View Full Version : Calculating probabilities of passing a re-rollable save?

Entreri Bloodletter
03-02-2013, 08:47
I was curious how to go about getting exact numbers for whether or not it is better to have a 2+ armor vs. say a re-rollable 3+? Or any other values for that matter, does anyone know of a formula that can calculate the odds of passing a re-rollable save?

My apologies if this isn't in the right section.

Lord Solar Plexus
03-02-2013, 09:03
(x * 2/3) * 2/3 for a 3+ or 3++ rerollable save I think, with x = number of wounds.

03-02-2013, 09:11
I find it easier when working out rerollables to calculate the chances of failure and reversing it.
If you just multiply the odds of success on the first dice by the odds of success on the second dice you only get the odds that both succeed.
eg. 4+ with reroll: The chance of failure on the first dice (3/6) x the odds of failure on the second dice (3/6) = 9/36 or 0.25
reversing it (basically 36/36 - 9/36 or 1 - 0.25 = 0.75 (if you want a percentage multiply by 100 for 75%)

for your example 2+ armour save is 5/6 or 0.833
3+ armour save is 2/6 x 2/6 = 0.111
1 - 0.111 = 0.888
Results are
2+ armour save is 5/6 or 30/36 or 0.833 or 83.333%
3+ with reroll is 32/36 or 0.888 or 88.888%
so 3+ reroll is better by 5.555%

03-02-2013, 09:17
(x * 2/3) * 2/3 for a 3+ or 3++ rerollable save I think, with x = number of wounds.

That will only give you the odds of both dice rolling a 3+.
4/6 * 4/6 = 0.444 success
2/6 * 2/6 = 0.111 failure
thats only 0.555 your missing another 0.444.
That is why I deducted the chances of failure from 1.

04-02-2013, 23:06
To do the calculation forward (for a rerollable save) you do the following:
-Determine the probability of success on one roll, e.g., 1/6 for a 6+, 2/6 for a 5+, and so forth.
-Also determine the probability of failure on one roll (subtract the above success value from 1).
-Plug those values into the following formula: [chance of success on first roll] + [chance of failure on one roll]*[chance of success on first roll]

The reason the above formula is correct is because there are two ways to succeed on a rerollable save: first, pass it on the initial roll, and second, fail the initial roll and pass the reroll.

The first condition is simply the probability of passing the first roll, which is fairly simple to calculate, as we're only working with 6-sided die.

The second condition is dependant on failing the first roll before you pass the second, hence you multiply the chance of failing the first roll by the chance to succeed on the second.

The probability of success is then the sum of the probabilities of both possible forms of success occuring (since they do not overlap).

Here are the relevant saves calculated out:


Prob. Success if Rerollable

Prob. Success if not Rerollable
















Hope that helps.

Lord Solar Plexus
05-02-2013, 05:01
True, forgot that 1/3 of failure.