Effect of Reroll on Probability

[MaxMahem]

Lately it has seemed to me that allowing a player to "reroll" a dice has become a popular mechanic for granting an advantage, as opposed to a flat a bonus (like +2 ect). I like this, rerolls give the perception of an advantage and are easy to keep track of and calculate.

My problem is that I do not understand their effects on the probability of a situation. A flat bonus is pretty easy for me to calculate. When rolling a D20 against a target number of 10 a player has a 50% change of success and every +1 gives a player a 5% advantage. It's fairly easy for me to scale this mechanic as appropriate for different situations.

But how does giving a player a chance to reroll that dice effect the situation? I'm assuming it increases his odds of success, but by what factor/percent? How would I calculate this.

Also some systems (Saga's in particular) make a distinction between forcing a player to keep the new roll, or allowing a player to keep the old one if it was better. Does this distinction mater? What is it's effect, statistically?

Lastly, some systems allow a player to reroll a die after they know the result (pass/failure which D&D generally does not). Does this effect the situation?

I don't worry about a reroll's effect on player strategy since even when the GM makes an effort to keep DC/Target numbers secret players rapidly figure them out themselves, and thus plan their reroll strategies accordingly, though this may affect the statistical situation somewhat.

If one of the more mathematically inclined members of the board could help me out with this, I would be much obliged.

[Lyinginbedmon]

Well, if I have to roll a 10 on a d20, it's a 50% chance. Then I have 2 d20 to roll and aim for that same 10. That means that, given the 50%, half of my rolls should be equal to or greater than 10. So we multiply 50% (The flat value without rerolls) by 50% (The value of probability on the reroll) equalling a 25% that we will fail. Basically, a 75% chance that we will succeed.

Let's try something else then (Because I obviously suck at explanations of mathematics):

I have 1d4, and in order to kill a troll, I need to roll a 4. 4/4 is 1, so I have a 1 in 4 chance, or 25%.
Let's say I get 2 rerolls of that d4 for whatever reason. Each d4 has a 25% chance of rolling a 4.
So, we have 25%*125%*125% of sucess, equalling a 39% chance of rolling at least 1 4.

[Jack_Simth]

Depends on the reroll mechanism, honestly.

Two chances, pick the better (e.g., Blind-Fight, which, while you don't always roll twice, is equivalent, as you always roll again whenever you miss):
PNF = Probability (Normal) of Failure
PRS = Probability (Rerolling) of Success
PRF = Probability (Rerolling) of Failure
PRF = (PNF * PNF) = The chance that both rolls fail
PRS = 1 - PRF = The chance that at least one roll succeeds

Pride Domain Power - reroll all saving throw results of 1, a single time:
Changes the distribution a little - a roll of 1 goes from 1 in 20 to 1 in 400; with all other specific roll results changing to 21 in 400 to compensate.

Chosen rerolls before results are known has an unknown effect ... until you get down to specifics (rules of the reroll). If you choose to reroll whenever you get a result in the lower 50% (1-10), then the distribution changes; you've got a 50% chance that any roll valued at 50% or below is actually a higher roll; you cut the chance of a reduced roll in half, and the result of anything higher is increased proportionally:
{table=head]d20 roll|Probability
1|1/40
2|1/40
3|1/40
4|1/40
5|1/40
6|1/40
7|1/40
8|1/40
9|1/40
10|1/40
11|3/40
12|3/40
13|3/40
14|3/40
15|3/40
16|3/40
17|3/40
18|3/40
19|3/40
20|3/40[/table]
Hmm... this'll be easier with d4's.
If you reroll 3's, 2's, or 1's, you get:
{table=head]d4 roll|Probability
1|3/16
3|3/16
2|3/16
4|7/16
[/table]
If you reroll 2's or 1's, you get:
{table=head]d4 roll|Probability
1|2/16
3|2/16
2|6/16
4|6/16
[/table]
Reroll 1's
{table=head]d4 roll|Probability
1|1/16
3|5/16
2|5/16
4|5/16
[/table]

A little work with a spreadsheet application will give you a d20 distribution for whatever you like.

