tag:blogger.com,1999:blog-5371395110910067088.post4920518915076426455..comments2023-11-06T02:38:32.169-06:00Comments on The Warmaster of Fun: Math by Request: An Introduction to CombinatoricsAndrewhttp://www.blogger.com/profile/11932657815610384115noreply@blogger.comBlogger15125tag:blogger.com,1999:blog-5371395110910067088.post-19461941179398849482015-01-14T02:30:58.883-06:002015-01-14T02:30:58.883-06:00THANK YOU. I should do some exerciseTHANK YOU. I should do some exerciseAnonymoushttps://www.blogger.com/profile/16647041183957580808noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-74461164998869747952015-01-13T09:29:58.037-06:002015-01-13T09:29:58.037-06:00The first situation is correct.
The second situat...The first situation is correct.<br /><br />The second situation should be<br /><br />4 * (2/3)^3 * (1/3) = (32/81) = 0.395<br />Andrewhttps://www.blogger.com/profile/11932657815610384115noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-66993177375522632172015-01-13T03:36:38.590-06:002015-01-13T03:36:38.590-06:00Ok it's not the same. But it is wrong. Why?Ok it's not the same. But it is wrong. Why?Anonymoushttps://www.blogger.com/profile/16647041183957580808noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-70199887502153011002015-01-13T03:25:33.215-06:002015-01-13T03:25:33.215-06:00Yes! Thank you!
So.... the probability that 3 SM ...Yes! Thank you! <br />So.... the probability that 3 SM hit with 2 LC is...<br />(2/3)^2*(1/3)*3=12/27= 0,44<br />Wrong? Right?<br /><br />And...4 SM that hit exactly with 3 LC is <br />6* (2/3)^3(1/3)=16/27 AAAAAHHhhhhhhhh! It's not possible! that was the other result! Cant'be the same.... <br />Were are my mistakes?<br />Anonymoushttps://www.blogger.com/profile/16647041183957580808noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-64817363449939848762015-01-12T15:18:35.228-06:002015-01-12T15:18:35.228-06:00I will try to simplify a bit:
We are always looki...I will try to simplify a bit:<br /><br />We are always looking for the chance of success. Let's pretend we have 4 lascannons. If we want to know the probability of EXACTLY two of them hitting and two of them missing we would calculate this as follows.<br /><br />Combinations * (Chance that two hit) * (Chance that two miss)<br /><br />With the combinations:<br /><br />There are six ways to pick two objects from four. If we number the lascannons 1-4 we can look at the following combinations of hits.<br /><br />1 and 2<br />1 and 3<br />1 and 4<br />2 and 3<br />2 and 4<br />3 and 4<br /><br />So, our number of combinations is six.<br /><br />With the successes and failures<br /><br />To find the probability of multiple events we multiply the probabilities of the individual events. <br /><br />Since a space marine has a 2/3 chance to hit the chances of two hits would be (2/3)*(2/3) = (2/3)^2 = (4/9)<br /><br />The chances that the other two space marines would miss would be as follows: (1/3) * (1/3) = (1/3)^2 = (1/9)<br /><br />So to find the overall probability that exactly two of the lascannons hit we would do:<br /><br />6 * (4/9) * (1/9) = (24/81) = 0.296<br /><br />This means that there is a 29.6% chance that four space marines shooting lascannons will hit exactly two times.<br /><br />Did this help clarify things?Andrewhttps://www.blogger.com/profile/11932657815610384115noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-22962192613773700972015-01-12T14:53:59.212-06:002015-01-12T14:53:59.212-06:00Once again it wouldn't change the formula, onl...Once again it wouldn't change the formula, only the chances of success.<br /><br />With roll 2 pick the highest you have the following probabilities to roll the number<br /><br />1: 1/36<br />2: 3/36<br />3: 5/36<br />4: 7/36<br />5: 9/36<br />6: 11/36Andrewhttps://www.blogger.com/profile/11932657815610384115noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-26223735663581672322015-01-12T11:53:47.670-06:002015-01-12T11:53:47.670-06:00How would the formula change if you had a "ro...How would the formula change if you had a "roll 2 dice, take the highest" scenario, like with an ordanance hit against armor value.Anonymoushttps://www.blogger.com/profile/01456095380931565889noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-79562180971107451662015-01-12T01:50:44.834-06:002015-01-12T01:50:44.834-06:00WOW, thank you!
