Spell Selection Odds

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Spell Selection Odds

#1 Post by SpellArcher »

Does anyone know what the rough odds are of getting a particular spell if you roll three selection dice at a Lore? Obviously you have a straight 50% of rolling it but I'm unsure how to factor in the chances of rolling a double here.
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Re: Spell Selection Odds

#2 Post by jwg20 »

Works out to be a 72.22% chance (if I assume you pick the spell in question in the event of any double).

Of course, if its the signature spell, its 100% :D .
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Re: Spell Selection Odds

#3 Post by Andrew_uk »

Sure well if you roll the same spell multiple times then you can swap it for one of your choice. So the chance of a wizard of each of the following levels gaining a specific spell is as follows

1 0.166666667
2 0.444444444
3 0.722222222
4 0.907407407
5 0.984567901

Edit : so yeh I agree with jwg
Second Edit: Included the probability if you are using a silver wand (so effectively level 5), just proves how unlucky AiE is
Last edited by Andrew_uk on Sat May 21, 2011 4:36 pm, edited 1 time in total.
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Re: Spell Selection Odds

#4 Post by SpellArcher »

Thanks guys that's very helpful.

Though I'm still not sure whether Silver Wand or a Scroll is better...
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Re: Spell Selection Odds

#5 Post by AiEthimar »

It's funny, I have been reading all of the dice calculations that have been flying around the past couple of days, and I s@#$ you not, at my last game, I was using shadow and rolled a 1,2,3,4,5... Perfect straight; My lvl4 had the Wand so thus the five spells. God how I hate probabilities 8)

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Re: Spell Selection Odds

#6 Post by SpellArcher »

I guess the thing is to write your list and pick your Lore so that you can compensate when you roll badly for spells.

Easier said than done though.
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Re: Spell Selection Odds

#7 Post by AiEthimar »

I understand how important it is to have a list that is assisted by magic, not dependent on it, I was merely remarking at the irony of my situation, where everyone is discussing how likely it is to get certain results and I got what I got. Strange is all.

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Re: Spell Selection Odds

#8 Post by jwg20 »

Haha, ouch. Yeah, the outcome of that is 1.54% (so 3x in every 200 attempts).

Though, I guess it only has to happen once to ruin your day...
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Re: Spell Selection Odds

#9 Post by Andrew_uk »

also can I just raise a point which SA mentioned? And this isn't to have a go at you or make you feel stupid, actually the point I am going to make is a misconception most people seem to hold and I felt raising it might help more people understand probabilities better...
SpellArcher wrote:Does anyone know what the rough odds are of getting a particular spell if you roll three selection dice at a Lore? Obviously you have a straight 50% of rolling it but I'm unsure how to factor in the chances of rolling a double here.
You actually don't have a 50% chance of rolling it, let me explain... The total of all mutually exclusive and exhaustive probabilities have to add up to 1 (mutually exclusive means there is no chance of one die showing a 3 and a 4 at the same time, exhuastive just means I've considered everything).

Now the probability of you rolling the number you wanted (lets say it was 6) when you roll 3 dice is actually the probability of you rolling 666 + the probability of you rolling two sixes + the probability of you rolling one six and two others... now thats a lot to consider and rather awkward so it's easier to do 1-p(no sixes at all)

This works because you're first working out the likelihood that you roll no sixes and then doing 1 minus that

So this is 1-(5/6)^3 = 0.4213 and not 50% as you initially assumed

It's just like with rolling six dice there is no guarantee that you will definitely roll a six, yet the chance of you rolling a six on any one of those dice is 1/6 so it would seem intuitively that it would be... there are a lot of misconceptions surrounding probabilities but providing you always try to be exhaustive and mutually exclusive it should be easier
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Re: Spell Selection Odds

#10 Post by Curu Olannon »

The misconception you're talking about Andrew is simply the difference between probability and averages.

On average, 3 dice will score at least on 6 50% of the time. The probability though, is less, as you put it.

Likewise, 10 Swordmasters will kill roughly 15 Skavenslaves on average. The probability however that they'll kill at least one Slave is <1 (though it is very, very close to 1).