Edit:
Hmm.... d20 results:

Reroll anything below This|Chance of any given result below X|Chance of any given result of X or above
1|0|20/400|
2|1/400|21/400
3|2/400|22/400
4|3/400|23/400
5|4/400|24/400
6|5/400|25/400
7|6/400|26/400
8|7/400|27/400
9|8/400|28/400
10|9/400|29/400
11|10/400|30/400
12|11/400|31/400
13|12/400|32/400
14|13/400|33/400
15|14/400|34/400
16|15/400|35/400
17|16/400|36/400
18|17/400|37/400
19|18/400|38/400
20|19/400|39/400

[RS14]

For the case in which the players know when they fail, and can then take a reroll. Note that this is equivalent to rolling the number of dice given and picking the best result.
Let P be the probability of a success.

P+(1-P)P+(1-P)(1-P)P etc gives the probability after a certain number of rolls; the first term corosponds to one roll; the sum of first and second to two rolls, etc. This can be rewritten Sigma{i,1,n} (1-P)^(i-1)*P (Sorry about the notation; the component in brackets is the range of i, the rest is each term)

This can be simplified to P*Sigma{i,1,n} (1-P)^(i-1) and then P*((1-P)^n-1)/(-P). Our final simplification yields

-((1-P)^n - 1)

For example, for P=.5
1: .5
2: .75
3: .825
4: .9375
5: .96875

or P=.3
1: .3
2: .51
3: .657
4: .7599
5: .83193

To the chance of the player not succeeding after a given number of rolls, adding another roll adds P*(1-that chance).

[Jack_Simth]

