What's the effective bonus of a double d20 roll take the highest?

[jaldaen]

What's the effective bonus to an attack roll or skill check (d20) where you roll two dice and take the highest?

In short, I'm looking for how much the reroll affects the end result on average as expressed as a +<insert number>.

I'm trying to reduce rolls and just want to know what bonus is close to/equivalent to a double roll.

Also what about rerolls where you take the lowest/highest?

Thanks for your help

 

[Alzrius]

Well, the average result of a d20 roll is 10.5, with the .5 being a representation that half the time it will be 10, and the other half it will be 11. Given that, I'd say that, on average, rolling twice would mean you'd get a roll of 10 and 11, so you'd take the 11, being higher.

Ergo, the effective bonus would be +11.

 

[Matt Black]

I did a quick computer simulation, and the average roll of 2d20 - keep highest is 13.7(02 +/- 0.04). So that's 3.2 above the 1d20 mean of 10.5.

For 2d20 - keep lowest, the mean value is 7, so it's a -3.5 penalty.

So granting a +3/-3 bonus/penalty is pretty much equivalent.

 

[Particle_Man]

I think the bonus depends on what number you need to hit.

So if you need an 11 to hit, then two rolls gives you a +5 bonus.

If you need a 20 to hit, then two rolls gives you less than a +1 bonus.

If you need a 2 to hit, then two rolls gives you less than a +1 bonus.

 

[DM-Rocco]

I think the bonus depends on what number you need to hit.

So if you need an 11 to hit, then two rolls gives you a +5 bonus.

If you need a 20 to hit, then two rolls gives you less than a +1 bonus.

If you need a 2 to hit, then two rolls gives you less than a +1 bonus.

First, I understand the original posters concern about trying to get rid of as much rolling as you can. However, a bonus to hit instead of a second roll doesn't get rid of a bad roll, which is what I think the main purpose of such an ability is for. It doesn't matter if you give the PC a +11 on the roll and he rolls a 2 but only needed a 14 to hit. All you have really done is weak an ability.

If you are bound and determined to do it though, I agree with the basic math from Matt Black, a +3 is about right. However, Particle_Man brings up an excellent point, depending on what number you need to hit, the bonus would be higher. I would suggest, if the PC needed to roll a 1-4 to succeed, the bonus would be +1, a 5-8 a +2, a 9-12 a +3, a 13-16 a +4 and a 17-20 a +5.

For the time it would to take into account varibles, I suggest just letting the PC have his second roll. He is only going to use it if he misses the first time or really needs a critical result (which your bonus to hit won't hep) and it only take a few seconds to re-roll a die. Taking that away from the player actually weakens the end result.

 

[Vurt]

The question of how to calculate the expected result of rolling a d20 twice and choosing the best result came up a couple of times before, and here's what I worked out. At first I thought it was 75% of the die, but when trying to work out the probabilities again I realized that this is technically incorrect. What you get is actually a 75% chance to roll above average.

Proof: For each roll of a die, there is a 50% chance of rolling less than average. Rolling twice, there is a 25% chance (0.5 x 0.5) of both rolls being below average, and thus a 75% chance that one of them will be above average.

To find the expected result of this process, however, it looks like you need to add the sum of each outcome multiplied by its relative probability (i.e. the weighted sum). From my doodling, there doesn't appear to be an easy algebraic formula to do this. The best I could come up with is summation notation (i.e. sigma notation) or pseudocode.

First the sigma notation:

y = Sigma [x=1..n] (x(2x-1)/(n^2))

Where y is the expected result, n is the number of sides on the die in question, and x is just a counter that runs from 1 to n.

Now the pseudocode, using the same variables:

y = 0; //initialize

for x = 1 to n do {
y = y + x * (2 * x - 1) / (n ^ 2)
}

Crunching the numbers, you get:

Rolling a d2 twice and choosing the best result should average 1 and 3/4 (and not 75% of 2, or 1.5 as I had originally surmised).

Rolling a d4 twice and choosing the best result should average 3 and 1/8th (or 3.125).

Rolling a d6 twice and choosing the best result should average 4 and 19/36ths (or 4.52777...).

Rolling a d8 twice and choosing the best result should average 5 and 13/16ths (or 5.8125).

Rolling a d10 twice and choosing the best result should average 7 and 3/20ths (or 7.15).

Rolling a d12 twice and choosing the best result should average 8 and 35/72 (or 8.486111...).

...

Rolling a d20 twice and choosing the best result should average 13 and 33/40ths (or 13.825).

~~

So to answer the OP's original question, take the average value of rolling twice, subtract the average value for just rolling normally, and round up or down as you see fit. That should be the bonus you're looking for.

 

[Nifft]

When using a maximization function, mean ≠ mode, so there's no effective bonus, because you've distorted the (previously linear) PDF.

For example, you've almost doubled the chance of ending with a result of 20, while you've made a result of 1 twenty times less likely.

Cheers, -- N

 

[Matt Black]

However, Particle_Man brings up an excellent point, depending on what number you need to hit, the bonus would be higher.

I agree. My result is only valid if you commit to making the second roll before learning the result of the first roll.

 

[Pielorinho]

If I understand correctly, there are only 400 possible results for this roll. How hard would it be to set up a spreadsheet to check it out? I'm gonna go try now.

Daniel

 

[Nifft]

If I understand correctly, there are only 400 possible results for this roll. How hard would it be to set up a spreadsheet to check it out? I'm gonna go try now.

Graph it. That's the illustrative part IMHO.

Cheers, -- N