# Converting this action resolution mechanic from total rolled to highest die rolled

Under a standard action resolution mechanic, a player rolls 3d6 and compares the total rolled to a Target Number (TN) assigned by the gamemaster, succeeding if the total meets or exceeds the TN. For example, if they rolled 1+3+2=6 or 1+3+1=5 against TN 5, they would succeed because their total was at least 5 in either case.

Under an alternative mechanic, a player rolls 3d6 and compares the single highest die rolled to a Die Rating (DR) assigned by the gamemaster, succeeding if the single highest die meets or exceeds the DR. For example, if they rolled max(1, 3, 4)=4 or max(1, 3, 3)=3 against DR 3, they would succeed because their highest die was at least 3 in either case.

Suppose the common TN's used for the standard mechanic are 5, 10, and 15. What DR value best approximates the probability of success for each common TN? In other words, if the player has a certain probability of success when rolling the total of 3d6 versus TN 5 (and so on) then they would have roughly the same probability of success when rolling the highest of 3d6 versus what DR (respectively)?

For reference, the standard mechanic described is from Risus for "professional"-ranked cliches using only the TN's that 3d6 can beat (the game actually allows rolling as many as 6d6 and the highest listed TN is 30), and the alternative mechanic described is an attempt to convert the "Best of Set" rule variant from the Risus Companion to work for unopposed rolls. I focused on only 3d6 to get the gist for the typical number of dice as a baseline.

• Since the question is about Risus mechanics, it's important that it can be found by Risus experts and future Risus players, and I've tagged it accordingly and removed the explanation for leaving off the tag. Jul 12, 2018 at 19:28
• Would it be acceptable to manipulate the number of dice rolled for the new system, or are you set on 3d6? Jul 12, 2018 at 22:14
• Keep it fixed at 3d6. (The actual system uses anywhere from 1d6 to 6d6, and I'm using 3d6 as the baseline because it's effectively the standard roll.) Jul 12, 2018 at 23:27

## 2 Answers

From the output of this basic AnyDice script, we can see that:

• 3d6 vs. TN 5 gives a 98.15% probability of success (i.e. a 1.75% probability of failure).
• 3d6 vs. TN 10 gives a 62.5% probability of success (i.e. a 37.5% probability of failure).
• 3d6 vs. TN 15 gives a 9.26% probability of success (i.e. a 90.74% probability of failure).

Also, the probabilities of rolling at least a given DR from 1 to 6 on the highest roll of 3d6 are:

• DR 1: 100% probability of success (0% probability of failure).
• DR 2: 99.54% probability of success (0.46% probability of failure).
• DR 3: 96.30% probability of success (3.70% probability of failure).
• DR 4: 87.50% probability of success (12.50% probability of failure).
• DR 5: 70.37% probability of success (29.63% probability of failure).
• DR 6: 42.13% probability of success (57.87% probability of failure).

Comparing these probabilities, we can see that the closest approximation to TN 5 is either DR 2 or DR 3. Neither of these is a very good approximation, though: DR 2 gives a failure rate only about a quarter of that of TN 5, while the failure rate with DR 3 is about twice that of TN 5.

TN 10 is probably best approximated by DR 5, although it's not a particularly close approximation either. And there simply is no good approximation for TN 15 using the highest-of-3d6 vs. DR mechanic; even DR 6 is more than four times as likely to succeed as TN 15.

While matching the original probabilities of the TN mechanics exactly is probably not really necessary (or even desirable), the fact that you can't get a success rate less than 40% using the highest of 3d6 vs. any DR is a bit of an issue. This suggests that, if you want a mechanic that works roughly the same as the original, you'll need to use something other than the highest of 3d6 for difficult challenges.

One possible option, off the top of my head, would be to always use a DR of, say, 4, but vary the number of dice in the 3d6 roll that have to meet the DR for the roll to be considered a success. The probability of rolling 4+ on all three d6's is exactly 1/8 = 12.5%, while the probability of rolling 4+ on at least two out of 3d6 is exactly 50% (and the probability of rolling 4+ on at least one die out of 3d6 is 7/8 = 87.5%, as already noted above). Of course, you can also vary the DR and the total size of the pool to further expand the available range of success probabilities.

• That last paragraph describes the resolution mechanic of the Burning Wheel. Jul 13, 2018 at 0:06

TLDR: Due to the mathematics involved, you will probably not be able to replace a sum of 3d6 mechanic with a best of mechanic using any die type. Sorry.

I do have some suggestions at the bottom.

If you would indulge me here a bit, I am going to teach you how to fish so you can work these figures out for yourself in the future. With these tools you should be able to do this with different die types, and a different number of die and tweak it to get close to what you want.

So the best was to figure these types of probability problems is to figure out what the chance of NOT rolling your target number or higher. Then, since, each die roll is independant you can simply multiply that probability by the number of die you roll.

In this case lets pick the target number of 6. 5 times out of 6 you will not roll a 6 or higher, so your probability is 5/6. That is true for each die you roll. In this case you rolled 3, so we get

5/6 * 5/6 * 5/6 which in this case is 125 / 216 or ~42%

That translates to a 0.46% chance of NOT rolling a 6 or higher on at least one of the three dice. Subtract the from a percentage of 100% and you have your chance of rolling that target number or higher.

Similarly if your TN is 5, then you will NOT roll that TN or higher 2/3 of the time on one die. On three die you just cube 2/3 (as above) and you get ~70% chance.

Here are the other in a chart:

But now you should be able to work out what other TNs will get you:

Want to know what id the Probability on a best of 3d8 with a TN of 7?

1. Probability of not rolling TN on 1 die is 6 out of 8 = 3/4 = 75%
2. 0.75 * 0.75 * 0.75 = ~42%
3. 1 - .42 = ~58%

An analysis of a TN of 8 on 3d8 pick best yields ~33% chance of success. You are approaching the TN of 10 in your original mechanic.

Some trends are starting to shake out of this analysis. First off, increasing the number of dice makes the TN easier to achieve. Makes sense if you think about it. Conversely, It does look like larger dice and higher TN are harder to reach. BUT, you will also find that even using larger dice your probability does not increase linearly. In fact it will drop off exponentially: ie larger and larger die have smaller and smaller increases; making a probability of 90% hard to ever reach.

All this means that you probably can not effectively replace a best sum of dice mechanic with a best of type mechanic.

You might try a best 2 out of three mechanic, if you like. In other words, if your TN is 6, you have to roll a 6 or higher on 2 of the 3 dice.

I will work the %ages for you tomorrow - the math is a bit harder on that one.