All right... assuming you have to keep the result of a reroll, even if it's worse, the following table should cover the results of a d20; it gives the probability of any given d20 die result, presuming that you keep a roll of X or better, where X is the leftmost number
{table]||||Chance|of|a|final|die|result|of|:||||||||||||
||Reroll chance|Average Result:|1|2|3|4|5|6|7|8|9|10|11|12|13|14|15|16|17| 18|19|20
Keep|1|0%|10.5|5.00%|5.00%|5.00%|5.00%|5.00%|5.00% |5.00%|5.00%|5.00%|5.00%|5.00%|5.00%|5.00%|5.00%|5 .00%|5.00%|5.00%|5.00%|5.00%|5.00%
a|2|5%|10.975|0.25%|5.25%|5.25%|5.25%|5.25%|5.25%| 5.25%|5.25%|5.25%|5.25%|5.25%|5.25%|5.25%|5.25%|5. 25%|5.25%|5.25%|5.25%|5.25%|5.25%
roll|3|10%|11.4|0.50%|0.50%|5.50%|5.50%|5.50%|5.50 %|5.50%|5.50%|5.50%|5.50%|5.50%|5.50%|5.50%|5.50%| 5.50%|5.50%|5.50%|5.50%|5.50%|5.50%
of|4|15%|11.775|0.75%|0.75%|0.75%|5.75%|5.75%|5.75 %|5.75%|5.75%|5.75%|5.75%|5.75%|5.75%|5.75%|5.75%| 5.75%|5.75%|5.75%|5.75%|5.75%|5.75%
this|5|20%|12.1|1.00%|1.00%|1.00%|1.00%|6.00%|6.00 %|6.00%|6.00%|6.00%|6.00%|6.00%|6.00%|6.00%|6.00%| 6.00%|6.00%|6.00%|6.00%|6.00%|6.00%
or|6|25%|12.375|1.25%|1.25%|1.25%|1.25%|1.25%|6.25 %|6.25%|6.25%|6.25%|6.25%|6.25%|6.25%|6.25%|6.25%| 6.25%|6.25%|6.25%|6.25%|6.25%|6.25%
better|7|30%|12.6|1.50%|1.50%|1.50%|1.50%|1.50%|1. 50%|6.50%|6.50%|6.50%|6.50%|6.50%|6.50%|6.50%|6.50 %|6.50%|6.50%|6.50%|6.50%|6.50%|6.50%
|8|35%|12.775|1.75%|1.75%|1.75%|1.75%|1.75%|1.75%| 1.75%|6.75%|6.75%|6.75%|6.75%|6.75%|6.75%|6.75%|6. 75%|6.75%|6.75%|6.75%|6.75%|6.75%
|9|40%|12.9|2.00%|2.00%|2.00%|2.00%|2.00%|2.00%|2. 00%|2.00%|7.00%|7.00%|7.00%|7.00%|7.00%|7.00%|7.00 %|7.00%|7.00%|7.00%|7.00%|7.00%
|10|45%|12.975|2.25%|2.25%|2.25%|2.25%|2.25%|2.25% |2.25%|2.25%|2.25%|7.25%|7.25%|7.25%|7.25%|7.25%|7 .25%|7.25%|7.25%|7.25%|7.25%|7.25%
|11|50%|13|2.50%|2.50%|2.50%|2.50%|2.50%|2.50%|2.5 0%|2.50%|2.50%|2.50%|7.50%|7.50%|7.50%|7.50%|7.50% |7.50%|7.50%|7.50%|7.50%|7.50%
|12|55%|12.975|2.75%|2.75%|2.75%|2.75%|2.75%|2.75% |2.75%|2.75%|2.75%|2.75%|2.75%|7.75%|7.75%|7.75%|7 .75%|7.75%|7.75%|7.75%|7.75%|7.75%
|13|60%|12.9|3.00%|3.00%|3.00%|3.00%|3.00%|3.00%|3 .00%|3.00%|3.00%|3.00%|3.00%|3.00%|8.00%|8.00%|8.0 0%|8.00%|8.00%|8.00%|8.00%|8.00%
|14|65%|12.775|3.25%|3.25%|3.25%|3.25%|3.25%|3.25% |3.25%|3.25%|3.25%|3.25%|3.25%|3.25%|3.25%|8.25%|8 .25%|8.25%|8.25%|8.25%|8.25%|8.25%
|15|70%|12.6|3.50%|3.50%|3.50%|3.50%|3.50%|3.50%|3 .50%|3.50%|3.50%|3.50%|3.50%|3.50%|3.50%|3.50%|8.5 0%|8.50%|8.50%|8.50%|8.50%|8.50%
|16|75%|12.375|3.75%|3.75%|3.75%|3.75%|3.75%|3.75% |3.75%|3.75%|3.75%|3.75%|3.75%|3.75%|3.75%|3.75%|3 .75%|8.75%|8.75%|8.75%|8.75%|8.75%
|17|80%|12.1|4.00%|4.00%|4.00%|4.00%|4.00%|4.00%|4 .00%|4.00%|4.00%|4.00%|4.00%|4.00%|4.00%|4.00%|4.0 0%|4.00%|9.00%|9.00%|9.00%|9.00%
|18|85%|11.775|4.25%|4.25%|4.25%|4.25%|4.25%|4.25% |4.25%|4.25%|4.25%|4.25%|4.25%|4.25%|4.25%|4.25%|4 .25%|4.25%|4.25%|9.25%|9.25%|9.25%
|19|90%|11.4|4.50%|4.50%|4.50%|4.50%|4.50%|4.50%|4 .50%|4.50%|4.50%|4.50%|4.50%|4.50%|4.50%|4.50%|4.5 0%|4.50%|4.50%|4.50%|9.50%|9.50%
|20|95%|10.975|4.75%|4.75%|4.75%|4.75%|4.75%|4.75% |4.75%|4.75%|4.75%|4.75%|4.75%|4.75%|4.75%|4.75%|4 .75%|4.75%|4.75%|4.75%|4.75%|9.75%

[/table]

[Murderous Hobo]

I did some playing around in excel and I found the following.


a d20 + 5 is always better then a d20 reroll except for DC 11 when both have an equal chance. Though chances are pretty close (0.02 difference) around DC [8..13].

a d20 + 4 is better on DC [2..5] (auto success) & [6] & [16..20] while a d20 reroll is better at DC [7...15]

a d20 + 3 is better on DC [2..4] (auto success) & [18..20] while a d20 reroll is better at DC [5...17]

a d20 + 2 is better on DC [2..3] (auto success) & [19..20] while a d20 reroll is better at DC [4...18]

a d20 + 1 is better on DC [2] (auto success) & [20] while a d20 reroll is better at DC [2...19]

a d20 reroll is always better then a d20.+ 0


In the middle band of the DC, re rolling is better as long as the bonus is small.

