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Using stun to reduce effectiveness of reflexive actions

I've used my own homebrew system a lot, and recently began making a spellcrafting system. One effect I want aspiring mages to be able to load essentially stuns the target while still allowing them to do any activities they usually do, but making them only half as likely to succeed on reflexive actions (such as defending and dodging) they take under this effect.

Assuming I roll d20s to make these checks, how can I make them half as likely to succeed in most circumstances? (See below for details.)

I don't just want to subtract from the target's rolls or raise DCs because that might make taking the action at all impossible (though it can work with other methods, and ultimately may be required). Using coins isn' t the answer I am looking for. Anything else, though, barring random decision by a human, is fair game. Bonus points for some sort of flavor.

Conditions to be satisfied:

  1. When large penalties are in play that result in an automatic failure, the stun mechanism shall also result in an automatic failure.
  2. When moderate bonuses or penalties are in play that result in neither automatic success nor automatic failure, the stun mechanism shall reduce the chance of success by one half (and as a consequence, result in neither an automatic success or automatic failure.)
  3. When large bonuses are in play that result in a automatic success, the stun mechanism shall result in a chance of success not less than 50% (but may be larger, including automatic success. The transition from stunned 50% chance of success to stunned automatic success should be gradual.)
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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – mxyzplk Nov 30 '16 at 8:27

11 Answers 11

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Flip a coin, subtract 20 if tails

Here's the basis of this: the chance of throw less than or equal to a certain number halves when you double the die size. However, we don't want to halve the chance of throwing less, we want to halve the chance of throwing more!

So let's throw some math in there.

The chance of success is actually: \$ \frac {20 - DC + 1} {20}\$

We want the chance to become: \$ \frac {20 - DC + 1} {40}\$

The 40 in the denominator would suggest we need a d40

However, using a d40, the chance becomes \$ \frac {40 - DC + 1} {40} \$

So we subtract 20 from the roll: \$ \frac {40 - 20 - DC + 1} {40} = \frac {20 - DC + 1} {40}\$

Now, you can simulate a d40 by rolling a d20 and then flipping a coin, adding 20 to the result of the first roll if the coin comes up heads. However, \$n + \left(20\;\text{half of the time}\right) - \left(20\;\text{all the time}\right)\$ is the same as \$n - \left(20\;\text{half of the time}\right)\$. So, we can instead roll a d20, flip a coin and if the coin comes up heads subtract 20 from the first roll.

Of course, a flip of a coin can be simulated by rolling a die with an even number of sides and considering the even numbers as head. (Or the even numbers tails. Or, you could go anything above \$ \frac {die size} 2 \$ as heads/tails.)

If odd, negative // If off, subtract 20

If you want to get rid of the coin flip (or second die roll) you could tie the two together. This sacrifices some of the precision, but leaves you with only one die to roll.

Let's take a d20, but add one to each of its odd sides. That d40 would go 2, 2, 4, 4, 6, 6..40, 40. Now, we know from above that what we need is d40-20. If we rolled that with out "double numbered d40", the possible results would be -18, -18, -16, -16..20, 20. If we continue to modify our die by removing one from all sides <= 0 we get: -19, -19, -17, -17..-1, -1, 2, 2, 4, 4..20, 20.

Of course, sides with the same number can be merged, giving: -19, -17..-1, 2, 4..20. There are two ways you can get exactly this result: subtracting 20 if the number is odd or multiplying odd numbers by -1 (i.e. make them negative). The first way keeps the sum of opposite sides constant (giving numbers that are slightly closer to truly random) but the second way may be easier on you and the players. Pick whichever one you feel more comfortable with.

Of course, we manipulated the chances a bit when we changed its sides. But how much did we change the chances? Negative even values can't be rolled. That's why, if the the shift will be slightly difference between the true half of the original chance. The same goes for odd positive values. Basically, in each of those cases the \$DC - bonus\,\$ is the same as either its value plus one. The chance will never be shifted more than 1 out of forty = 2.5%, so that's your inaccuracy. No check that was previously possible to make has disappeared, because the only chance that was at risk of that (DC-skill = 20) has the 2.5% shift in the player's advantage, staying at 5% instead of halving.

