Probability Myopia/Skill Analysis for Opposed Tests

[Decurio]

Something occurred to me while I was reading the 'oposed' (sic) skill test thread.

There have been several posters who have mentioned concerns with the scenario that arises when an opposed skill test wherein both parties fail their skill checks. (pertinent text is located on p.20, MRQ Corebook)

Mathematically, as has been pointed out elswhere, this scenario favors the person with the lower skill percentage. However, what noone seems to be taking into account (unless I've missed it) is the overall probability of this scenario arising in the first place; having both parties fail their skill checks assumes that even the more highly skilled of the two participants also failed their initial roll-which would, logically, favor the participant with the higher skill. I am more than willing to let the lesser skilled person have a fighting chance against there more highly-skilled opponent in the event that both parties fail their initial checks; part of the charm of RQ was always the chance that Jippy the Cripple would take out Arpellius the Runelord with a rock. A small chance, but luck sometimes favors the foolish...

It seems to me that certain parties are intent on "breaking" the system by myopically isolating one element of an integrated whole, and I'm beginning to wonder about the reasons why some of these criticisms are being levelled, other than MRQ isn't RQ 2...

No game I've ever played-and I've been gaming since the late 70's folks-has been devoid of quirks or design flaws. If I were to take the same scrutiny that MRQ has been subjected to, and apply that same scrutiny to other game systems, I'd place a pretty wager that I could find you islands of mathematical wonkiness.

I have found plenty to like about MRQ, and am looking forward to seeing how the overall system looks/runs/feels with the Companion, Glorantha, Magic of Glorantha, etc., in my hands.

 

[King Amenjar]

Something occurred to me while I was reading the 'oposed' (sic) skill test thread.

There have been several posters who have mentioned concerns with the scenario that arises when an opposed skill test wherein both parties fail their skill checks. (pertinent text is located on p.20, MRQ Corebook)

Mathematically, as has been pointed out elswhere, this scenario favors the person with the lower skill percentage. However, what noone seems to be taking into account (unless I've missed it) is the overall probability of this scenario arising in the first place; having both parties fail their skill checks assumes that even the more highly skilled of the two participants also failed their initial roll-which would, logically, favor the participant with the higher skill.

I know what you mean and I wondered about this at first as well. I can assure you, though, I very carefully worked it out with a spreadsheet, and in cases where you have both contestants with a low skill, so that the bulk of cases will indeed be lose-lose, this scenario does take a big enough slice of the "result cake" to outweigh the advantage the higher skill contestant has anyway. In a contest of 30% vs 15% I used, the 15% character wins 50.75% of the time and that's in all possible cases. I wasn't looking for an opportunity to say, "Har har - this is teh b0rken! Mongoose sucks!" at all.

 

[atgxtg]

A couple of points:

First off, two rolls are not used in combat oppostion. That was discussed and bluejay did up a online tool that can show why that's not a good idea.

The second thing is that the big contention with the two roll system is over the skill halviing situations.

For instance, if Sam the Sneak (at 150% sneak) is trying to slip past Groggy the Guardsman (Perception 20%), Sam's suffers from the skill halving, putting him at 75% vs Gorggy's 10%.

This gives the follwing percentages:
Attacker and defender tie 0.35
Attacker wins 77.45
Defender wins 22.20


If Sam only had a 90% Sneak he'd have these percentages:
Attacker and defender tie 0.3
Attacker wins 88.35
Defender wins 11.35


So, with everything else being equal, Sam has a lower chance of success due entirely due to his improved skill.

 

[King Amenjar]

In RQ2 didn't they used to subtract the % over 100 from the opposing skill of an opponent? So if you had Sneak 125%, anyone opposing you would get a -25 perception penalty.

 

[atgxtg]

In RQ2 didn't they used to subtract the % over 100 from the opposing skill of an opponent? So if you had Sneak 125%, anyone opposing you would get a -25 perception penalty.

I believe so, but RQ2 hadd a few other factors too.

For starters no everyone could get a 125% sneak skill. In RQ2 skillsover 100% reflected divine favor. Rune Lord could get over 100% with combat skills, Lankhaor Mhy sages could get over 100% with knowledge skills. You would have to be a Rune Lord in a thief cult (Lodril?) to get over 100% with sneak.

Secondly, in RQ2 a successful listen roll overrode the move silently roll. SO the -%>100 for listen roll was really the only way to improve you chance of sneaking.

 

[Ravage]

In RQ2 skillsover 100% reflected divine favor. Rune Lord could get over 100% with combat skills, Lankhaor Mhy sages could get over 100% with knowledge skills. You would have to be a Rune Lord in a thief cult (Lodril?) to get over 100% with sneak.

Hilariously our group used to argue long into the night every time we played about the maximum skill of DEX X 5 for Rune priests as the theory went they were too busy to gain new skills - which contravened with the tick a skill and learn thereby it on skill checks.
Still a stupid rule (the DEX X 5) but does go to show that RQ2/3 had crapola too

 

[King Amenjar]

Ah yes - RQ3 did the same for Sorcerers. It's all part of the gaming legacy that whoever is a spellcaster primarily must be a wimp in robes and a conical hat who can't fight their way out of a paper bag. RQ2 and 3 then basically abolished this rule by creating a horde of warrior cults to whom the rule wasn't applied.