Generally speaking, IF the bonus is less then +5 AND the DC is less or equal too 20-bonus, then you should use reroll.


a d20 + 8 is always better then a d20 double reroll. Though chances are pretty close (0.02) around DC [11..14].

a d20 + 7 is better on DC [2..8] (auto success) & [9] & [16..20] while a d20 double reroll is better at DC [10...15]

a d20 + 6 is better on DC [2..7] (auto success) & [17..20] while a d20 double reroll is better at DC [8...16]


The same as above basically, in the middle-band of DC's the double reroll is better then a bonus unless the bonus is large enhough.

edit:

I ignored critical failure and critical hits.
I assumed you only rerolled in case of a failure on the original roll and you stick with the result of the new roll.

RS14:

Why not use P(success) = 1- P(failure) = 1 - ((DC-1)/20)^2 ?

It's allot easier.

edit:

A bonus moves the probability curve to the right without altering it's shape while rerolling makes the slopes of curve steeper and the top flatter.

[Cyclone231]

The reroll of one particular roll multiplies the odds of a failure by themselves (assuming that failure is known). In other words, if I have a 10% chance to fail, with the aid of a reroll, I have a 1% chance to fail. Just convert the percent into a decimal and square it. Oh, and you might want to subtract it from one to see what the percent chance of success is, though with the percentage chances of a twenty-sided dice it should be pretty obvious (I mean, the longest ones are 4 digits, and only with a .25 at the end - ex: 35% chance of failure squared is 12.25%).

Of course probability mechanics are worth noting here:
• This statistical benefit does not apply if the player will not, or can not, spend the reroll.
• Once the initial roll is made (and failed), the odds go directly back to their original ones - previous rolls have no effect on future rolls.


Also some systems (Saga's in particular) make a distinction between forcing a player to keep the new roll, or allowing a player to keep the old one if it was better. Does this distinction mater? What is it's effect, statistically?Statistically, your probability of success and failure are precisely the same as without it (assuming known win/lose). But when success is gradient, it is more complex. The odds of improving a past roll are equal to the odds of rolling anything higher.



Lastly, some systems allow a player to reroll a die after they know the result (pass/failure which D&D generally does not). Does this effect the situation?If the rerolls automatically counts, the probability of rolls below the player's personal "guesstimation" area reduces (in other words, if the player rerolls anything below an eleven, his odds of rolling something below an eleven are 25%; if he rerolls anything below a six, his odds of rolling something below a six is 6.25%) - the effect upon success depends upon what the DC actually is in comparison to the player's guess. If you have no tells, and the rolls can be chosen between, then the player's odds of success are increased as normal, though he'll feel bad about wasting a reroll.

[MaxMahem]

Many thanks for all the replies. I was afraid that the statistics would be considerably more complex then predicting the outcome of a simple die roll. Let me see If I am understanding it correctly then.

Example:
D20 vs DC 5 = 5/20 failure = 75% of success.
with reroll 5/20 failure * 5/20 failure = 25/400 failure = ~94% of success.
or a ~20% increase (equivalent to a +4 modifier)

D20 vs DC 15 = 15/20 failure = 25% success.
with reroll 15/20 failure * 15/20 failure = 225/400 failure = ~44% of success.
or a ~20% increase!?! (equivalent to a +4 modifier)...
hmm thought it should be more... let me try the extreams.

DC20 vs DC 2 = 1/20 failure = 95% of success.
with reroll 1/20 failure * 1/20 failure = 2/400 failure = ~100% of success.
or a ~5% increase (+1 modifier)

DC20 vs DC 19 = 19/20 failure = 5% of success.
with reroll 19/20 failure * 19/20 failure = 361/400 failure = ~10% of success.
or a ~5% increase (+1 modifier)

I think I've figure it out! The effect must be on a bell curve with DC 10 giving the best results (25% increase), governed by:
y=(x/20)-(x^2/400) where y is the percent increase and x is the DC.
though this only applies to situations where the player know where/when to make the reroll, otherwise the odds will be effected by how correctly he plays his strategy.

------

Hmm... it seems you all have lead me down the path to wisdom, my thanks! But this brings another question to mind. Is there a way to calculate the the statistical effects of player strategy on this decision? IE, I know the effects of "perfect play" on the odds of the situation, but how does a player making incorrect decisions effect these odds?

Also something I am still unclear about is the situation of keeping the best roll vrs keeping the second. It seems to me that if the player knows the DC already and plays correctly then there should be no difference. However, this could still be important in situations where the odds are variable, such as possibly when rolling against another player.
OTOH, now that I think about it, so long as the players rolls are not secret, the player already know who rolled higher and a player with the ability to take a reroll will know if he should take it, regardless if he has to keep the second roll or not.

