Maths guys - hey! (Probabilities)

[Morrus]

I've co-opted it for death saves and for suffocation. It works great there!

 

[Morrus]

Sorry! Forgot the extra shot on all except the "all sixes" and recursive cases. Results Should be:

Dice    Shots
0    1
1    6
2    8.727272727
3    10.55544456
4    11.92669625
5    13.02366151
6    13.9377967
7    14.7213416
8    15.40694348
9    16.01636737
10    16.56484886

So isn't the average number of times you need to roll 1d6 till you get a 6 just 3.5, not 6? I mean, I know each roll is a 1-in-6 chance, but surely the chance of it taking all 6 rolls (avoiding 6 five times in a row) is low?

Edit - a did an intrawebs search and found this, which seems to say it's just under 4 - which seems to match the 3.5. I don't know whether it's good reasoning though:

A seemingly simple but tricky question. The odds of getting a 6 are 1 in 6 but this is NOT equal to the average number of rolls needed to get a six. To find that, let's look at the odds of NOT getting a six which are 5/6. In one role the odds are:

(5/6) = 5/6 = 83.3%

For 2 rolls:

(5/6) x (5/6) = 25/36 = 69.4%

3 rolls:

(5/6) x (5/6)x (5/6) = 125 / 216 = 57.9%

4 rolls:

(5/6) x (5/6)x (5/6)x (5/6) = 625/1296 = 48.2%

So after 4 rolls you are slightly more than half likely to have rolled a six. Thus half the time you will roll a six in 4 or fewer rolls and half the time it will take more rolls which mean the average is 4 rolls.

I did some experimentation using a 3-dice pool. Admittedly, not a big statistical sample - I only did it 10 times. But not once did it hit the predicted 11 turns to deplete. The closest it came was once when it took 9 turns. The average number of turns it took over the 10 attempts turned out to be 5 or 6.

Might have been a fluke, of course, with only 10 attempts, but I don't know how to do those massive simulations!

That said, this document totally agrees with your figures.

 

[Morrus]

The mechanic doesn't work well for its stated purpose, as Morrus said, but there's a certain elegance to it, and it seems to me that there really ought to be somewhere it could apply.

It occurred to me that it might be useful for modeling combat fatigue. In a hand to hand fight, fatigue is a really big deal, but I've never seen a system that modeled it both well and efficiently.

Start with a common "buckets of dice" dice pool system. Characters have a combat rating (say 20) and in an attack that number of d6 are rolled and all 5's and 6's are counted as a "success." Depending on the number of successes, maneuver choices, and defender's actions, things happen.

Now add the fatigue mechanic. All dice showing a 1 are removed from the dice pool and set aside to model fatigue. They become the "fatigue pool." Obviously, the character becomes less effective over time. If a character chooses to not attack in a round (but still defend against any incoming attack) a small number of dice are returned from the fatigue pool to the attack pool. If the character is able to completely rest for a round, a larger number of dice are returned.

Seems like a slick system, but I'd need to actually play test it to see if it's good.

A possible further refinement: the fatigue pool can never be larger than the attack pool. If moving dice from the attack pool to the fatigue pool would result in this happening, the dice are instead set aside, modeling serious fatigue. A long rest is required to restore them.

A possible simplification: rather than using numbered d6's, use special colored dice. Two faces of each are green for success, one is red for fatigue, and the others are white for "nothing."

Thoughts? Anything out there like this now?

As far as I can make out, that makes having a large attack dice pool a disadvantage. The ideal dice pool is 1. Less chance of getting fatigued, and your fatigue pool can't go above 1.

 

[Mishihari Lord]

As far as I can make out, that makes having a large attack dice pool a disadvantage. The ideal dice pool is 1. Less chance of getting fatigued, and your fatigue pool can't go above 1.

Within the common dice pool system I described, I don't see how having an attack pool of 1 could be an advantage over having, say, 20. More successes is always better. Sure the 20 gets worse over time, but it's still always better than the 1. Unless there's some subtlety to your system I don't know, not having seen it.

 

[Morrus]

Within the common dice pool system I described, I don't see how having an attack pool of 1 could be an advantage over having, say, 20. More successes is always better. Sure the 20 gets worse over time, but it's still always better than the 1. Unless there's some subtlety to your system I don't know - I haven't seen it.

I was talking about your fatigue system, not mine. I may have misunderstood it, but if a 1 is a fatigue, having a 20-dice attack pool will earn you 3.5 fatigues per turn on average, while having a 1-dice attack pool will earn you 0.6 fatigues per turn on average. Also, with an attack pool of 20 you could have 19 fatigues, but with an attack pool of 1 you'll never have more than 1 fatigue. Though I'm not clear on what the mounting fatigue dice do? Are they deducted from the attack dice pool?

I think it's probably a great little mechanic, and I'm interested in it - but I don't think I'm fulling grokking it yet!