Conclusion

So that's it, you can either use an extra coin flip (or dice throw) and get complete accuracy on halving the chances or use only one dice throw and accept a rounding error of 2.5% in half of the required values to throw.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – SevenSidedDie Nov 25 '16 at 17:53
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To halve success probabilities make the roll as normal; if the result indicates success then flip a coin,* and if it comes up tails the roll fails.

This works for any combination of dice, with any modifiers, for any probability of success.

* Or equivalent.


Why this works:

The probability of failing any given roll is:

$$P(\text{failure}) = 1 - P\left(\text{success}\right)$$

The coin flip halves the probability of success (i.e. "make it 50% less likely to succeed"), thereby increasing the probability of failure:

$$P(\text{failure}) = 1 - \frac{P\left(\text{success}\right)}{2}$$

Note: that "make it 50% less likely to succeed, or in other words, twice as likely to fail?" is actually a non-sequitur because success and failure probabilities are bound by the relationship:

$$P\left(\text{success}\right) + P\left(\text{failure}\right) = 1$$

Therefore halving the probability of success cannot mean doubling the probability of failure for all probabilities of success. For example:

$$\text{If } P\left(\text{success}\right) = 0.2; P\left(\text{failure}\right) = 0.8\\ \phantom{0}\\ \text{but } \frac{P\left(\text{success}\right)}{2} = 0.1\\ \phantom{0}\\ \text{so } P\left(\text{failure}\right) = 1 - 0.1 = 0.9 \text{, not } 0.8\times 2 = 1.6\\ \text{ (and a probability} = 1.6 \text{ is meaningless anyway.)}$$

The only probability of success for which "make it 50% less likely to succeed, or in other words, twice as likely to fail" can be true is:

$$P\left(\text{success}\right) = \frac{2}{3} \text{, so } P\left(\text{failure}\right) = 1 - \frac{2}{3} = \frac{1}{3}.\\ \phantom{0}\\ \text{Because then } \frac{P\left(\text{success}\right)}{2} = \frac{1}{3} \text{, so } 2\times P\left(\text{failure}\right) = \frac{2}{3}\text{, and }\\ \phantom{0}\\ \frac{1}{3} + \frac{2}{3} = 1.$$

For all other success probabilities, the sum of halving success and doubling failure probabilities will not sum to 1.

* Or equivalent.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – mxyzplk Nov 30 '16 at 8:27
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Roll a d40 instead of a d20, and subtract 20. That way you double the space you're selecting from, while leaving the number of successful slots the same.

(You can use a d4 and a d10 if, for some strange reason, you don't have a literal d40 lying around)


From the comments, success is defined as \$mods+d20 \ge DC\$. Generalizing for a roll with arbitrary min and max values (inclusive), we can define a value \$x = \frac{max + 1 - target}{max + 1 - min}\$.
The chance of success is then: \$\begin{cases} 0 & x < 0 \\ x & 0 \le x \le 1 \\ 1 & x > 1 \\ \end{cases}\$

With a d20, \$x = \frac{21 - target}{20}\$; this method simply halves \$x\$ to \$\frac{21 - target}{40}\$.
If \$x\$ was originally between 0 and 1, this also halves the chance of success; if \$x\$ was originally between 1 and 2, this transitions the new chance of success gradually from 50% to 100%.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – SevenSidedDie Nov 25 '16 at 17:53
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terminology

First of all, we need to look at the success rate, as we can't multiply a factor 2 towards the failure rates: you will easily acknowledge, that multiplying a 70% failure chance \$f\$ by two gives a 140% chance of failure, which is not a valid argument.

So the question is: How to reduce the chance for success \$s\$ to half? This will also tell us the failure rate because $$f+s=100\%$$

starting at the dice: looking at success chances

Assuming a d20, each face stands for 5%. So if you would need to roll a 15 or better to succeed would give a 30% chance to do so, a 16 gives 25% and so on. Having \$s=30\%\$ means \$f=70\%\$, \$s=25\%\$ means \$f=75\%\$.

