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That is to say, the value of a second roll, how much of a bonus value is it worth.

For example: when you make a saving throw, attack or check roll, when does +bonus to the roll and rolling twice equal the same worth.

I believe the value of a second roll would be different depending on if it was 'Choose the Highest', 'Choose the Lowest', 'Choose any Result' and 'Reroll on Failed'. Though in this case I'm thinking for 'Choose the Highest'.

Edit in response to comments; For the intentions of this question the dice is a d20.

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2 Answers 2

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The answer depends on what you're rolling, what your target number is, and what modifiers are already in place.

For D20s, the answer is about five, but less if you need to roll really high or really low. See this answer.

In general, the formula for probability of success is:

100% - (chance of failure per roll)^(number of attempts).

This applies to Choose the Highest, Choose any Result, and Reroll on Failed (all of which are mechanically identical, unless the PC wants to fail).

Suppose you're attempting a task with a 60% success rate. With two attempts, you end up with an 84% success rate.

A task with a 25% success rate ends up with a 43.75% chance of success.

A task with a 75% success rate ends up with a 93% chance of success.

How does this correspond to static modifiers?

It depends entirely on what you roll! A +1 modifier means very different things with a D20 than, say, 3D6.

This is an example of a question that is ridiculously complicated to answer in a system agnostic fashion, and fairly easy to answer for a specific system.

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    \$\begingroup\$ It's not "about +/-5, it's "at most +/-5". It's +5 under optimal circumstances (you need an 11 to succeed. If you need a 2 or a 20, it can be a +/-1 or even a +0.05. It also depends whether you get the reroll automatically (roll twice and take the best/worst), or you spend some resources on the reroll only if you fail the first time. On average, not knowing the target number, it's probably about +/-2.5. \$\endgroup\$
    – mcv
    Commented Mar 6, 2014 at 8:22
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    \$\begingroup\$ @mcv Target numbers have a tendency to fall into the middle range most of the time. You don't roll a lot of checks with very low or high probabilities of success, and modifiers to these rolls matter a lot less. Counting them as equal to the more common scenarios is less useful. As to resources being spent, the question makes no mention of them. They're not important to this answer. \$\endgroup\$
    – AceCalhoon
    Commented Mar 6, 2014 at 13:58
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The mean of 2d20k1 is 13.82 - roughly equal to +3, +4 if rounding up.
The median is 15 - equal to +5
The mode is 20...

The problem being that, if the target number ("TN") is greater than 10, the impact is more, while if less than 10, less. For a TN 15+ save, you go from 30% chance to 51% - slightly better than a +4, while on a TN 5+, you go from 80% to 96%, roughly a +3. On an 11+, you go from 50% to 75%, equal to a +5.

The distribution is

  N     p(N)    p(N)    p(N...20) p(N...20)
===========================================
  1    1/400    0.25%   400/400   100%
  2    3/400    0.75%   399/400    99.75%
  3    5/400    1.25%   396/400    99%
  4    7/400    1.75%   391/400    97.75%
  5    9/400    2.25%   384/400    96%
  6   11/400    2.75%   375/400    93.75%
  7   13/400    3.25%   364/400    91%
  8   15/400    3.75%   351/400    87.75%
  9   17/400    4.25%   336/400    84%
 10   19/400    4.75%   319/400    79.75%
 11   21/400    5.25%   300/400    75%
 12   23/400    5.75%   279/400    69.75%
 13   25/400    6.25%   256/400    64%
 14   27/400    6.75%   231/400    57.75%
 15   29/400    7.25%   204/400    51%
 16   31/400    7.75%   175/400    43.75%
 17   33/400    8.25%   144/400    36%
 18   35/400    8.75%   111/400    27.75%
 19   37/400    9.25%    76/400    19%
 20   39/400    9.75%    39/400    9.75%

N: number on the die
p(N) probability of rolling N
p(N...20) Probability of a number between N and 20, inclusive.
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