 

[Mishihari Lord]

Yeah, you just set them aside, they don't do anything else. Here's an example:

Round 1 - attack - attack pool is 20, result is 8 successes and 3 fatigue.
Round 2 - attack - attack pool is now 17, result is 6 successes and 2 fatigue.
Round 3 - attack - attack pool is now 15, get 5 successes and 3 fatigue.
Round 4 - attack pool is now 12, so the player chooses to defend only, recovers 4 fatigue.
Round 5 - attack - attack pool is now 16 ...

So your combat power reduces slowly over time unless you pace yourself. Obviously, such a mechanic isn't necessary for combat - most do without, but for sim folks like myself who like modeling reality it's nice. It also adds another meaningful choice to combat, which is nice for those who like such things. It can also be expanded into the system, frex, a power attack that let's you roll 4 extra dice that round but automatically costs 1 fatigue.

 

[Balesir]

So isn't the average number of times you need to roll 1d6 till you get a 6 just 3.5, not 6? I mean, I know each roll is a 1-in-6 chance, but surely the chance of it taking all 6 rolls (avoiding 6 five times in a row) is low?

Well, there's actually three types of "average", and this seems to be a confusion between them.

The "average" I used (assumed) was the mean - and the mean for a single d6 is definitely 6. That is, if you do the "experiment" (roll 'til you get a 6 and count the rolls) umpteen times, the total nomber of rolls divided by the total number of experiments will be 6. But the distribution is very skewed; most of the experiments will have 4 or fewer rolls - but the odd one will have an infeasibly big number (because you can go on rolling 1-5 forever, in theory).

The median is more what you are thinking of in your post. The median is the number that 50% of the results fall above and 50% of the results fall below; if the distribution is skewed (and this one definitely is) then the mean and the median will be different. In this case it looks like the median is between 3 and 4 (and, since you can't get "half-a-roll" results, it's not really relevant where, precisely, it falls between 3 and 4 - calling it 3.5 is as good an answer as any). As a complete aside (bit of Christmas trivia), there will in principle be pressure for wealth redistribution in a democratic society only as long as the mean income is greater than the median income, since the median vote always wins...

The final type of average is the mode, which is the specific result which occurs more times than any other. My guess is that for the dice scheme you proposed this would be 3, but I haven't done any actual calculation to get that.

 

[Morrus]

Well, there's actually three types of "average", and this seems to be a confusion between them.

The "average" I used (assumed) was the mean - and the mean for a single d6 is definitely 6. That is, if you do the "experiment" (roll 'til you get a 6 and count the rolls) umpteen times, the total nomber of rolls divided by the total number of experiments will be 6. But the distribution is very skewed; most of the experiments will have 4 or fewer rolls - but the odd one will have an infeasibly big number (because you can go on rolling 1-5 forever, in theory).

The median is more what you are thinking of in your post. The median is the number that 50% of the results fall above and 50% of the results fall below; if the distribution is skewed (and this one definitely is) then the mean and the median will be different. In this case it looks like the median is between 3 and 4 (and, since you can't get "half-a-roll" results, it's not really relevant where, precisely, it falls between 3 and 4 - calling it 3.5 is as good an answer as any). As a complete aside (bit of Christmas trivia), there will in principle be pressure for wealth redistribution in a democratic society only as long as the mean income is greater than the median income, since the median vote always wins...

The final type of average is the mode, which is the specific result which occurs more times than any other. My guess is that for the dice scheme you proposed this would be 3, but I haven't done any actual calculation to get that.

Yeah, in terms of a game, the useful figure is stuff like "how many rounds do we expect a character counting down from 5d6 to death to survive?"

So would it be a reasonable approximation to simply halve all the figures above? It doesn't need to be spot on, but within a point would be good.

 

[Balesir]

Yeah, in terms of a game, the useful figure is stuff like "how many rounds do we expect a character counting down from 5d6 to death to survive?"

So would it be a reasonable approximation to simply halve all the figures above? It doesn't need to be spot on, but within a point would be good.

Don't think so. I'm giving up for the festivities, now (Merry Christmas, by the way), but the first three medians I get are 3.x, 6.x and 9.x...

 

[Nagol]

Yeah, in terms of a game, the useful figure is stuff like "how many rounds do we expect a character counting down from 5d6 to death to survive?"

So would it be a reasonable approximation to simply halve all the figures above? It doesn't need to be spot on, but within a point would be good.

This is why I plot multiple points on the curve. I find it hard to visualise the population probabilities otherwise.

For 5d6:
8% will deplete in 5 rounds or less (closest I could get to 5%).
26% will deplete in 8 rounds or less.
34% will deplete in 9 rounds or less.
55% will deplete in 12 rounds or less.
67% will deplete in 14 rounds or less.
75% will deplete in 16 rounds or less.
91% will deplete in 22 rounds or less.
95% will deplete in 26 rounds or less.

So about half the characters will survive to the 12th round.