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Post by jonas on Aug 29, 2013 3:06:59 GMT -8
Next week, I'm going to run my first Savage Worlds game and trying to learn the rules right now.
I think I heard it mentioned on the podcast, but since the dice explode, does it mean that D4s are actually better than D6s?
I'm not a math guy (I'm more of a 'This isn't my glasses, it's a broken lightbulb again'-guy), so can somebody explain if this is true or not?
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maxinstuff
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Post by maxinstuff on Aug 29, 2013 3:55:17 GMT -8
This is a common misconception about SW. Statistically yes - it is true that a d4 explodes more often. However the shape of the curve is a bit different, and you need to keep the target number in mind (4). The statistical frequency of simply hitting or exceeding 4 is actually better on a d6. Also note that the wild die significantly diminishes the importance of the difference between a d4 and a d6 anyway so it isn't actually that important. 1d4 will explode 25% of the time, and a d6 will only explode 17%. To keep it simple I'll leave out the complicated math the explosiion rules create, because you will soon see it isn't relevant - remember all you want to do it hit 4 or more. It doesn't matter what the exponential impact of that 25% vs 17% explosion chance is because if your dice explodes you've hit the target number - on ANY die. 1d4 = 25% chance of success (reaching or exceeding 4) 1d6 = 50% chance of success (reaching or exceeding 4) So really a D6 is TWICE as good at hitting the systemetized target of 4. You can have your die explosions, give me 'old reliable'
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Post by jazzisblues on Aug 29, 2013 10:49:44 GMT -8
In aggregate when you include both the probability of succeeding and the probability of exploding the balance point is a d8. Thus statistically a d8 is the best overall die to roll if you're concerned about exploding. Of course bear in mind that the TN is typically 4 so I personally place a little more value on success rather than exploding.
Your mileage may of course vary.
JiB
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maxinstuff
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Post by maxinstuff on Aug 29, 2013 15:03:43 GMT -8
In aggregate when you include both the probability of succeeding and the probability of exploding the balance point is a d8. Thus statistically a d8 is the best overall die to roll if you're concerned about exploding. Of course bear in mind that the TN is typically 4 so I personally place a little more value on success rather than exploding. Your mileage may of course vary. JiB That is only true if you are calculating the 'average roll' on each die. This is the mistake that people make. When you include explosions the smaller dice curve upwards with every explosion you count, and because they are more likely you eventually get higher average numbers rolled. The problem is that the average roll has nothing to do with the frequency of a simple success. In the d8 example - without explosions the 'average roll' is 4.5, and this goes up with every explosion chance you count. Thus a d8 is the die with the highest chance of exploding that also succeeds on average without them. HOWEVER, when you roll that d8, the explosion chance does not effect your chance of success in game terms. Your chance of rolling 4 or better is still 62.5%. I suspect that the target numbers required to make smaller dice 'better' are so high as to be meaningless in relation to the game. If I get time I will work it out.
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Post by greatwyrm on Aug 29, 2013 16:17:01 GMT -8
Even on a bigger target number, you're always better off rolling bigger dice. Each successive die roll means you can explode, but just keeps making it harder to get the proper results.
Let's look at a target number of 20. On the dice you use in SW, any would have to explode at least once to hit that.
That's less than 0.1% on a d4, but 3.47% on a d12. Even the middle ground of the d8 still has less than a 1% chance.
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maxinstuff
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Post by maxinstuff on Aug 29, 2013 17:45:43 GMT -8
OK - so I have tabled likelyhood of success for d4 - d12 for TN's up to 100, and at no point is ANY smaller die better than a bigger die in terms of likelyhood of success. At all. In fact the differences get farther apart as the TN rises. You are correct greatwyrm - *tips hat*For anyone who is interested in my math I summarised the success chance based on: x = faces on the die y = the target number Success chance = (1/x)^(y/x) This 'chops off' the last roll and instead summarises it as a fraction of an explosion chance. This means every xth result is exactly correct and there is a smooth line drawn between them. The margin of error should go in favour of ther smaller dice as result.
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Post by greatwyrm on Aug 29, 2013 18:22:39 GMT -8
The professors tried to tell me all the statistics classes were for optimizing assembly lines and using failure rates to estimate production costs and crap like that. I knew the truth. I was getting six credit hours in gambling and rpg odds.
