A "fudge die" is a die with equal probability to result in -1, 0, or +1. The commercially produced fudge dice are generally 6-sided dice with two "–", two "+", and two blank sides.

Fudge Dice Example

In the FATE game system, players roll 4 fudge dice for any interaction, and then add a skill to the result. The player also has "Fate Points" he or she can use to modify the roll in the following ways:

  • Add +1 to a roll (in any circumstance)
  • Add +2 to a roll (if an Aspect is applicable)
  • Reroll all four dice (if an Aspect is applicable)

Fate Points are generally considered a valuable commodity, so the first option is usually ignored. Aspects describe a game object (character, item, environment, etc.) in some way; for example, a player character might have the Aspect Chivalry Is Not Dead, Dammit which could be invoked on a roll to assist a damsel in distress. A scene might have the Aspect Poorly-Lit, invoked to assist in a stealth roll. (Aspects can also be "compelled" to introduce negative complications for a character and earning additional Fate Points, but that's not relevant to the math question.)

The latter two options for spending Fate Points can be used multiple times on a single roll (at the cost of multiple Fate Points) if and only if multiple Aspects apply to the roll.

Naturally, the average value of a roll is 0, the minimum value is -4, and the maximum value is +4. What I'm not certain on when it comes to the probabilities is when a Fate Point for +2 is superior to a Fate Point for a reroll. Getting +2 on a roll of +4 is obviously better than rerolling a +4 (you've got nowhere to go but down). The same is true for a +3 roll, as even if the reroll gives you +4 (which is a small chance on its own), that's only a +1 increase instead of a +2. The other 7 possible results are more murky for me, however. In short:

For a given roll of 4dF, when is a reroll going to increase the result by +2 or more 50% of the time or more?


1 Answer 1


The expected value of 4dF (or any number of such die rolls) is $0$. Therefore, if you roll a $-1$ or higher, you should take the $+2$ bonus rather than re-rolling. Similarly, for the first bonus you mentioned (i.e., the "$+1$"), you should take it if you roll a $0$ or higher.

Put another way, if the bonus isn't enough to give you $0$, then roll again.

If it helps, here's a distribution of the rolls:

# ways: 1 4 10 16 19 16 10 4 1

roll of 4dF: +4 +3 +2 +1 0 -1 -2 -3 -4

You can see the chances of rolling within the range from $+4$ down to $0$ is $\frac{1+4+10+16+19}{81} = \frac{50}{81}$, which is better than $50\%$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .