Diminishing returns? The more dice you throw, the more likely you'll get a 6 and end up dropping at least one die?
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Maths guys - hey! (Probabilities)
- Thread starter Morrus
- Start date
Balesir
Adventurer
Hmm, I should list assumptions:How did you calculate it? My brain's a bit tired, but I got as far as:
g(x) = expected number of rolls starting with xd6
g(0) = 0
Then for n >= 1, solve for g:
g
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My explanation: g(0) = 0 because you've got no dice.
With n dice, then you could end up with between 0 and n dice next time.
So g
where h(i,n) is the number of ways to get exactly n-i sixes on n dice.
When [MENTION=1]Morrus[/MENTION] initially explained he said if you have no dice left you "just took your last shot". I took this to mean that you first resolve the shot, then check ammo; this has the nice effects that (i) you always get at least one shot after you reload and (ii) a single shot weapon is just a special case of the general (i.e. it has zero ammo dice to begin with).
Given this, the number of shots for zero dice is 1. For one die it's pretty obviously 6 (although this actually comes out of the general equation, too). For two dice it's the probability of getting all sixes (=1/36) times the number of shots for zero dice (1), plus the probability of getting exactly one six (=10/36 = {2C1}*5/6^2) times the number of shots for one die (6) plus one (this was the bit I missed first time around) for the shot you already took, plus the probability of getting zero sixes times one plus the number of shots for two dice (in other words, the number of shots you are trying to calculate). This can be written as follows:
S(2) = P(2).S(0) + P(2-1).(1+S(1)) + P(2-2).(1+S(2))
Where S(2) = shots with two dice, S(1) = shots with one die (= 6), S(0) = shots with zero dice (= 1), and
P(2) = probability of getting 2 sixes, P(2-1) = probability of getting exactly one six and P(2-2) = probability of getting no sixes.
This can be written:
S(2).(1 - P(2-2)) = P(2).S(0) + P(2-1).(1+S(1)) + P(2-2)
or: S(2) = [P(2).S(0) + P(2-1).(1+S(1)) + P(2-2)]/[1 - P(2-2)]
...and, if you already know the P and S numbers for the numbers of dice lower than the current case this can be solved quite easily.
The same basic principle can be generalised for any number of dice >2.
It occurs to me, of course, that this doesn't actually save the player any work. In fact it increases it. I've come to the conclusion that this is a Bad Rule (I can say that because I invented it - though now somebody's bound to point out that it's been done before somewhere).
Basically, you're rolling an extra bunch of dice every time you take a shot to see if you have to remove some dice from a pool. If you're doing that, you might as well just have a pool of dice (or counters) equal to your ammo and remove one each time you fire. You still have to remove a dice/token/whatever - the dice roll is extra work, and adds the confusion that you can reload a weapon and still only get a couple of shots out of it.
Thanks for the help guys! It was really interesting to see how the maths didn't work out how I expected them to.
Basically, you're rolling an extra bunch of dice every time you take a shot to see if you have to remove some dice from a pool. If you're doing that, you might as well just have a pool of dice (or counters) equal to your ammo and remove one each time you fire. You still have to remove a dice/token/whatever - the dice roll is extra work, and adds the confusion that you can reload a weapon and still only get a couple of shots out of it.
Thanks for the help guys! It was really interesting to see how the maths didn't work out how I expected them to.
I don't. I've decided it's a rubbish rule! I wanted to reduce record-keeping, but this doesn't really accomplish that well.
Balesir
Adventurer
Yeah - it's as much work as tracking ammo and less "realistic", so...
I can see some point in random ammo if it's much simpler than tracking. Something like "roll a 1 on the d20 to hit roll and your quiver (that was bought as 20 arrows) is empty". That would just generate arrow depletion as a by-product of the to hit roll (that you're making anyway) instead of having extra record keeping.
I can see some point in random ammo if it's much simpler than tracking. Something like "roll a 1 on the d20 to hit roll and your quiver (that was bought as 20 arrows) is empty". That would just generate arrow depletion as a by-product of the to hit roll (that you're making anyway) instead of having extra record keeping.
Viking Bastard
Adventurer
I encountered a rule on some OSR blog a while back that I liked (but didn't have any reason to implement) that tackles this.
Your ammo is represented by a single die (d4-d12) that you roll alongside your to-hit roll.
The more ammo you have, the bigger the die.
When you roll a '1', you step down the die.
Once you roll a '1' on a d4, that was your last shot.
Your ammo is represented by a single die (d4-d12) that you roll alongside your to-hit roll.
The more ammo you have, the bigger the die.
When you roll a '1', you step down the die.
Once you roll a '1' on a d4, that was your last shot.
Mishihari Lord
First Post
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?
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?
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