Skip to Content

WoW Insider has the latest on the Mists of Pandaria!
  • rdjuanto
  • Member Since Oct 21st, 2009

Are you rdjuanto? If So, Login Here.

BlogComments
WoW7 Comments

Recent Comments:

Win a ticket to BlizzCon 2010 from WoW Insider {WoW}

Oct 12th 2010 5:52PM Give me a ticket!

Totem Talk: Enhancement 201, spell selection {WoW}

Feb 12th 2010 11:37AM If the macro has 1 FS and 2 ES (as shown in the article), I don't see the point of having a time reset at all. The CD for 1 FS and 2 ES accounts for the 18 sec duration of the debuff itself. I agree with Sephy on the 8sec. I'd probably have it at 4-5, even.

Also, "target" would be another good conditional reset.

Math problem: Average winning roll {WoW}

Dec 21st 2009 3:19PM EDIT:
When I said "WINNING NUMBER", I really meant "AVERAGE WINNING ROLL".

Math problem: Average winning roll {WoW}

Dec 21st 2009 3:11PM Out of curiousity, I did a little spreadsheet after reading this entry.

The premise is that a "WINNING NUMBER" is a number that will guarantee more than 50% chance of winning. I know that there are 4 (or 9 or 24) other people that will be rolling for the piece of item, but really all we care about is whether we are going to "WIN" or "LOSE" the roll, hence, the 50%.

Since I simplified the problem by assuming that rolls are integers, I introduced the probability of getting a "DRAW" which would be (1/100)^"N" with "N" being the number of rollers - 1. This is really an insignificant number in 5-men and even more so in others, but it exist nonetheless.

For a roll value of "X", the probability of any roll higher than X is (100 - X)/100. This is of course the probability of "LOSING" to any roll at all.

If event "L1" is the probability of "LOSING" against 1 other roller while rolling an "X", then:
L1 = (100 - X)/100

If event "L2" is the probabilty of "LOSING" against 2 other rollers while rolling an "X", then:
L2 = L1 + (1 - L1) * (100 - X)/100
Instinctively, we might want to go ahead and say that L2 = 2 * L1, but such is not the case, because it only take 1 roll higher than "X" for "X" to be "LOSING".
(1 - L1) * (100-X)/100 in the above calculation is the probability that the second roll is not higher than "X", and the third roll is.

L3, the probability of "LOSING" against 3 other rollers, is then:
L3 = L2 + (1 - L2) * (100 -X)/100

Using the above progression, the probability of "LOSING" is then L4 for a 5-men, L9 for 10-men, and L24 for 25-men.

Probability of "WINNING" is 1 - ("LOSING" + "DRAW").

Using the above, I made a spread sheet and produced the following "WINNUNG NUMBER"s:
for 5-men 85
for 10-men 93
for 25-men 98

I suspect my rounding the rolls to integers caused the discrapencies, but these numbers are a pretty good estimation, nonetheless.

A WoW player's guide to microtransactions {WoW}

Nov 5th 2009 11:55AM Exactly. People are willing to pay lots of extra G's for pets that are sold by NPC, just for convenience. Imagine what they would pay for a "premium" pet.

A WoW player's guide to microtransactions {WoW}

Nov 5th 2009 11:36AM The problem I have with the new feature is that people can start selling these pets for in-game money. Which realy is an indirect way of buying gold with real money. I am pretty sure that people already are selling them.

The only way to make sure that microtransactions will not affect the game is to get rid of the option of sending the items as gifts.

Enter to win a wireless headset from Creative and WoW.com {WoW}

Oct 21st 2009 6:19PM I'd love to have one!