I have watched both video, but I ...WOW, thank you! <br />I have watched both video, but I guess I have to watch them some more times to remember something....<br />One question: it would have been useful if the formula was = to something. I mean, reading it, it's difficult to understand/remember what we are looking for. <br />And some practical example would have been helpful, at the beginning. I understand that you can not talk about WH (or any specific game) but,for instance, you could use dice... <br />Thank you again!Anonymoushttps://www.blogger.com/profile/16647041183957580808noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-86287746700030334732015-01-09T14:35:18.836-06:002015-01-09T14:35:18.836-06:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/14218434317360757240noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-86120430567588913032015-01-09T10:41:07.765-06:002015-01-09T10:41:07.765-06:00Sorry, I switched from talking about shooting atta...Sorry, I switched from talking about shooting attacks to talking about melee attacks, and I didn't really make that clear.<br /><br />Yes, you are correct with your numbers.Andrewhttps://www.blogger.com/profile/11932657815610384115noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-54927864556068109952015-01-09T10:34:44.884-06:002015-01-09T10:34:44.884-06:00you have a mistake:
the space marine have 2/3 prob...you have a mistake:<br />the space marine have 2/3 probability to hit and not 1/2<br />the chance of success is <br />(2/3)*(1/2)*(1/3)=1/9 without reroll<br />(8/9)*(1/2)*(1/3)=4/27 with reroll<br /><br />Anonymoushttps://www.blogger.com/profile/14218434317360757240noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-77935883799158266232015-01-07T11:14:45.871-06:002015-01-07T11:14:45.871-06:00It will just affect the chance to succeed and fail...It will just affect the chance to succeed and fail.<br /><br />The chance for a space marine to hit with a shooting attack is 2/3 but if the attack were twin linked it would be as follows:<br /><br />2/3 + (1/3)*(2/3) = 8/9 because there is a 2/3 chance of hitting the first time and a 1/3 chance you miss the first opportunity and then 2/3 chance again to hit the second opportunity.<br /><br />So with combinatorics. Lets say your space charges into combat to punch a chaos marine. You would have a 1/12 chance of success because (1/2)*(1/2)*(1/3) = (1/12). <br /><br />Now if you were able to re-roll your misses your calculation would be (3/4)*(1/2)*(1/3) = (1/8)<br /><br />So, if you were doing calculations with combinatorics you would use the 1/12 for regular situations and the 1/8 when you could re-roll your misses.Andrewhttps://www.blogger.com/profile/11932657815610384115noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-79560907827990182802015-01-07T11:07:16.515-06:002015-01-07T11:07:16.515-06:00How does reroll of failures affect the calculation...How does reroll of failures affect the calculations?Anonymoushttps://www.blogger.com/profile/01456095380931565889noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-38493208418735097062015-01-07T03:11:21.652-06:002015-01-07T03:11:21.652-06:00WOW... I will watch this video soon....! TY :-)
(I...WOW... I will watch this video soon....! TY :-)<br />(I hope I will understand something...!!)<br />Anonymoushttps://www.blogger.com/profile/16647041183957580808noreply@blogger.comtag:blogger.com,1999:blog-5371395110910067088.post-18933900005872840382015-01-06T18:18:49.774-06:002015-01-06T18:18:49.774-06:00Excellent! Its cool to see the math beyond the ha...Excellent! Its cool to see the math beyond the hammer :)Anonymoushttps://www.blogger.com/profile/01456095380931565889noreply@blogger.com