In this case, averages make more sense than probability. On average, you will get a least one 6 50% of the time when you roll 3 dice.
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Re: Spell Selection Odds

#11 Post by SpellArcher »

Andrew_uk wrote: this isn't to have a go at you or make you feel stupid,
I do a good enough job at that myself!

:)

Wow, talk about mind-expanding. I think I'll just take the 72 and run with it!
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Re: Spell Selection Odds

#12 Post by hewhorocks »

Andrew_uk wrote:also can I just raise a point which SA mentioned? And this isn't to have a go at you or make you feel stupid, actually the point I am going to make is a misconception most people seem to hold and I felt raising it might help more people understand probabilities better...

You actually don't have a 50% chance of rolling it, let me explain... The total of all mutually exclusive and exhaustive probabilities have to add up to 1 (mutually exclusive means there is no chance of one die showing a 3 and a 4 at the same time, exhuastive just means I've considered everything).

Now the probability of you rolling the number you wanted (lets say it was 6) when you roll 3 dice is actually the probability of you rolling 666 + the probability of you rolling two sixes + the probability of you rolling one six and two others... now thats a lot to consider and rather awkward so it's easier to do 1-p(no sixes at all)

This works because you're first working out the likelihood that you roll no sixes and then doing 1 minus that

So this is 1-(5/6)^3 = 0.4213 and not 50% as you initially assumed

It's just like with rolling six dice there is no guarantee that you will definitely roll a six, yet the chance of you rolling a six on any one of those dice is 1/6 so it would seem intuitively that it would be... there are a lot of misconceptions surrounding probabilities but providing you always try to be exhaustive and mutually exclusive it should be easier
Actually you need to add in the chance of rolling any doubles as they let you select. Well doubles that dont include double 6es as they have been accounted for or doubles anything else with a third die as a 6 as those results are also aready accounted for.
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Re: Spell Selection Odds

#13 Post by Brian Mage »

hewhorocks wrote:Andrew_uk wrote:
also can I just raise a point which SA mentioned? And this isn't to have a go at you or make you feel stupid, actually the point I am going to make is a misconception most people seem to hold and I felt raising it might help more people understand probabilities better...

You actually don't have a 50% chance of rolling it, let me explain... The total of all mutually exclusive and exhaustive probabilities have to add up to 1 (mutually exclusive means there is no chance of one die showing a 3 and a 4 at the same time, exhuastive just means I've considered everything).

Now the probability of you rolling the number you wanted (lets say it was 6) when you roll 3 dice is actually the probability of you rolling 666 + the probability of you rolling two sixes + the probability of you rolling one six and two others... now thats a lot to consider and rather awkward so it's easier to do 1-p(no sixes at all)

This works because you're first working out the likelihood that you roll no sixes and then doing 1 minus that

So this is 1-(5/6)^3 = 0.4213 and not 50% as you initially assumed

It's just like with rolling six dice there is no guarantee that you will definitely roll a six, yet the chance of you rolling a six on any one of those dice is 1/6 so it would seem intuitively that it would be... there are a lot of misconceptions surrounding probabilities but providing you always try to be exhaustive and mutually exclusive it should be easier
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Re: Spell Selection Odds

#14 Post by Andrew_uk »

hewhorocks wrote:
Andrew_uk wrote:also can I just raise a point which SA mentioned? And this isn't to have a go at you or make you feel stupid, actually the point I am going to make is a misconception most people seem to hold and I felt raising it might help more people understand probabilities better...

You actually don't have a 50% chance of rolling it, let me explain... The total of all mutually exclusive and exhaustive probabilities have to add up to 1 (mutually exclusive means there is no chance of one die showing a 3 and a 4 at the same time, exhuastive just means I've considered everything).

