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rhane
Underboss
1412 Posts
    
 | | 03/02/2006 3:56 PM |
| Okay, so 1000 years ago, I used to have a fair grasp of statistics, but 4 kids have driven out anything that isn't Dora or SpongeBob related.
If I have 2 separate events, and I know the probability of success on each, I can derive the overall probability of both occurring by multiplying the two probabilities together, right?
So, what are the chances of getting a 1 on BOTH rolls of a 4 sided die?
1/4 * 1/4 = 1/16
Easy.
Now for the part that I've either forgotton, or just don't know...
What are the chances of getting a 1 on either roll?
I can certainly work it out by figuring the total possible results, and the number of positive results. In this case, there are 16 possible outcomes, 7 of which will result in the desired goal of getting AT LEAST one 1:
1 1
1 2
1 3
1 4
2 1
2 2
2 3
2 4
3 1
3 2
3 3
3 4
4 1
4 2
4 3
4 4
so 7 in 16 or about 44%.
Writing out the possibilities for this example are fairly easy. But what if instead of having 2 events with a probability of 1/4 each, I had 1 event with a probability of 3/632 and another with a probability of 77/928. Personally, I don't want to write down 586000+ different outcomes to figure out the total number of possible successes.
Is there a formulaic way to achieve this answer without going through the gyrations?
Thanx all... | |
Rhane
"The focus is sharp in the city..."
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ShadowLord XT
Commander
2646 Posts
Plane of Shadow
| | 03/02/2006 4:53 PM |
| | It's on the tip of my toungue! I know this, but can't remember. I'm trying, but I think all of the mucus has seeped into my brain blurring my train of thought. | |
Disipline is the only way to overcome chaos.
Champion of Half-Golems
Knight of Golems
"This world is made for love and peace" - Trigun
"anyway..shadow..you've figured women out. KUDOS." - raye_kino16 | |
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Avatar of the Irrelevant
Diomedes
Commander
3185 Posts
| | 03/02/2006 5:34 PM |
| quote:
Originally posted by rhane
Okay, so 1000 years ago, I used to have a fair grasp of statistics, but 4 kids have driven out anything that isn't Dora or SpongeBob related.
If I have 2 separate events, and I know the probability of success on each, I can derive the overall probability of both occurring by multiplying the two probabilities together, right?
So, what are the chances of getting a 1 on BOTH rolls of a 4 sided die?
1/4 * 1/4 = 1/16
Easy.
Now for the part that I've either forgotton, or just don't know...
What are the chances of getting a 1 on either roll?
I can certainly work it out by figuring the total possible results, and the number of positive results. In this case, there are 16 possible outcomes, 7 of which will result in the desired goal of getting AT LEAST one 1:
1 1
1 2
1 3
1 4
2 1
2 2
2 3
2 4
3 1
3 2
3 3
3 4
4 1
4 2
4 3
4 4
so 7 in 16 or about 44%.
Writing out the possibilities for this example are fairly easy. But what if instead of having 2 events with a probability of 1/4 each, I had 1 event with a probability of 3/632 and another with a probability of 77/928. Personally, I don't want to write down 586000+ different outcomes to figure out the total number of possible successes.
Is there a formulaic way to achieve this answer without going through the gyrations?
Thanx all...
In the case you cited, it's just %chance 1 + % chance 2 - % chance both.
1/4 + 1/4 - 1/16 ~= .44
so for the other case you mentioned, it'd be:
3/632 + 77/928 - (3*77)/(632*928) ~= .083
I *think* that you might find this interesting reading: http://en.wikipedia.org/wiki/Binomial_coefficient
I'd be more specific, but I was just glancing at maxminis on my way out of work.
Sorry to be in a rush, I hope that helps.
-Diomedes | |
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Terentia
Sneak
62 Posts
| | 03/02/2006 6:04 PM |
| I just tested my kids on this section.
Since it is a compound event one of the possible formulas is:
P(A or B) = P(A) + P(B) - P(A and B)
If they are mutually exclusive events then the formula is:
P(A or B) = P(A) + P(B)
You do the 1st one if a double count is going to occur.
Example would be if you are rolling a D6 and you want to know the probability of rolling a multiple of a 3 or 5.
P(A)=2/6
P(B)=1/6
P(A or B)= 2/6 +1/6 = 1/2 or .5
If you want to know the probability of rolling a multiple of a 2 or 3 on a D6 then:
P(A)=3/6
P(B)=2/6
P(A and B)=1/6 {This one accounts for 2 and 3 being a multiple of 6)
P(A or B)=3/6 + 2/6 - 1/6=2/3 or 0.67
Probably more than you needed but may come in handy when the youg'uns get to Algebra. [)] | |
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rhane
Underboss
1412 Posts
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Gunthar
Commander
2938 Posts
| | 03/03/2006 9:57 AM |
| | It depends: If you are swinging and hit with two non-magic daggers at a monstrous brute with 3 HP left, high attack bonus, multiple attacks and a large axe when you only have 5 HP left yourself, about 97% on both rolls. [:D] | |
Champion of Prit(Wemic vindication is here)
Minneapolis/St. Paul area
Completed trades: Aspect of Cheese (Love that moniker), Tickparasite, Elderthing, Lalato, Sodj, Grimoire, SmilinIrish, Zeb, RWarehall,Link, wikkawikkawa, Auramancer, Rommers, HK, Ivid5,Qillan_dvra, Puggins, Arcabius, Ironfist Boulderbender, Robby, Corim Danex, monster_slayer, DNDJUNKIE, Kelemvor, Krush, ckissee, Massawyrm, hockey fan, Wish, Uninspiring Lieutenant, vtloon x2, Vrecknidj, Darthpoke, WakeXX, AnarionZell, lycusmike, papabear5 and umpteen local trades with board members
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ShadowLord XT
Commander
2646 Posts
Plane of Shadow
| | 03/03/2006 10:05 AM |
| quote:
Originally posted by rhane
Okay, so 1000 years ago, I used to have a fair grasp of statistics, but 4 kids have driven out anything that isn't Dora or SpongeBob related.
Hey what's wrong with Spongebob? | |
Disipline is the only way to overcome chaos.
Champion of Half-Golems
Knight of Golems
"This world is made for love and peace" - Trigun
"anyway..shadow..you've figured women out. KUDOS." - raye_kino16 | |
| *censored*
glumag
Warlord
5968 Posts
| | 03/03/2006 6:33 PM |
| | 42 | |
Trades >> Completed: 49 | Bad: Ø | Pending: 0 | Trade & talk Live on IRC! SERVER: irc.psionics.net CHANNEL: #maxminis | |
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