 

[bluejay]

Don't forget that you can use my probability calculator for Opposed Tests and it definitely does seem that in low skill situations the higher skilled character has a minimal advantage (which would be not noticeable in a game).

For instance in skill 30% versus skill 15% situation the actual outcome is only 2% off 50/50.

 

[King Amenjar]

But your calculator includes a "tie" result, which wouldn't happen in the game: in actual play, one opponent or the other will have rolled higher or lower and win (except in the rare case when they do actually both roll the same number where this gives them the same result, which would presumably call for another roll to see who won). When you resolve the "tie" in favour of one or the other, a 30% vs 15% contest means a win for the worse contestant 50.75% of the time.

 

[bluejay]

But your calculator includes a "tie" result, which wouldn't happen in the game: in actual play, one opponent or the other will have rolled higher or lower and win (except in the rare case when they do actually both roll the same number where this gives them the same result, which would presumably call for another roll to see who won). When you resolve the "tie" in favour of one or the other, a 30% vs 15% contest means a win for the worse contestant 50.75% of the time.

As you say, a tie is certainly possible on the first roll as they could roll the same number but both succeed or both fail.

Your assumption is just that, an assumption. Currently there has been no official word on how a tie roll is resolved which is why I left the result as it stands.

Having said that, my mathematics on this disagrees with yours regarding the probabilities involved. Not sure who is right on this but you're free to check my working in my script, just 'view source' on the calculator page. I've tried to comment everything as clearly as possible.

 

[King Amenjar]

Okay, here's how I calculated it. I'd love it if someone pointed out that I was wrong and that the system isn't broken at all, but...

Character B has skill x. Character A is better than Character B and has skill x+y. (Here x and y are both fractions less than 1 although in actual RQ they'd be %s - same thing, though)
We've got four possible scenarios, ignoring the rare possible tie:
Case 1 - Both characters succeed: probability of this = (x+y)*x
Case 2 - Both characters fail: probability of this = (1-(x+y))*(1-x)
Case 3 - A succeeds, B fails: probability of this = (x+y) * (1-x)
Case 4 - B succeeds, A fails: probability of this = x * (1-(x+y))

Now, in cases 3 and 4 we have an obvious winner straight away. In case 1, the higher roll wins. In case 2, the lower roll wins. The formulae below are the bits I'm less sure about...
The probability of A winning in a Case 1 is (0.5+y)*((x+y)*x), since if A's roll falls within the range which is greater than x but less than x+y, A must have rolled higher. If not, it's obviously 50/50.
The probability of B winning in a Case 2 is (0.5+y)*(1-(x+y))*(1-x), since if B's roll falls within the range which is greater than x but less than x+y, B must have rolled lower (because A would succeed with such a roll). If not, again, it's obviously 50/50.

Therefore, A's chance of winning increases the better both contestants are, because as well as his chance to win outright, he gains from Case 1 scenarios. B's chance of winning increases the worse both contesants are, because he gains from Case 2. We run into trouble if Case 2 scenarios occur so often (because both contestants suck, but one more than the other) that (0.5+y)*(1-(x+y))*(1-x) added to the other probabilities of B winning are greater than 0.5. At this point, the worse contestant has the advantage.

 

[King Amenjar]

The more I think about it, the more I feel I must have gone wrong with those last two...
EDIT... hang on, it's not 0.5+y - it's 0.5+y^2... because they must BOTH roll within those ranges. Aha...

 

[bluejay]

Let's look at Case 1 again.

Let's say that if A and B both succeed and roll the same they tie. The chance of this happening is (x/100) * (x/100). Thankfully we can ignore this result.

If A rolls above x but less than x+y, A wins (because B cannot roll higher than A and still succeed). The chance of this happening is y/100. It is irrelevant what B rolls in this instance as they simply cannot win.

If both A and B roll x or less (but do not roll the same number) then A must roll higher than B.

What is the chance of this happening? Well if A rolls n (which is less than x), he wins if B rolls any number less than n. The chance of this is (n-1)/100. However you have to incorporate the chance of A rolling the number n in the first place (which is simply 1%).

So we now need to sum all of these chances which I guess is the sum of

(n-1)/100 as n ranges from 1 to x. It would be nice to use a proper Sigma sign here...

This is the way my algorithm works.

Make any sense?

 

[atgxtg]

Hilariously our group used to argue long into the night every time we played about the maximum skill of DEX X 5 for Rune priests as the theory went they were too busy to gain new skills - which contravened with the tick a skill and learn thereby it on skill checks.
Still a stupid rule (the DEX X 5) but does go to show that RQ2/3 had crapola too

The DEX5% rule got dropped in one of the supplmenets too. Essentially Chaosium explained that it was an artifical contraint to repsenent the reduced time such people would have to devote to weapon skills.