calculating the half success chances

Now, our chances are (as anything below a certain number is a failure): $$f_{(n)}={5\%\times n}\text{ with }n\in\{1,2,3,...,19,20\}$$ $$s_{(n)}=100\%-f_{(n)}=(100-(5\times n))\%$$ To get half the success chance, we have to divide \$s\$ by two, which is equal to multiplying with 0.5 or 50%. So, \$s'\$ is: $$s_{(n)}'=(100\%-f_{(n)})\times 50\%=\frac {[100-(5\times n)]\%} 2$$ With a bit math magic (we know the exact result of \$s'=s/2\$ and can work with that) we can create an alternate formula for this: $$s_{(n)}'=[100-(5\times n')]\%\text{ with }n'=\frac n 2\text{ with }n\in\{1,...,20\}$$ Now, we just got into trouble: our die can only show results of 1 to 20, with each side having 5% chance. However, we get s'=2.5% for n=19. That would be a 19.5 on a 20 sided die, which is not a full number and thus not possible. So we'd need a die, that has each side with 2.5% chance to show up. That is a 40 sided die (100%/40=2.5%). Those we don't have access to usually, so we need different ways to possibly get a perfect result.

approaches to fixing

approximation

The most easy way could be approximation: just round up the DC. This gives us a nice thing to show in a graph: we expect the light blue graph, but we get the orange one. Maybe this is good enough, but it is a shift not favorable for the players - they loose 0.5 all the odd rolls. approximation method

multiplication

Taking a look back at the formula above we see there a very interesting one:$$s_{(n)}'=s_{(n)}\times 50\%$$ This can be very easily modeled, as "\$\times 50\%\$" is nothing different than a coin. So just rolling an additional die with a DC11 and checking this one if the other die is voided would would give an exact remodeling of the deired DC'. Also, this simulated DC' can be extended for rolls that would be trivial otherwise, as the system does work with degrees of success: a task that would be graning the 200% class usually still gives a "no need to roll" 100% class. exact method: d20 x d2-1

However, this is not compatible with the tenet "no altering of the DCs".

switch the dice!

We are already going to mathland, so we could very easily switch to a better suitable dice: multiples of d10. These can model anything that can be expressed in %:

The success chance is 150%. So roll 1d100(=1d10*10+1d10) and add 100% to determine real success grade (or some other method to determine extra success). With modification, the new success chance is 75%, any roll of 75 or below is a success.

If the success chance is awkward to split like 25%, add a d10/10, and the 12.5% chance can be calculated perfectly.

That is better, but maybe still somewhat awkward... So, maybe a better solution:

switch the base die, so there is a half die that is available

Ok, we started to throw all the rules into the wind, so let's try to go back and get at least one rule back to the board: we don't want to need to recalculate the DC!

Ok, so we need a die, that has some chance to show this or that side with a reliable outcome (so no d7, sorry sevensideddie), and of which a die exists that has exactly twice the number of sides, as then each side has half the chance to show up. Commonly available ate the following dice: $$\text{d}4\text{, d}6\text{, d}8\text{, d}10\text{, d}12\text{, d}20$$ These can be grouped in 3 doubles, each having exactly the right relation: d4 to d8, d6 to d12, d10 to d20.

Now, we want to keep the DC. To do this we have to change a very large detail: Success is not determined by rolling over a specific number, it has to be rolling under a specific target number. The best result would be a 1 now, the worst a 20.

Now, DC5 has become "roll 5 and lower" and usually is a 50% chance on a d10. If we wnat to make rolling on this half as successful, we take the d20, on which rolling a 5 or lower is 25%.

keep the dice, change the DC and fiddle with "coins"

We know for sure, that the only possible modified DCs should be results that can be modeled as \$DC'=n\text{ or }DC'=n+0.5\$, because we only allow dividing by 2 once.

So, we could reasonably provide a table like the one we already saw (Table 2), and alter the DC accordingly. Now, instead of rolling 1d20 with DC11 (which simulates a coin) first to check if we succeed, we roll the DC against the DC'. If DC' is a Natural number, proceed as normal.

If DC' is n+0.5, we have exactly one case, where we have a problem: When the d20 shows exactly the number n. Now the most simple solution is a coin: we just roll the die again with DC11, taking the total success or failure based on that last roll. This simulates adding a 0 or .5 to the only result where this matters: the breaking point. And it reduces unnessecary clutter on the table.

