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The original rules in TROS use varying pools of d10s to attempt rolling one or more successes against a varying Target Number (e.g., commonly a TN=7). Roll the pool of dice, count up the successes that meet or exceed the TN. Fair enough.

But the original rules state that a fumble occurs when "you fail a roll AND have 2 or more 1's showing on any die." That is, none of the dice meet or exceed the TN and 2 or more dice result in a score of 1.

Obvious to us amateur game designers that the more dice in your pool (the more skilled the character) the MORE likely you are to roll a fumble. A poor design choice IMHO, and one of the few broken rules in an otherwise great game system.

So I tweaked the rules with House Rules, variances, more House Rules, etc., until the problem was solved, but in my solution the rules became unwieldy and un-user-friendly.

How can the likelihood curve for fumbles be straightened or slightly reversed for TROS? How can I undo this problem of more competent people being more likely to fumble?

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As Dale M correctly points out, the probability of a fumble actually goes down for sufficiently large pools. This is because you can only fumble when you don't succeed, and with a large enough pool of dice, success becomes overwhelmingly likely.

Here's a simple AnyDice program to calculate the fumble rate for various target numbers and pool sizes:

function: roll ROLL:s target TARGET:n {
  if (ROLL >= TARGET) { result: 1 }  \ success \
  if (ROLL = 1) >= 2 { result: -1 }  \ fumble  \
  result: 0                          \ failure \
}
loop T over {6..8} {
  \ optimization: use a custom d10 that can only roll 1 (fumble?), 2 (no success) or 10 (success) \
  DIE: {1, 2:(T-2), 10:(10-T+1)}
  loop N over {2..10} {
    output [roll NdDIE target T] named "[N]d10 vs. [T] (-1 = fumble, 0 = fail, 1 = success)"
  }
}

(The only non-obvious part of this program, besides the general trick for "freezing" dice in AnyDice by passing them to a function as a sequence, so that we can examine the result of a specific roll, is the use of a custom die to "relabel" the sides of the d10. This is strictly an optimization; the program would give the exact same results with Nd10 instead of NdDIE, but would run much slower, and would likely time out unless you reduce the maximum size of the pool.)

This program gives the following fumble probabilities for various target numbers and pool sizes:

 Pool | vs. 6 | vs. 7 | vs. 8
------+-------+-------+-------
 2d10 | 1.00% | 1.00% | 1.00%
 3d10 | 1.30% | 1.60% | 1.90%
 4d10 | 1.13% | 1.71% | 2.41%
 5d10 | 0.82% | 1.52% | 2.55%
 6d10 | 0.54% | 1.23% | 2.43%
 7d10 | 0.33% | 0.92% | 2.17%
 8d10 | 0.19% | 0.66% | 1.85%
 9d10 | 0.11% | 0.46% | 1.52%
10d10 | 0.06% | 0.31% | 1.21%

Ps. The reason these don't precisely match Dale's numbers is that his formula seems to have an error; specifically, it double-counts cases where one ends up rolling more than two ones (and no successes).

The correct formula can be derived by first calculating the probability of not succeeding on any roll, which is simply:

$$ P({\rm fail}) = \left(\frac{T - 1}{10}\right)^N $$

where \$N\$ is the number of dice rolled, and \$T\$ is the target number. Now, given that one has not succeeded (i.e. all rolls are less than \$T\$), the conditional probability of a fumble is equal to the probability of rolling 2 or more ones on \$N{\rm d}(T-1)\$. This is equal to 1 minus the probability of rolling either 0 or 1 ones on \$N{\rm d}(T-1)\$, i.e.:

$$ P({\rm fumble} \mid {\rm fail}) = 1 - \left(\frac{T-2}{T-1}\right)^N - \frac{N}{T-1} \times \left(\frac{T-2}{T-1}\right)^{N-1} $$

Combining these, we get:

$$ \begin{align} P({\rm fumble}) & = P({\rm fumble} \mid {\rm fail}) \times P({\rm fail}) \\ & = \left( 1 - \left(\frac{T-2}{T-1}\right)^N - \frac{N}{T-1} \times \left(\frac{T-2}{T-1}\right)^{N-1} \right) \times \left(\frac{T - 1}{10}\right)^N \end{align} $$

which indeed yields numbers matching the AnyDice results.

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  • \$\begingroup\$ Thank you all so much for your help on this! I appreciate the explanations, graphs and suggestions given, and have studied them in depth. As I continue to experiment with my homebrew variant, I do a lot of probability graphs on paper and spreadsheets, but sometimes I need your professional opinions to get past a formulaic brick wall. Thanks again! \$\endgroup\$ – Everett Steed Mar 20 '16 at 7:18
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Obvious to us amateur game designers that the more dice in your pool (the more skilled the character) the MORE likely you are to roll a fumble. A poor design choice IMHO, and one of the few broken rules in an otherwise great game system.

Except ... you're wrong - the more dice you roll the less likely you are to get a fumble.