Now the probability of you rolling the number you wanted (lets say it was 6) when you roll 3 dice is actually the probability of you rolling 666 + the probability of you rolling two sixes + the probability of you rolling one six and two others... now thats a lot to consider and rather awkward so it's easier to do 1-p(no sixes at all)

This works because you're first working out the likelihood that you roll no sixes and then doing 1 minus that

So this is 1-(5/6)^3 = 0.4213 and not 50% as you initially assumed

It's just like with rolling six dice there is no guarantee that you will definitely roll a six, yet the chance of you rolling a six on any one of those dice is 1/6 so it would seem intuitively that it would be... there are a lot of misconceptions surrounding probabilities but providing you always try to be exhaustive and mutually exclusive it should be easier
Actually you need to add in the chance of rolling any doubles as they let you select. Well doubles that dont include double 6es as they have been accounted for or doubles anything else with a third die as a 6 as those results are also aready accounted for.
I did do earlier, I just wanted to raise the misconception about having a 50% chance of actually rolling it.. in SA's words
SpellArcher wrote:Does anyone know what the rough odds are of getting a particular spell if you roll three selection dice at a Lore? Obviously you have a straight 50% of rolling it but I'm unsure how to factor in the chances of rolling a double here.
This is a common misconception
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Re: Spell Selection Odds

#15 Post by hewhorocks »

Ok let us see the actual total chance to get your one spell.

With 3 dice there are 216 distinct results. 156 of these either have a 6 or have a double (or both.) With 3 dice the chance of getting the single spell you want is a little better than 72% :oops:
Last edited by hewhorocks on Mon May 23, 2011 9:34 pm, edited 1 time in total.
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Re: Spell Selection Odds

#16 Post by John Rainbow »

+1 for the maths lesson.
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Re: Spell Selection Odds

#17 Post by Andrew_uk »

hewhorocks wrote:Ok let us see the actual total chance to get your one spell.

With 3 dice there are 216 distinct results. 116 of these either have a 6 or have a double (or both.) With 3 dice the chance of getting the single spell you want is a little better than 53.7%
I'm sorry, you're just wrong. The total probability that you get the spell you wanted is the probability of you rolling it added to the probability of you rolling none of the number you want but at least one double from the rest of the dice now the actual maths involved in this second bit are a bit more complicated but I can demonstrate to you how to work them out

1st part p(that you roll at least one six) = 1 - (5/6)^3

2nd part p(roll a double not including a six) = (5/6)^3 * (1-(5/5)*(4/5)*(3/5))

Now I need to explain how I got that second bit... First it's the probability of you rolling no sixes multiplied by "1 minus the probability of you rolling no doubles" here I'm doing the same trick as earlier as it makes things much simpler

So the probability that I roll no doubles without rolling a six is 5/5 (because the first dice has nothing else it could be the same as

multiplied by 4/5 because the second dice cannot be equal to the first dice

multiplied by 3/5 because the third dice cannot be equal to either of the first two

this simplifies for rolling n dice to give (5/6)^3 * (1-FACT(5)/(FACT(5-n)*5^n)

Factorials are a mathematical way of writing 5*4*3*2*1 for example fact(100) would be 100*99*98.... all the way down to 1 so they can be used (especially it probabilities) to simplify a lot of really horrible looking maths
John Rainbow wrote:+1 for the maths lesson.
Were you referring to me or hewhorocks? I actually trained as a maths teacher though giving a lesson over the internet is never easy...

I'm not sure if Jwg20 worked it out in the same way there are other ways of working it all out... like for instance you could do 1-(5/6)*(4/6) * (3/6)

This is basically worked out by saying the probability of the first dice NOT giving me either a six or a double is 5/6 the probability of the second dice NOT giving me a 6 or a double is 4/6 and the probability of me not getting a double or a six from the third dice is 3/6 both these methods of working it out result in the same answer at the end, which more than anything confirmed for me that I haven't made a mistake... they should both simplify to give exactly the same thing so general result for n dice here would be written as

1-FACT(5)/(fact(5-n)*6^n)

Hope this helps
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Re: Spell Selection Odds

#18 Post by hewhorocks »

For giggles rolling 4 dice there are 1296 combinations with 1176 doubles or 6es (or both) for about a 91% chance.
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Re: Spell Selection Odds

#19 Post by hewhorocks »

I'm not doing math per se. I'm doing excel and its counting the occurrences. Long Hand matrix for the win!

Edit: Opps I did 156!
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Re: Spell Selection Odds

#20 Post by jwg20 »

I did it similarly (well, technically its the same with a different way of working it out. It works out the same way, obviously).

p=probability of 6 or doubles. therefore -p=probability of neither 6 or a doubles.

