\frac{10}{81}$$


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