This is probably a rediculous question, but some forum flaming to me is better than no closure xD
When rolling, do your chances of better stats increase with the amount of time spent rolling? If the connection drops due to being afk or I click the open space accidentally when a math problem pops up, I feel like I lost all of the potential for good stats that I have built up from hours of rolling.
Basically what my question boils down to, is if I rolled unsuccessfully for 2 hours, if my client closes and I reload the roller, are the odds still the same when rolling?
Thanks
Rolling
Started by Sentry, Jan 05 2009 10:10 PM
7 replies to this topic
#1
Posted 05 January 2009 - 10:10 PM
#2
Posted 05 January 2009 - 10:13 PM
Odds are constantly the same, no matter how long or if you answer a problem incorrectly
Piggy Or Derlok ingame
#3
Posted 05 January 2009 - 10:17 PM
/nod piggy.. ill search for the exact odds and post sec..
#4
Posted 05 January 2009 - 10:22 PM
"Hmm, interested myself actually, so let's see if i can work it out. The roller works by having base stats (set by race), then a random number from 1 to 12 added on.
This is based on the binomial distribution, with the probability of getting a max stat being 1/12, and therefore the probability of a stat not being max is 11/12.
To get a 1-stat crit, whereby 1 of the 6 stats is 12 and the rest are not 12. We have (1/12) * (11/12)^5, but we have 6 stats, so we multiply this number by 6 to get .32, which is about 1 in 3.
To get a 2-stat crit, whereby 2 of the 6 stats are 12 and the rest are not 12. We have (1/12)^2 * (11/12)^4, but we have 6 stats and it could be any 2 of them that are max which means we have to work out all ways of getting 2 of the stats max out the 6 available (ie, str-int, str-dex, str-con etc). Or using nCr or the formula n!/r!(n-r)! we get 6!/2!4! which is 6*5/2 = 15 combinations. So 15*(1/12)^2*(11/12)^4 = .0735, which is about 1 in 14.
To get a 3-stat crit, (6!/3!3!)*(1/12)^3*(11/12)^3 = about 1 in 112.
4-stat crit, (6!/4!2!)*(1/12)^4*(11/12)^2 = about 1 in 1645.
5-stat crit, (6!/5!1!)*(1/12)^5*(11/12)^1 = about 1 in 45,242.
6-stat crit, (6!/6!0!)*(1/12)^6*(11/12)^0 = 1*(1/12)^6*1 = 1 in 2,985,984.
For interest, a 0-stat crit is (6!/0!6!)*(1/12)^0*(11/12)^6 = 1*1*(11/12)^6 = 0.59, which is about 1 in 1.7" -JLH
had to search old forum.. wasnt on this one lol....
This is based on the binomial distribution, with the probability of getting a max stat being 1/12, and therefore the probability of a stat not being max is 11/12.
To get a 1-stat crit, whereby 1 of the 6 stats is 12 and the rest are not 12. We have (1/12) * (11/12)^5, but we have 6 stats, so we multiply this number by 6 to get .32, which is about 1 in 3.
To get a 2-stat crit, whereby 2 of the 6 stats are 12 and the rest are not 12. We have (1/12)^2 * (11/12)^4, but we have 6 stats and it could be any 2 of them that are max which means we have to work out all ways of getting 2 of the stats max out the 6 available (ie, str-int, str-dex, str-con etc). Or using nCr or the formula n!/r!(n-r)! we get 6!/2!4! which is 6*5/2 = 15 combinations. So 15*(1/12)^2*(11/12)^4 = .0735, which is about 1 in 14.
To get a 3-stat crit, (6!/3!3!)*(1/12)^3*(11/12)^3 = about 1 in 112.
4-stat crit, (6!/4!2!)*(1/12)^4*(11/12)^2 = about 1 in 1645.
5-stat crit, (6!/5!1!)*(1/12)^5*(11/12)^1 = about 1 in 45,242.
6-stat crit, (6!/6!0!)*(1/12)^6*(11/12)^0 = 1*(1/12)^6*1 = 1 in 2,985,984.
For interest, a 0-stat crit is (6!/0!6!)*(1/12)^0*(11/12)^6 = 1*1*(11/12)^6 = 0.59, which is about 1 in 1.7" -JLH
had to search old forum.. wasnt on this one lol....
#5
Posted 05 January 2009 - 10:23 PM
i've gotten a four stat a couple times within 10 rolls..other times i can roll for a half hour and get nothing better than a three stat with the wrong stats
Page me on Heritage
#6
Posted 05 January 2009 - 10:30 PM
ya.. its luck more than math.. lol
but what if you could calculate luck in a formula ...
but what if you could calculate luck in a formula ...
#7
Posted 05 January 2009 - 10:31 PM
So hopefully 112 clicks gets you a 3stat
Logos.
#8
Posted 05 January 2009 - 10:39 PM
Great to know the calculations haha and kudo on the quick replies guys. Off I go in hopes of a decent 4-stat so I can finally start playing again
-crosses fingers-
-crosses fingers-
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