It works like this: To get a fumble you have to roll a 1 on any 2 dice and less than the target number on all of the others. This probability increases for a while as dice are added to the pool but after a certain point it reaches a maximum and then falls away very rapidly.

The exact formula for a failure is

$$ nC2 \times 0.1^2 \times \left(\frac{T-1}{10}\right)^{n-2} $$

where n=number of dice, C=Combination function, T=target number.

Here is the data for dice pools from 2 to 10 and target numbers of 6 to 8.

\begin{array}{r|ccc} & 6 & 7 & 8 \\ \hline 2 & 0.010 & 0.010 & 0.010 \\ 3 & 0.015 & 0.018 & 0.021 \\ 4 & 0.015 & 0.022 & 0.029 \\ 5 & 0.013 & 0.022 & 0.034 \\ 6 & 0.009 & 0.019 & 0.036 \\ 7 & 0.007 & 0.016 & 0.035 \\ 8 & 0.004 & 0.013 & 0.033 \\ 9 & 0.003 & 0.010 & 0.030 \\ 10 & 0.002 & 0.008 & 0.026 \end{array}

For 7 target number the peak is at 4d10 and 5d10 at 2.2% and it falls after that. At 9d10 it is below what it is at 2d10.

It looks like this.

Now, I am not familiar with the game so I do not have any idea what size dice pools or target numbers are realistic but if 4-5+ are typical then you don't really have a problem; more dice mean less fumbles and a harder target number mean more fumbles.

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  • 2
    \$\begingroup\$ After looking at your graph, I decided to have a look at where this break-even point is, i.e. I checked which n yields the highest fumble chance for any given t. This number seems to rise exponentially: [t: N] 6: 3, 7: 4, 8: 6, 9: 9, 10: 19. This means that the problem OP presents exists for a large enough number of cases to warrant a fix. \$\endgroup\$ – MrLemon Aug 11 '15 at 8:22
  • \$\begingroup\$ Your general conclusion is correct, but I think your numbers are a bit off, presumably because your formula double-counts some combinations (since "rolling a 1" is a sub-event of "rolling under the target"). Here's an AnyDice script that calculates the correct numbers by brute force (and an optimized version). The table labeled "1" gives the probability of a fumble; the one labeled "0" gives the probability of no fumble. \$\endgroup\$ – Ilmari Karonen Aug 12 '15 at 14:02
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I don't completely agree with Dale M's answer. Yes, the overall chance of fumbling might be less because the chance of failure decreases. I believe you should look at the chance of fumbling given that the character has failed. An expert has less chance of failing, but when he fails he has a bigger chance of failing spectacularly, which seems indeed unrealistic.

As notes in the comments a specialist might have a higher chance of fumbling because he does complicated things. However, this should depend on the target number and not on the number of dice rolled.

Possible solution 1: have the number of ones depend on the number of dice

The probabilities for a fumble for a target number of 7 as a function of \$n\$ dice rolls and \$k\$ one's for a fumble, given that the character has failed. (It is assumed all dice are below the target number) are given in the following table:

\begin{array}{r|rrrrr} & k \\ n & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 0.17 & - & - & - & - \\ 2 & 0.31 & 0.03 & - & - & - \\ 3 & 0.42 & 0.07 & 0.00 & - & - \\ 4 & 0.52 & 0.13 & 0.02 & 0.00 & - \\ 5 & 0.60 & 0.20 & 0.04 & 0.00 & 0.00 \\ 6 & 0.67 & 0.26 & 0.06 & 0.01 & 0.00 \\ 7 & 0.72 & 0.33 & 0.10 & 0.02 & 0.00 \\ 8 & 0.77 & 0.40 & 0.13 & 0.03 & 0.00 \\ 9 & 0.81 & 0.46 & 0.18 & 0.05 & 0.01 \\ 10 & 0.84 & 0.52 & 0.22 & 0.07 & 0.02 \end{array}

Assuming you want approximately a 10% chance on a fumble (given a failure). You should require the following number of ones depending on the number of dice:

\begin{array}{rl} n & k \\ \hline 1 & 1 \\ 2 & 2 \\ 3 & 2 \\ 4 & 2 \\ 5 & 3 \\ 6 & 3 \\ 7 & 3 \\ 8 & 3 \\ 9 & 4 \\ 10 & 4 \end{array}

However, this is only approximate and since this also depends on the target number, this becomes quite complex.

Possible solution 2: use a separate die for fumbles

The simplest solution I can think of is to use a separate die for fumbles. For example, use a differently coloured d10 in your die pool. A one on that roll indicates a fumble (when it is a failure). In that case you have a flat probability of 10% given failure, and since an expert has less chance for failure a smaller overall chance for a fumble for experts.

Using more than one die and fumble on both a one or two, you can tweak the fumble chance.

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  • \$\begingroup\$ Note that, for small die pools (1d10 or 2d10), a 10% chance of fumble on failure is significantly higher than what following the official rule gives. This may or may not be what you want. \$\endgroup\$ – Ilmari Karonen Aug 12 '15 at 22:36

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