IMO giving them an peanlty on impreovemnt rolls or to the d6 roll would have accomplished the same thing without placing a cap.

 

[King Amenjar]

Let's look at Case 1 again.

Let's say that if A and B both succeed and roll the same they tie. The chance of this happening is (x/100) * (x/100). Thankfully we can ignore this result.

A very common mistake made in probability: actually, that's the chance of them both rolling a particular number e.g. the chance of them both rolling 14s. The chance of them rolling the same is a straight 1%, but of course, that's not quite what we're after - we only want the same number within the spread where this gives them both the same result - e.g. if it's 30% v 15% we can exclude results where either of them rolls 16-30, because that's a fail for one a success for the other. The chance in that case would be 0.85% (1-y*0.01). Anyway, this isn't really important, because a tie will have to be resolved by GM fiat, such as making the players do it again, or just rolling a d6, highest wins etc.

If A rolls above x but less than x+y, A wins (because B cannot roll higher than A and still succeed). The chance of this happening is y/100. It is irrelevant what B rolls in this instance as they simply cannot win.

If both A and B roll x or less (but do not roll the same number) then A must roll higher than B.

What is the chance of this happening? Well if A rolls n (which is less than x), he wins if B rolls any number less than n. The chance of this is (n-1)/100. However you have to incorporate the chance of A rolling the number n in the first place (which is simply 1%).

Well, given that it's outside the spread where A must win, it's simply 50/50. Either player has an even chance of rolling higher.

So we now need to sum all of these chances which I guess is the sum of

(n-1)/100 as n ranges from 1 to x. It would be nice to use a proper Sigma sign here...

This is the way my algorithm works.

Make any sense?

I think in places you're getting too complicated, but sort of. My original formula is wrong, so the system is less "broken" but whether it's still possible for the worse to be better I'll have to look at again.

 

[nennafir]

If both A and B roll x or less (but do not roll the same number) then A must roll higher than B.

What is the chance of this happening? Well if A rolls n (which is less than x), he wins if B rolls any number less than n. The chance of this is (n-1)/100. However you have to incorporate the chance of A rolling the number n in the first place (which is simply 1%).

Well, given that it's outside the spread where A must win, it's simply 50/50. Either player has an even chance of rolling higher.

So we now need to sum all of these chances which I guess is the sum of

(n-1)/100 as n ranges from 1 to x. It would be nice to use a proper Sigma sign here...

This is the way my algorithm works.

Make any sense?

I think in places you're getting too complicated, but sort of. My original formula is wrong, so the system is less "broken" but whether it's still possible for the worse to be better I'll have to look at again.

No, bluejay has it right here. It's a summation. Of course, as we all know, summation i=1 to n of i is n(n+1)/2.

-nennafir

 

[bluejay]

You're right about the ties KA, I had meant to write (x * 1/100 * 1/100). Don't forget I wasn't referring to all ties, just ties within the success range.

Ties within both success or failure would rightly be (100-y) * 1/100 * 1/100.

 

[King Amenjar]

Was right about being wrong, but wrong again about what was right... nnh... the probability that both succeed and A beats B is the chance both succeed times 0.5 (x/(x+y)) since this represents the range where both have rolled in the same range to give a 50/50 chance. Probability both fail and B beats A is chance both fail times 0.5 (1-(x+y))/(1-x))
Reworking the spreadsheet, it doesn't look as if the worse contestant can ever have the better chance of winning. Phew. Not broken after all (although I'm still not keen on this system...) :wink:

 

[bluejay]

Was right about being wrong, but wrong again about what was right... nnh... the probability that both succeed and A beats B is the chance both succeed times 0.5 (x/(x+y)) since this represents the range where both have rolled in the same range to give a 50/50 chance. Probability both fail and B beats A is chance both fail times 0.5 (1-(x+y))/(1-x))
Reworking the spreadsheet, it doesn't look as if the worse contestant can ever have the better chance of winning. Phew. Not broken after all (although I'm still not keen on this system...) :wink:

Sorry, 0.5 (x/(x+y)) doesn't indicate where both have rolled within the same range.

For both to roll in the same range they both have to roll x or less. If A rolls more than x but less than (x+y) then B cannot win at all. It's only when they both roll x or less that this contest about who rolls higher comes into play.

So ignoring ties, the chance that A wins might be expressed as: -

Chance A rolls higher than x but less than (x+y) = y%

plus

Chance A and B both roll x or less but A rolls higher 0.5 * (x/100) * (x/100)

This doesn't take into account situations regarding automatic success, automatic failure or the High Skill rule (like my calc) but is sufficient for this example.

Final result is y + (x ^ 2) /200 as a percentage.

Which is similar to my result from the summation method which works out as

y + x(x+1) /200 as a percentage (as my version removes the tie chance).

So really we have the same idea overall!

 

[Melkor]

So the end result is that with all probabilities figured, it works out fine ?
Awesome.

Thanks Decurio for putting the thought into that and bringing up the issue!

Thanks Blujay and King for working up the equations.