Surprisingly, this will show exactly the same behavior as above, BUT it reduces the cases where extra die rolling is nessecary to exactly one case. However, again, we couldn't keep the DC the same. But at least the dice and no approximation!

conclusion

As the question doesn't disclose the exact method of the rolling system, it might not be possible to solve this problem better at this time. However, there are solutions based upon better-suited dice or other revisions of the dice system. Revising the rolling system to incorporate one idea is not pleasent, especially if some tenets can't be upheld in the process. One can only do so much:

  • Either keep the Dice and fiddle with the DC
  • Approximate for easy handling or be exact and have the need arise to roll extra (and probably different) dice
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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – mxyzplk Nov 30 '16 at 8:27
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This solution only works when the chance of success is less than 1.

Make every odd roll fail (or every even roll), regardless of modifiers. This roughly halves the chance of success when the chance of success is not very small to begin with.

That is, if you go with every odd roll being a failure: If the d20 comes up 3 or 17 or 11 or 1 (etc), the roll fails, regardless of modifiers. If the d20 comes up with 4, 20, or 2 (etc), then determine the success or failure as usual.

There are statistical anomalies if success is rare to begin with: If only results of 18, 19 or 20 would succeed, then this reduces the number of successes by one third, not half. In most cases these deviations are fairly small.

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    \$\begingroup\$ This method gives (still) an uneven distribution for even and odd DC: let's say the DC is 15 (or 30% chance to hit). so 15, 17, 19 and 16, 18, 20 make the odd and even set. Declaring all odds failures eliminates 3 numbers, leaving 3 (for 15% resulting). On DC16 though the same formula will eliminate only two-fifth (17 and 19), leaving 3 successes for 15% effective chance instead of 25% for the not-fixed roll - As proven an odd DC becomes essential the same as the next up even DC. \$\endgroup\$ – Trish Nov 24 '16 at 16:58
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    \$\begingroup\$ @Trish Yes, it only approximately halves the original chance of success, but the difference from exactly halving is not large in most cases. \$\endgroup\$ – Thanuir Nov 24 '16 at 17:30
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    \$\begingroup\$ also, does not comply to "a flat 50% fail chance regardless of the actual check is NOT what I'm looking for." \$\endgroup\$ – Trish Nov 25 '16 at 13:55
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Instead of using a d20, use a d100. (Use two d10s if you need to). To transfer the DC:

  1. Multiply the DC by 5. This is now what the DC would be if you did not have the probability reduction.
  2. Take half of the difference between the DC and 100 and add it to the DC. Effectively, move the DC twice as close to 100.

Faults:

  • Does not work with decimal DCs, although that can be resolved by combining different dice.
  • If the probability of success is more than 100%, does not make the new DC under 50.
  • Does not comply with not changing the DC.

Examples:

Bob is making a check with a success probability of 50%. You change the DC to 76, a 25% probability.

Joe is making a check with a success probability of 20%. You change the DC to 91, a 10% probability.

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Let \$DC\$ be the difficulty of the original roll and \$bonus\$ be the bonus to it.

  1. The chance of failure (after dividing by 20 or multiplying by 5 %) is \$DC - bonus - 1\$.

    Hence, the chance of success (after dividing by 20 or multiplying by 5 %) is:

    $$20 - (DC - bonus - 1) = 21 + bonus - DC$$

  2. Half of the chance of success is $${{21 + bonus - DC}\over 2} = 10 + \frac 1 2 + \frac {bonus} 2 -\frac {DC} 2$$

  3. To halve the chance of success in the sense of the examples, we can either add half the chance of success to DC or subtract it from the bonus.
  4. Since there is a possible fraction, we need to round (or change the die or add additional random input etc.). Supposing we want someone who, before this modification, had a 1/20 chance of success to still have it, then the modifier needs to be rounded down.
  5. As a further complication, if the original chance of success is negative, we don't want to halve it in this sense, since it would make success easier. The simplest solution is to do nothing in that case.

Rules

If stunned, calculate \$(21 + bonus - DC)/2\$ and round down. This is the stun penalty.

  1. If the stun penalty is zero or negative, do nothing.
  2. If the stun penalty is positive, either add it to DC or subtract it from the bonus.

The calculation is not that difficult, but players adverse to arithmetic are likely to not enjoy it. It is trivial to automate if one has any programming experience. Note that it only depends on the difference \$bonus - DC\$, or (equivalently) on the difference \$DC - bonus\$ if one uses the formula \$(21 - (DC - bonus)/2)\$. Another equivalent way would be to calculate \$(20 - DC + bonus)/2\$ and round up.