So solving -p, first dice there are 5 outcomes that give you neither a double or a six. For the second die, there are 4 that give no doubles or a six, and likewise 3 (can't be a six OR any previous results. Total number of outcomes is 6^3=216. So

-p=5*4*3/216=(5/6)*(4/6)*(3/6) (as Andrew calculated)=0.2778

So as -p+p=1, p=1--p = 0.7222.

I find it easier to find instances where an event DOESN'T occur than it is to find all occurrences where an event does. Its too easy to forget certain outcomes (as HEWROCKS did: common error. If I had a dollar for every time I made a similar mistake on a test or something, I would be able to pay off my school loans).

Hooray for math! (though, I am a physicist and biomedical engineer, technically speaking).
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Re: Spell Selection Odds

#21 Post by John Rainbow »

Andrew_uk wrote:Were you referring to me or hewhorocks?
I meant you (Andrew)
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Re: Spell Selection Odds

#22 Post by Andrew_uk »

Lol yeh Jwg20 I know that feeling, but so long as you always ask yourself have I been mutually exclusive and have I exhausted all possible outcomes it's usually safe to assume you're right

Oh and hewhrocks 156/216=0.7222 so you got it right on the recount :P

is -p the american notation of the probability of something not happening? Over here we just write (1-p) or if you are looking at the probability of an event A happening you can simply write P(A) and the probability of it not happening is then P(/A)
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Re: Spell Selection Odds

#23 Post by Andrew_uk »

I decided as a challenge to extend this problem to pose the question "suppose I would like 2 specific spells, what is the chance I would get them from rolling n dice"? Well after considering all possibilities and resorting to highlighters and stuff to check I had considered all possible ways of rolling tripples and multiple doubles ect...

1 0
2 0.111111111
3 0.388888889
4 0.722222222
5 0.938271605

So taking a L4 if you have your eye on 2 specific spells you still have a 72% chance of getting it; not enough to rely on it but enough to plan around it
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Re: Spell Selection Odds

#24 Post by Kulgan »

One alternative, I think, how to get your Archmage with a certain spell, without having to spend your arcane item slot is the following:

You take a Lvl 1/2 mage of the same lore, roll his spells first.

If he hits the spell you want for the Archamage, you switch over to the signature spel.
If he hits any other two spells, your Archmage will automatically have the 4 remaining spells left in the Lore.

( Is this even right? if not then I'm ashamed now :oops: )
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Re: Spell Selection Odds

#25 Post by SpellArcher »

Kulgan wrote:Is this even right?
Yes.

Problems with it include not getting the flexibility of a second lore and having to spend 100pts+ on another character.

But if your army's built right it can be a smart move.
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Re: Spell Selection Odds

#26 Post by jwg20 »

-p is the logic game equivalent notation that is used here to write proofs. It technically isn't a negative sign, there is no equivalent sign on the keyboard to annotate it, it looks more like a sideways L. It is often used for statistics in america, but I have also seen professors in either stat or quantum mechanics use 1-p to show probability of an event not occurring too, so I am not sure if it is the "official" notation method in America.

And about the second character: as SA says it is legal. I have done that with my daemons in the past. Keeper of Secrets LV 4 that I want to have Phantasmagoria, so I take a herald (which you pretty much have to have anyway with daemons), and make her a level 1 then roll for her spell first. Nice thing about the old lores: The "signature spell" was the "1" spell, so trading out for signature they get a spell that can be rolled, which is nice.
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Re: Spell Selection Odds

#27 Post by Demosthenes »

Speaking of how spells get generated, my gaming pals and I went over this discussion and agreed upon a method we would use until we got more data about how other people interpreted it.

Do you roll for the spells for *all* of your wizards, and then resolve duplicates, and then swap for signatures?
Or do you do that process for *each* of your wizards in turn?

The first gives a much-improved chance of getting the spell(s) you want, on the caster(s) you want them; the RAW doesn't seem explicit enough to force one method or the other. So I'd love to hear how others do spell generation.

(sorry to get away from the mathematical discussion a bit, I do find it entertaining and enlightening)
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Re: Spell Selection Odds

#28 Post by grantmepower »

One wizard at a time. This means that your first guy to role on a lore likely will not have duplicates.
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