Anyways, thanks for helping me out!

[Murderous Hobo]

Example:
D20 vs DC 5 = 5/20 failure = 75% of success.
with reroll 5/20 failure * 5/20 failure = 25/400 failure = ~94% of success.
or a ~20% increase (equivalent to a +4 modifier)

If you roll a 5 you will beat the DC.

So it's d20 vs DC 5 = (5-1)/20 = 80% chance of success.
with reroll (5-1)/20 failure * (5-1)/20 failure = 16/400 failure = ~92% of success.

[Jasdoif]

Since I just worked the probabilities out, here's a list of your chances to meet or exceed a particular number on a d20 with a certain number of rerolls upon failure (or simply rolling that many d20s and taking the highest number):

{table=head]Chance[br]to meet[br]or beat|Only 1|Best of 2|Best of 3|Best of 4|Best of 5|Best of 6
2|95.0000%|99.7500%|99.9875%|99.9994%|100.0000%|10 0.0000%
3|90.0000%|99.0000%|99.9000%|99.9900%|99.9990%|99. 9999%
4|85.0000%|97.7500%|99.6625%|99.9494%|99.9924%|99. 9989%
5|80.0000%|96.0000%|99.2000%|99.8400%|99.9680%|99. 9936%
6|75.0000%|93.7500%|98.4375%|99.6094%|99.9023%|99. 9756%
7|70.0000%|91.0000%|97.3000%|99.1900%|99.7570%|99. 9271%
8|65.0000%|87.7500%|95.7125%|98.4994%|99.4748%|99. 8162%
9|60.0000%|84.0000%|93.6000%|97.4400%|98.9760%|99. 5904%
10|55.0000%|79.7500%|90.8875%|95.8994%|98.1547%|99 .1696%
11|50.0000%|75.0000%|87.5000%|93.7500%|96.8750%|98 .4375%
12|45.0000%|69.7500%|83.3625%|90.8494%|94.9672%|97 .2319%
13|40.0000%|64.0000%|78.4000%|87.0400%|92.2240%|95 .3344%
14|35.0000%|57.7500%|72.5375%|82.1494%|88.3971%|92 .4581%
15|30.0000%|51.0000%|65.7000%|75.9900%|83.1930%|88 .2351%
16|25.0000%|43.7500%|57.8125%|68.3594%|76.2695%|82 .2021%
17|20.0000%|36.0000%|48.8000%|59.0400%|67.2320%|73 .7856%
18|15.0000%|27.7500%|38.5875%|47.7994%|55.6295%|62 .2850%
19|10.0000%|19.0000%|27.1000%|34.3900%|40.9510%|46 .8559%
20|5.0000%|9.7500%|14.2625%|18.5494%|22.6219%|26.4 908%[/table]

Note that the odds of getting 2 or better isn't really 100% with five or six dice. However, the odds of not getting a 2 or better are 1 in 3,200,000 with five dice, and 1 in 64,000,000 with six dice. It'd take a lot of percentage spaces to display that....

[Yakk]

Reading the above...

I'd value a reroll at about a +3.5 modifier. :)

The advantage rerolls have is that they are very binary and require less math. You either reroll, or you do not. Knowing if you succeeded/failed is easy -- but I find most people take a number of seconds to work out "if I add in my one-off bonuses, can I win? What one-off bonuses should I save and which should I spend?"

Rerolls also keep you away from "modifier hell" -- past +/- 10, d20 probabilities break down pretty badly.

[Jasdoif]

The advantage rerolls have is that they are very binary and require less math. You either reroll, or you do not. Knowing if you succeeded/failed is easy -- but I find most people take a number of seconds to work out "if I add in my one-off bonuses, can I win? What one-off bonuses should I save and which should I spend?"Another feature is that rerolls don't allow you to do something otherwise impossible, like an added modifier does. If you need to roll a 21 on a d20 (in the absence of automatic natural 20 successes), no amount of rerolls is going to make it happen. A +3 modifier, on the other hand....

Basically, it keeps the linear modifiers relevant. If those linear modifiers are solely from your ability score and/or feats you've taken, it keeps your own personal ability relevant even if you have a lot of reroll chances.