I would recommend using some simpler solution; exactly halving the chance of success is unlikely to be that important when compared with playability.

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  • \$\begingroup\$ Is this solution not correct? \$\endgroup\$ – Thanuir Nov 27 '16 at 10:38
  • \$\begingroup\$ What is "b" in the second section? I want to upvote this answer but I'm not quite clear on what I should be doing exactly. The instructions directly pertinent to me seem to be in the second section with the theory groundwork in the first (I'd swap those around), and those instructions aren't entirely clear for me for what "b" is. \$\endgroup\$ – doppelgreener Dec 7 '16 at 11:33
  • \$\begingroup\$ I've just done an edit to replace it with the full word, since I think that's more immediately comprehensible and doesn't come at any significant cost. \$\endgroup\$ – doppelgreener Dec 7 '16 at 15:36
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The simple solution:

  1. Make your roll as usual.
  2. If the roll would be a success but the number on the die appears "facing away" (relative to roller), it's a failure.... unless the roll is less than MOD-DC. In that case, it's a success no matter the facing.
  3. Continue playing.

I.e. use the rotation of the visible face as a 50% failure rate on your successful rolls, but account for the fictitious "greater-than-100%" success rate you want to preserve.

Worked examples:

  1. DC=15, MOD=+4: "normal" success rate should be 50%, as any roll 11-20 succeeds. But half of those are failures due to the number facing, leaving us with 25% successes.
  2. DC=10, MOD=+12: "normal" success rate is 100%, as it's not possible to roll low enough to fail. But you want to call it 110% success rate for... reasons. So all rolls naively succeed, then half of them are discounted due to number-facing. But MOD-DC=2, so rolls of 1 or 2 are exempted from this rule, making this scheme's success rate the 55% you want.
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  • \$\begingroup\$ 1. People have already suggested very similar things. I am NOT looking for a flat 50% chance of failure. 2. God, that relies WAY too much on SO many things. Just... no. No. \$\endgroup\$ – Papayaman1000 Nov 25 '16 at 3:44
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    \$\begingroup\$ @Papayaman1000 edited to account for your above-100% accounting. \$\endgroup\$ – nitsua60 Nov 26 '16 at 16:07
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Perhaps I am missing something, but it seems to me that this can be resolved by applying disadvantage to the rolls in question and making them "flat-footed", to use 3.5E terminology. I would word it as follows:

The target gains disadvantage on all saving throws and opportunity attacks, and cannot benefit from their Dexterity bonus to AC. If the target attempts to perform an action that would otherwise apply disadvantage to their roll, that action fails automatically.

It doesn't actually make the success rate a straight 50%, but this would fit within the existing framework of rules and it's simple enough that your players wouldn't have to spend five minutes learning what is happening. Moreover, it won't require the coins that some other (very good) answers are recommending.

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No extra roll/coin/whatever required.

Roll the die. It can only succeed if the number rolled is even

1/20 is the lowest partition of probability you can have on a d20, you wouldn't be able to divide that up further anyway. Basically you're halving their odds of success, rounded up to the nearest 20th.

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Use another roll, such as percentiles, before the check is made, to repesent something beyond their control

For example, the spell can randomly "shock" the affected creature - before doing a check, make a percentage roll - have a 50% chance that you they a shock and the check auto-fails.

Otherwise its unfair - a character who is "good" at that kind of save is being short changed.

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  • \$\begingroup\$ You seem to be making some assumptions about what the dice rolls mean that dont necessarily exist in the homebrew system being used, and emotionally appealing to those assumptions about unfairness and what a character being "good" should mean. Those aren't universal assumptions. The character may well be good, but that doesn't mean things can't hamper their success, nor that their "goodness" should dictate the effects of their dice roll or is even in question in that roll. \$\endgroup\$ – doppelgreener Nov 25 '16 at 3:08
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    \$\begingroup\$ I'm not sure why people keep duplicating answers -- a flat 50% fail chance regardless of the actual check is NOT what I'm looking for. \$\endgroup\$ – Papayaman1000 Nov 25 '16 at 3:09

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