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Max Damage
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Post by Max Damage on Aug 30, 2013 15:17:47 GMT -8
Here is a discussion of the Savage Statistics: Savage Worlds Results Charts
I personally like this post because it provides graphs that allow you to see some of the anomalies of the Savage dies when adding in the exploding die and wild die. Anomalies such as: you have a slightly better chance of beating a target number with a on increment lower die than the die with the same number of sides as the target number. For instance, if your target number is 6. You have a slightly higher probability of meeting or exceeding a 6 with a d4 and the d6 wild die than you do with a d6 and a d6 wild die. Because of exploding, the d4 has a an 18.8% chance of reaching a 6 or above, while the d6 only has a 16.7% chance of reaching a 6 or above. The percentage chance for the d6 wild die is the same for both cases. The final result is: TN 6 d4 + d6 wild die = 32.3% chance of success TN 6 d6 + d6 wild die = 30.6% chance of success This is the only situation where it is actually better to have a d4 versus a d6. This same inversion happens in all of the following cases: TN 8 d6 + d6 wild die (27.0%) vs TN 8 d8 + d6 wild die (25.3%) TN 10 d8 + d6 wild die (18.4%) vs TN 10 d10 + d6 wild die (17.5%) TN 12 d10 + d6 wild die (11.5%) vs TN 12 d12 + d6 wild die (10.9%) While this is a curiosity of the system, unless you know you will always be trying to achieve a TN 8, then the higher die is superior. But for you true min/max'ers the real key to success is edges. A d4+2 is superior to a d6 for TN 4 through 9 and TN 11 to 13. It is superior to d8 for TN 4 and 8 and tied at TN 5,7, and 9. If you expect typical TNs a d4 with an edge that gives you a +2 to your roll is superior to a d8, and it costs the same and it doesn't require you to have as high of an associated stat, which also saves you points. Of course, this is why most edges that give you a +2 to a skill require you to purchase a minimum amount, but even with the minimum you are better-off taking the +2 than trying to buy up the die of the skill any further.
MD
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maxinstuff
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Post by maxinstuff on Aug 30, 2013 15:48:02 GMT -8
I had the effect of the fractional explosion chance ass backwards. The errors are going in favour of the bigger dice.... I'll have to see if I can fix it now.
Because its bothering me.
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Post by greatwyrm on Aug 30, 2013 16:06:42 GMT -8
This is the only situation where it is actually better to have a d4 versus a d6. Weird, but it does check out.
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Post by shadrack on Aug 30, 2013 16:14:23 GMT -8
Possibly because it's hard to land on 6 when you're rolling 2 d6's and they explode.
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Post by greatwyrm on Aug 30, 2013 16:51:42 GMT -8
No, the percentages check out at hitting at least the target number, not getting it exactly.
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jpk
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Post by jpk on Sept 7, 2013 11:56:58 GMT -8
For what it's worth, the general formula for hitting a Target Number of TN with a Die Type of DT is: DT - remainder((TN-1)/DT) DT^(quotient((TN-1)/DT)+1) So, for example, if you're looking for a TN 6 on a DT d6, the formula would give you 6 - remainder((6-1)/6) 6^(quotient((6-1)/6)+1) 6 - remainder(5/6) 6^(quotient(5/6)+1) 6 - 5 6^(0+1) 1 6^(1) 1 6 Now, for that example, the math seems to be an extreme waste of time to get to "one out of six, you pinhead." However, if you want to know what the likelihood of actually hitting a target number (TN) of 55 on a die type (DT) of d4 is, you get: 4 - remainder((55-1)/4) 4^(quotient((55-1)/4)+1) 4 - remainder(54/4) 4^(quotient(54/4)+1) 4 - 2 4^(13+1) 2 4^(14) 2 268,435,456 1 134,217,728 If you're sort of probability minded, you might already realize that only gets you the one die. To find you chance of success with either of two dice, you'd need to find your chance of failure with both and subtract that from 1 (for fractional folks) or 100 (for percent people). For instance, your chance of failing a TN 4 on a d6 (the Wild Die) is 1/2. Your chance of failing a TN 4 on a d4 is 3/4. Your chance of failing on both dice at the same time is 1/2 times 3/4, for 3/8. So, your chance of success on a d4 with a d6 Wild Die is 1-(3/8), which is 5/8. So, there you go. Math out! ;-)
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maxinstuff
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Post by maxinstuff on Sept 8, 2013 20:04:00 GMT -8
Yeah I got as far as incorporating the remainder and inverting probabilities but hit the same snag as previously, which is that every dXth result is the same as the next TN up (if you need a 1 you may as well not roll) But with different numbers as inputs (giving a different answer).
To incorporate this requires nested IF functions in my equation which I frankly can't be arsed to get working right now.
I am interested in looking at the probability curve of the various dice though to see if they DO converge/diverge though so I am sure I'll come back to it. Eventually.
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jpk
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Post by jpk on Sept 12, 2013 12:38:07 GMT -8
Well, if you're trying to write a general "find the probably of hitting target number TN with die type DT and Wild Die of type WD" sort of thing, you'd just need to do this:
1-((1-(Chance of hitting TN with DT))*(1-(Chance of hitting TN with WD)))
Use the formula above with whatever your particular script supplies for getting the quotient and the remainder. Also note that the chance referred to above is a 0-1 probability. If you want to work in percentages, you'd have to change the ones to one hundreds and alter the formula to generate percentages (or multiply the final answer of the probability version by 100).
But any way you slice it, it's simply true that the chance of hitting a target number of a specific die type (from 6-12) is about 1% higher with a die type one lower (from 4-10). It's a one-spot sort of thing and hasn't really impacted play that I've seen (barring obsessive math folks like, well, possibly me).
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